www.pi314.net


History
Mathematicians
All formulas
Num. approx.
Softwares
Misc. math.
Digits
Poetry
Papers/videos
Delirium !
 Pi-Day
Images
Music
Links
Bibliography



Boris Gourévitch
The world of Pi - V2.57
modif. 13/04/2013



Pi-Day in
Pi Day Countdown
Google
Home Version history Guestbook Who I am Some pictures (fr) Acknowledgements Last modifications Contact

Cette page en français This page in English


12 Other binomial coefficients

There's not just central binomial coefficients in life! We can also find some series with other combination to the denominator, Cn3n  for example. An other idea is also useful : The numerous series with Cn2n  can cleverly be accelerated by taking uniquely part of the terms in this series, for example by keeping the C24nn  . Here is a little lot of series giving p2  , or ln(2)  , and who are most likely not all known. Followed by the proof of a few of them. A few are among the fastest known rational series!

Unless otherwise stated, all the series are personal discovery, ignoring any reference I don't know of!

12.1 Primitive formulae

In the whole document, we will call primitive formula those that minimise the degree of considered polynomials in the series, and who minimise the number of members of series such as the BBP binomial considered a few paragraph lates. In fact, for a polynomial binomial formula  sum o o -P(n)-
  n=02knCn2n  , the decomposition into even and odd parts (sum on 2n  amd 2n+ 1  ) gives another serie  sum o o   Q(n)
  n=0 22knC24nn but which is still equivalent. This said, maybe all the formulae that follows are not primitive, it's sometime hard to see!

12.2 Factorial polynomials formulae

First of all, we are interested in series of the form  sum o o -P-(n)-
  n=02knCjnajn  , generalisation of the previously seen central binomial series  sum o o  P-(n)--
  n=0 2knCn2n  .

  • With   2n    1n
C 3n = C 3n    :

 sum  oo 
    -6n+-5n0n-= p
n=0  2 C3n
(485)

Such a simplicity, it's beautiful...

We have 2nCn3n  ~   a13 V~ .5n
      n--> oo     n  , which allows us to conclude that the previous serie gives just a bit more than one decimal per term, i.e. a speed of roughly 1.13n  .

  • With   2n
C 4n    :

- 22    sum  oo  (- 1)n2n
----+ 2   ----2n- (7- 30n) = p
  3    n=1  C 4n
(Guillera)

 oo  sum 
   --27+-270n = 4p V~ 3
n=0   C2n4n
(486)

whom I give two proofs after all those formulae.

 sum  oo  -405-+1134n-
       4nC2n    + 384 = 16 ln(2)
n=0       4n
(487)

 oo  sum  - 384+ 3072n(16 )n    sum  oo  - 1944+ 10692n( 16 )n
   ------2n----  --   +    -------2n-----  ---   + 2000 = 75p
n=0    C 4n       25     n=0     C 4n        100

this last fform is not random... as we will see in part 12.4. [link].

Note that   2n       16n-
C 4n n-->~o o   V~ n  which is not uninteresting from a calculation point of view! But wait to see the rest....

I remind you here that if I write 16n  , this means that we already gains more than one decimal by iteration. The exact calculation is done by using the decimal logarithm (because it's the reciprocal of the function 10x  hence this gives us x  the number of decimals, in base 10, equivalentes to the given number) Here,     (   )
log10  1 V~ 6nn- = 1.2n-  12 log10(n)  is the n  are very big in front of log10(n)  very quickly.

  • With C2n
  5n

I found nothing.... which does not mean there exist nothing !

  • With C3n
  6n    :

 sum  oo  972- 6561n+ 15309n2            V~ -
    --------C3n---------- 1296 = 40p  3
n=0          6n
(488)

Note that   3n       64n-
C 6n n-->~o o   V~ n

  • With   2n
C 6n    :

pict

The last formula is among those that make a ln(2)  appear but no p  for the same power in the denominator...

Note that C2n   ~  a (-66)n  ~~  a(4 V~ 6)n
  6n n-->o o    2244        n  .

  • With C2n
  7n    :

It's the order of Bellard's relation. It's a particular order, because 7  is an odd number ( !). Numeriquely, it's the only order j  (at least I  think so) for which we can find a coefficient nj-2  implied.

    ( 65979888- 1181293644n+ 6191776770n2- 17657815350n3)
  oo  sum               +18752083422n4- 5314039086n5
    ------------------------2nC2n------------------------- -86359168 = 740025p
n=0                             7n
(492)

but also

   (                                     )
         271906001328- 3918976142938n
      +21587589744115n2- 50172923799365n3
 oo  sum  ---+56223128657342n4 +-22573799287772n5--
                    2nC2n7n                  - 264159407778 = 124324200ln(2)
n=0
(493)

Concerning the speed of the convergence, we have 2nC2n   ~   a(-77-)n  ~~  a (13 V~ 2)n
   7n n--> oo    2255        n

  • With C2n
  8n    :

   (      172432908318- 2790899445213n    )
      +16668235200353n2- 42758901871730n3
 sum  oo   +50771457154396n4- 21833905721320n5
   ------------------4nC2n----------------- - 168573734400 = -147026880ln(2)
n=0                     8n
(494)

With  n 2n       (  88-)n    (360)n-
4 C8n n-->~o o  a 4 2266   ~~  a  V~ n

  • With   4n
C 8n    :

pict

We have   n 4n       (88)n          n
16 C8n n-->~o o  a 48    ~~  a(4096)

  • With C3n9n    :

   (                                         )
          2432748 - 47198830n + 314662272n2
 sum  oo ---958580892n3 +-1342529748n4--768151350n5-
                      8nC3n9n                    -2301696 = -15015p
n=0
(497)

This time, we have n  3n       ( -99-)n       n
8C 9n n~--> oo  a 83366    ~~  a2460

  • With   2n
C 10n    :

    (      250485147675 - 5936814044607n      )
       +52544973042581n2- 215675737481744n3
      +449152546743385n4 -444526162997758n5
 sum  oo    +34376346918444n6 +76601758034424n7                                V~ -
    --------------------2n--------------------- 30683221875 = -8079671600p  3
n=0                    C10n
(498)

At equivalent levels, we have            (  10)n
C21n0n n-->~ oo   a  102288-   ~~  a149n

  • With C3n12n    :

    (                                            )
           355802970745720- 8585041645767210n
       +81615477992592869n2-333211330576835622n3
      +655621008551811783n4- 352741368593950830n5
      - 556073914030962489n6+920192581802693862n7
  oo  sum  --------------379330413567463683n8--------------                                 V~ -
                         C3n12n                      -338812710016000 = -120725404800p  3
n=0
(499)

With  3n        (1212)n      n
C12n n~--> oo  a 3399    ~~  a852

  • With   4n
C 12n    :

    (      38076224256- 1074010706880n      )
       +11279342361792n2- 59037649865344n3
      +170466696380928n4- 278182243508224n5
 sum  oo    +241235982262272n6-88848355655680n7
    -------------------n-4n-------------------37679385600 = - 43648605p
n=0                  16 C12n
(500)

   (                                              )
           326440320157431 -28937516611612130n   3
       +89908515866103392n4-441885501076996544n  5
 oo  sum     +1177159045662595328n -6 1729916148121063424n7
   ---+1316777873347407872nn--44n04436261768519680n------326281074064200 = - 8380532160p
n=0                     16 C12n
(501)

   (                                                 )
          -14097558940862121+2 389204231048912622n  3
       -3953865455193366256n +19661803268086575136n
  oo    - 52846597819262937088n4+778039428452651376644n5
 sum  -----58535200950020276224n6+17424985780117307392n7---+359523048663 = - 268177029120 ln(2)
n=0                      256nC41n2n
(502)

The speed is                (      )
256nC4n   ~   a 2561212-n  ~~  531441n
     12n n--> oo       4488

  • With C6n
  12n    :

    (                           )
          1740852- 38174085n
       +264952863n2-815781618n3
 sum  oo --+1147203972n4--644815080n5----                 V~ -
                 C61n2n             - 1679616 = -6160p  3
n=0
(503)

    (    -14717727054+ 312245502345n    )
      - 2064710463597n2+5822757762786n3
 sum  oo   - 7306291257516n4+3343524913512n5
    ----------------6n---n---------------+14716428288 = 394240ln(2)
n=0                C12n64
(504)

the speed is               (    12)n
64nC61n2n n~--> oo  a 64162666-   ~~  262144n   :

  • With C6n18n    :

  oo 
 sum  P-(n)-14184611224629819300 = 32541813304000p V~ 3
n=0 C61n8n
(505)

with               14176156929732225900 - 632834547957106992000n
          +10870535340008572095495n2 - 97312752507652497923274n3
        +512280981936407241803853n4- 1677608112864248615276964n5
P (n) = +3464504827463767568699322n6 - 4389617823914659523895468n7
        +3047445608666357208880512n8- 662159753036907439330656n9
        - 423196182699119543819232n10 + 214663956171311152763712n11

 sum  oo  P(n)
   --n--6n+1895389046876422190318400 = 161731979240832000 ln (2)
n=064 C 18n
(506)

with                - 18953957365336602485173502+ 88718418961818586370893385n   3
          - 1640772073640628436250559957n 4+ 16266757793935182287966062690n -5
P (n) =    98166577084756978666140492660n 6+ 384563780652279067490128227120n  7
        -1009244981043062390325473240256n8 + 1789495235800681177225726017120n9 -
         2114731370352057790001239912320n10+1594631600862337977789113228160n1-1
          693512658595516062956146117632n  + 132257667454051420111105497600n

  oo 
 sum   -P-(n)--+1565767793815142400 = 52647128659125p
n=0 64nC61n8n
(507)

with                -1566259769242649600 +74380856960349239360n
          - 1404907786091424993952n2 + 14324220750452155291840n3
         - 89528655619328696225760n4 + 365931942892429890856320n5
P (n) = -1009918573918657336233216n6 + 1899526509357067346296320n7
        -2404490163304300329323520n8 + 1964360952134061230453760n9
        - 938411873139437499082752n10 + 200312172515602364313600n11

Note on the way that 64nC6n   ~   a(64-1818-)n  ~~  6053400n
    18n n--> oo      121266   ! ! The last formula adds a bit more than 6 new decimals at each iteration ! It is now better than the best rational series of Ramanujan  !

  • With C122n4n    :

    (                                                    )
             4231942624137060 - 222759564240763665n
         +4385614031516211939n2-45616240090032011370n3
       +288113577357095245620n4-1184133888632911121640n5
       +3276284359477677553152n6- 6164738823471380184960n7
       +7793201634519570416640n8- 6347445587443372677120n9
 sum  oo --+3017311363298550841344n10-638959960656262103040n11---                                 V~ -
                             C1224nn                          -4231494800064000 = -41644002400p  3
n=0
(508)

  12n     ( 12)n          n
C24nn-->~o o  2      ~~  16777000  . Impossible unfortunatly to find a formula for p  .

  • With   16n
C 32n    :

 sum  oo  P(n)                                         V~ -
    -16n--5751162662492062771200 = 935948953940000p  3
n=0 C32n
(509)

where                  - 5751172814242964199525+ 431521408590208193289330n
            - 12603727549612641868970688n2 + 201946668204590211911196960n3
          - 2047087065971276065760666112n4 + 14165961499688329616190676992n5
          -69997594901404885250357624832n6 + 253748666775695272192598999040n7
P(n) =   -685053118985113141250931818496n8 + 1384816706184068494846303666176n9
       - 2088250793463253952490909990912n10 +2315666621747475203456740884480n11
        -1834031550806791394823443054592n12 + 982464432301136793019405565952n13
         -319253432772702624859677523968n14 + 47590573803868948158291640320n15

 sum  oo  P (n)
   ---n--16n + 8923391361809421872736614400 = 2338547211.13.17.19.23.29.31p
n=0256 C 32n
(510)

where                   8923391346671538132476623725-2665206805283623996428461672720n     3
            +19234784845151225969596207764237n 4- 304151310790319471452952294968086n  5
          +3032958911347016158089740357038016n 6- 20576037941520000660127171571143328n 7
P(n) =    +99294811157156339256068440506994944n 8- 350018237892823871397833615112153088n 9
         +914268728549376302315970998665658368n10- 1777620930511060794594593827649224704n 11
       +2560116484483562303674928488404484096n 12- 2688153090709441777294170302212734976n13
        +1994646664772435098534901910513647616n14 - 987669694437093352756584653885800448n15
         +291530209843404894708880474005045248n  - 38563221903276982647173755070054400n

but also

 sum  oo   P(n)
    ---n-16n- 5618174347094280909397150924800 = 786017419110443520000 log(2)
n=0 256 C32n

where                   5618174347548089898981989892075 - 421098079220529602568327860126490n
            +12279158984607989711854217378240304n2 - 196320130925997777597664903245042912n3
          +1984645970323542311490123954102121472n4 - 13688177767013745425057397228460978176n5
          +67364519495207601370423942235494350848n6 - 243021071568401760035506416710784516096n7
P(n) =   +652276342765602853078565522814172921856n8 - 1309340663714878748772460572435460128768n9
       +1957782039035544577182285734185724280832n10- 2148775807107880796770359601282827681792n11
        +1680566813227110462017445252816984604672n12 -886352566529628907272207202135327113216n13
         +282456627503213365917892411798444834816n14 -41068230823970564113517731323366604800n15

And if we look at the speed of convergence, we get

   n 16n       ( 83232)n                n
256 C32n n-->~ oo   a 2 1632    ~~  1099511 627 776  which gives at least 12 new decimals at each step of the serie !

Even with it's apparent complexity, it's a rational serie, hence simple to put to work. The calculation of the central binomial coefficeint can as well be done in a recursive way from one step to the next.

  • With C244n8n    :

pict

where P(n) = - 3841403444306253083992509771327832500 +471148214634256669931181490304104333125n

                                          2                                           3
- 23480822326039365418691160664239300395775n + 666353782635656300780601331418642024797170n                                                4                                              5
- 12417264254262731471457553661953159748556036n + 164349539045865484455081586061158801910482824n                                                                                                                                        6                                                7                                                 8                                                 9
                                                                                       -1622809383811255876123664026332895835355126528n  + 12355975073531641668406952243794826844294207360n  - 74244199819829674727637631331612966036993780736n  + 3579438699
                                                  10                                                  11
- 1401040504596445070899138565913722649517902102528n   +4488300494138137535781366329518100596922844610560n
                                                  12                                                   13
- 11827398936905381920358994038925975320579657957376n   +25695792588467683854423928595051100389120969539584n
                                                  14                                                   15
- 46008127131406502564765669946976843765906667470848n   +67674182522091504564791351456373805642285433487360n
                                                  16                                                   17                                                   18                                                   19
- 81254118178422030960760468039239656346429891280896n   +78808966451353174037517246676769934231381227864064n  - 60775164646173353226062113514203716402153529540608n  + 36385314363927335868453330316855779017586439618560n
                                                  20                                                  21
- 16298284219960028553979877506702436175888783507456n   +5139260980144703889225203033782459240512648904704n
                                                  22                                                23
- 1017402895725449845366234974452573952900845273088n   +95133777262638723375339529534690070114850570240n

12.3 So, what to make of all of this ?

Several things need to be noted :

  • The big coefficients in front of the constants can be generally factorised into product of simple prime numbers, but this does not give much information on the nature of the coefficients.... the one inside the brackets they on the other hand contains big prime factors.

Exemple : In dront of the coefficient of p  of the last serie (denominator of the fraction in fron of the serie), we have 1535190 271 700085000 = 2338547211*13 *17* 19* 23 *29 *31  but on the other hand the first coeefient of the brackets is factorise as follow : 8923391346671538132476623725 = 35527213× 31× 47× 20 482 861248481× 77267  (which is a lot less simple !).

  • Second of all, we can already note that at each order there exists some relation with p  and the logarithm, we can find series wuth null sums like we did for the BBP series. I think that they are done "mechanically" and that there must exists a polynomial P  of degree of at least k  which cancel  sum 
 o o n=0 PC(jnn)
       kn  . But I don't have the proof.
  • Then, it seems that only well determined constants can be represented under this form. For my part, with this kind of exploration, I have never find anything other than     V~ -
p,p  3,ln(2)  . No   V~ - V ~    V~ -
p  2,p  5,p  6,ln(3)  or   ln(5)  at the horizon... I can understand for p  , but for the logarithm it's quite surprising!
  • We can also notice that the repartition of the signs is so regular (especially in this two by two presentation in the brackets!) to give a constant, that seems not to have an existance more justified than an other... unless if      p2  or ln(2)  , are very important... Furthermore, between the formulae for p  and the formulae for ln(2)  , the coefficients are of same order and same sign.
  • The coefficients themself are interesting. They are increasing according to the powers, up to a point, and then decreasing for the rest, as if they had reach a maximum. This phenominum is particularly visible with the presentation of the serie in C2448nn  where we can obswerve the maximal coefficient comes in front of n16  (and by coincindence, a power of 2  and a third of 48 !).
  • We can finnaly notice that apart from the case C27nn  , we have in general some formulae for combination of kind    Chnkn  , where h| k  . And thanks to the properties of combinations, we know that a formula with C3n12n  is also a formula with    C9n12n  since C31n2n = C9n12n  . We can already create a little table of formulae that have to naturaly appear. In fact, we will notice that between all those formulae there are a few charecteristics that come back. Hence, there systematicly exist some formula of kind  sum o o  P(knn)= a.pV ~ 3
  n=0 C2kn  where P  is of degree k  and a  (-  Z  . We also have formulae for  sum o o --P(n)4kn = a.p
  n=024knC 8kn  and  sum o o   P (n)
  n=0 24knC48knkn-= a.ln(2)  . This last point is confirm by the article [7]. For the Ck2nkn  we often have some 2knCk2nkn  but not always (k = 3  ). For the Ck4nkn  we always have some 2knCk4nkn  and sometimes some 23knCkn4kn  (because 3 = 4 - 1   !).For the    Ckn3kn  we have some series with 2knCk3nkn  uniquely it seems to me. In all cases, if there is no powers of 2  , we found a serie giving p V~ 3  . If there is a power of 2, we have p  or ln(2)  . Easy ! For all those formulae, with a     Cjknn  the polynomial P  function of n  used in the numerator has a degree of at most k -j  or most of the time max(j- 1,k - j- 1)  .

Note 24 Since I wrote this document, I found an article printed in octobre 2001 [7] which present the fundamental research principles of such formulae, and which shows that we can find some series as fast as we with with the central binomial coefficients of type  4kn
C8kn  .

12.4 Proofs

Those formulae seems difficult to proof because the functions they represent and hence the differential equations that they satisfy have few chances of being simples...

The differential equations are most likelt very complicated, but we be a bit cunning. Notice that we already know the sum

 sum  oo  22n-1    V~ x-arcsin( V~ x)
   ---n-xn = ---- V~ -------
n=1 nC2n          1- x
(513)

and that the serie  sum o o  24n-12n-x2n
  n=1 2nC4n  is the even composant of the previous serie. Then, we have two method that can seem natural. A third then appeared when following  [7].

  • The first consist to build differential equation according to the series using a few of their properties (even part of another serie...). Then, in certain particular cased those differential equations will gives some series:

Let us pay attention to the case of the serie  sum o o  -27+270n-= 4p V~ 3
  n=0   C24nn  .

First of all, we have                    V~       V~ 
 sum o o n=1 2C2nn-1xn = x ddx-x-ar V~ c1si-n(x-x)
        2n  . We need therefore to extract to the first serie 513 part of it's members (those in x2n  ). We introduce

       sum  oo  1  4n          oo  sum    1    4n+1
f(x) =   C2n-x  et g(x) =    C2n+1-x
      n=0  4n             n=0 4n+2
(514)

They are, more or less, the even and odd composant of  sum o o  22n-1 n
  n=1 Cn2n x  ... more precisely,

pict

Let us now find a relation between the functions f  and g   :

pict

from which if we derive the relation from 515, we get

                                    (               ( ))
d (x     )  1       d (1     )    d      x d xarcsin  x2
dx- 4f(x) + 4f (x) + dx- xf(x)  = dx- 1 + 8-dx V~ ----(x)2
                                               1 -  2
(518)

                                      (               )
      '  (x    1)       (1    1)    d   x d xarcsin (x2)
<====> f (x)  4-+ x- + f(x) 2 - x2  = dx-  8-dx V~ ----(x)2
                                              1 -  2
(519)

The right hand side is not very nice, but it contains aarcsin  . We just now need to let x = 1  to find

5       1       2   V~ -
- f '(1)-- f(1) =--p  3
4       2       27
(520)

let

   oo  sum              V ~ 
1    10n-2n1-= -2p  3
2 n=0  C 4n    27
(521)

and it is the serie we were looking for.

We can also let      V~ -
x =   2  to find

 oo 
 sum  n4n-= p-+ 2
n=0C24nn   4   3
(522)

this also allows to find nearly the same relations for g(x)  .

Finnaly, we can use the famous relation      ( V~ 1)       ( V~ 1-)  p
arcsin   5  + arcsin   10  = 4  to obtain

 oo  sum              (   )n    sum  oo              (    )n
   --384+-3072n  16   +    --1944+-10692n  16-   + 2000 = 75p
n=0    C2n4n       25     n=0     C2n4n        100
(523)

This case works we find ourself with the two equations for f  and g  , which allows us to come out with a quite useful differential equation. If we divide in the same way in three part so for example to find some series in C36nn  , nothing is garented ! In a general way, the differential equations will most likely be unsolvable, but we can use them to find a few relations.

  • The other method, which came from Gery Huvent , consist of explicitely calculating certain series which are pieces of other series, by hoping that simplifications will be possible. For this. we use the fact that if we denote w = e2ipp  the p  -th root of unity, we have 1+ w + w2 + w3 + ...+ wp-1 = 0  . So with f(x) =  sum o o n=0 anxn
                   (  2 )       ( p-1 )    oo  sum     n(     n    2n    3n        n(p- 1))
f (x)+ f (wx) + f w x  + ...+ f w   x  =    anx   1+ w  + w  + w   + ...+ w
                                      n=0
    (524)

when p  is prime, two cases are possible: n  is not a multiple of p  , so since the wkn  , k  (-  [[0,p- 1]]  are the p  -th distinct root of unity, 1+ wn + w2n +w3n + ...+ wn(p-1) = 0  . If n  is a multiple of p  , each wkn  is worth   1  , hence 1+ wn + w2n +w3n + ...+ wn(p-1) = p  . With all of this, we obtain the sum only on the multiples of p  , i.e.

               ( 2 )        ( p-1 )     sum  oo    np
f(x)+ f(wx) + f w x  + ...+f  w   x  = p   anpx
                                       n=0
(525)

Let us apply all of this without further ado ! But I will only detail one case, because it is quite a lot of work still...

Let us take the case p = 2  with the serie  sum o o  2n- 1     V~ x-arcsin( V~ x)
  n=1 2nCn2n xn =-- V~ 1-x--  .

We obtain on one hand

  oo                       oo                       oo 
 sum  22n-1xn (1+ e2ip2n)=   sum  22n-1xn (1+ (-1)n) =  sum  -24n-x2n
n=1 nCn2n               n=1 nCn2n               n=1 2nC2n4n
(526)

and on the other

pict

By bringing those two expression together and composing with x2  , we obtain

                                                       (     V~ -----)
 sum  oo -24n--4n   xarcsin-(x)-  xarcsinh-(x)-  x-arcsin-(x)  x-ln-x-+--x2-+-1-
    2nC2nx   =   V~ 1---x2 -    V~ 1-+-x2 =   V~ 1---x2 -      V~ 1-+-x2
n=1    4n
(528)

We can see that we can directly obtain with this form a serie giving p  since we can cancel thw term in front of the logarithm (which we then obtain a diverging series if we had the idea to multiply the expression by  V~ ----2
 1 - x  and take      x = 1  ...). We need to derive the serie 528, which after multiplication by x  gives:

pict

it's not very nice... but by doing    1
x = 2  , we get

 sum  oo         V~           V~    (     V~  )
   --12n=  p-3-+ -1 - 2-5-ln  1-+--5
n=1C 4n     27    15    25       2
(530)

and we can also use an extra differention so to make appear a serie in  sum 
 o o n=1Cn2n
      4n  , but that completly ugly !

pict

by letting     1
x = 2  , we get

 sum  oo         V~ -        V~ -  (     V~ -)
    -n2n-= p--3+ -8 - --5 ln  1-+--5
n=1 C4n    54   75   125       2
(532)

Starting form here, we can use 530 and 532 to obtain

pict

then

 oo  sum  --27+-270n      V~ 
      C2n     = 4p 3
n=0     4n
(534)

There we go ! Might as well say that the other series in   kn
C 2kn  can be deduced in the same way while the method seems more complicated to use for example for the series in   2n
C7n  . If we step back to look at the two method explained above, we note that the first, is based on differential equation, gives accurate information from wehre comes the coefficients of polynomial P(n)  in  sum o o  P(n)-
  n=0 Cjknn  by giving us directly the formuation formula of those coefficient through differential equation. The second gives more fredom in concerning the produced series since we have the exact expression of the serie.

  • The method presented now gives the start of a proof for each formulae above. It comes from the article [7].
    This method consist of representing the combination as an integral, then sum it and hence obtain an integral equivalent to the serie, which can be more easily solved. Explenations!

    We consider first of all the integral representation of the function Bêta, which gives to integers value the formula

                   integral  1
(1-)= (mn + 1)   xpn(1- x)(m-p)ndx
mnpn            0
    (535)

    We then sum on n  to obtain the represantation equivalent of the sum under integral form

                 integral                 (            )
 oo  sum  -S(n)--    1 sum  oo             xp(1--x)m--p n
   (mn)an =  0    (mn + 1)S(n)       a         dx
n=0 pn         n=0
    (536)

    The polynomial S  is the key to the problem. If it is of degree d  , then to obtain the sum of  sum o o  nkyn
  n=0  , we derive successively  sum o o  yn
  n=0  and we multiply each time by y  so to compensate the yn- 1  which appears with each derivation, whichallows us to get

      oo  sum              n   --T(y)---
   (mn + 1)S(n)y =  (1 - y)d+2,
n=0
    (537)

    where the polynomial T  is of degree d + 1  . By writing that 

        xp(1- x)m-p
y = -----a------
    (538)

    we have finaly

     sum  oo  S(n)    integral  1       P(x)
   (mn-)n-=     -p------m--p----d+2-dx,
n=0  pn a     0  (x (1- x)    - a)
    (539)

    where P(x)  is a polynomial in x  of degree m(d + 1)  . The roots of the polynomial xp(1- x)m-p -a  will be fundamental in what concern the apperance of the constants. In fact, if (1 + x)  divide this polynomial, we will have some  integral 1-1 dx = ln(2)
 01+x  , after decomposition into simple elements, which appears. If      2
1 + x  divides  p      m-p
x (1 - x)   - a  , we will have some p  . The authors did not seems to consider the explicite conditions on the values m  and p  to see any remarkable constants appear, which we will do, us, for the factorial BBP formulae !

12.5 Fast combinations: Factorial formulae BBP

In the same way as the 12.2, we can find some factorial formulae having the BBP series form. There are as fast! We are therefore interested in series with the form  sum        vn ( sum         )
 o o n=0 (2-k1n)Cqnrn  pm-=11 pn1+m- , generalisation of BBP series.

Here's a little anthology, among which the proof for Guillera's formula is given. We will work out during this that the method is not that different from the second one presented for non BBP fast factorial formulae.

Then a method for a more general proof inspired by the idea [7] will have it's own special section !

Note that on the contrary to polynomial factorial formulae, here it's is very fundamental to find some series that we will call "primitive" because we can easily see that a formula in -1----1--
Cn2n2n + 1  will mechanicly give birth to a formula in --1-( --a---  --b---  ---c--)
C2n4n  4n+ 1 + 4n + 2 + 4n + 3 by division of the sum into even parts (2n  ) and odd parts (2n+ 1  ) then bringing together the terms.

To a formula Cn3n  is also associated a formula in C26nn  , etc... But it is not sure that I have alway written here uniquely some "primitive" formulae ! Doesn't matter, they are so beautiful :-) We are at least sure that for a sum            (           )
 sum o o -(-k 1n)vnqn  sum p -1-1--
  n=02  Crn   m=1 pn+m , if q  /\ r = 1  , we have a primitive formula...

An example yet again to understand this deception in the case of binomial BBP formulae: to artificially accelerate the series we can reorganise the terms in such a way so to obtain for example  oo  sum         sum  oo   sum 6
   f (n) =     f (7 n+ j)
n=0      n=0j=0  but this can be seen when we factorise the integers which then have the annoying tendence of being quite simple... Typical example, has Jesus Guillera reminded me. The formula

pict

which seems very fast, is essentialy reducable to the formula  sum  oo       sum  oo    n (                       )
   f(n) =   (-1)-  ---2-- + --2---+ ---1--
n=0      n=0 22n   4 n+ 1   4n + 2  4 n+ 3 since this breaks down to take  sum  oo   sum 6
      f (7n + j)
n=0j=0  pratically. We can see that the integers coefficients that are used are very simple (    2a j+b  mostly).

  • With Cn3n    :

  sum  oo      (              )
1   --1--- ---8--+  --2--- = p
3n=02nCn3n  3n + 1   3n+ 2
(Guillera)

          (               )
1  oo  sum  --1--- --7---  --2---
9    2nCn3n  3n+ 1 - 3n +2   = ln(2)
 n=0
(541)

1-  oo  sum  (-1)n-(--23--   --5--)
36   4nCn   3n + 1 + 3n+ 2  = ln(2)
  n=0    3n
(Guillera)

  • With   2n
C 4n    :

   sum  oo  (- 1)n2n( 1     2  )
-4    ---2n--  ---+ ------ = p
  n=1   C4n    2n   2n- 1
(Guillera)

   oo     n n (              )
1 sum  (-1)-2-  ---7---  --1--- = p
2n=0  C24nn    4n + 1   4n + 3
(Guillera)

 sum  oo -1-( --34--  --32--)   4  V~ -
   C2n   4n+ 1 + 4n+ 3   = 2p  3
n=0  4n
(542)

 sum  oo  1   ( - 5.32     3   )
   -n--2n  ------+ ------ = - 26ln(2)
n=04 C 4n   4n+ 1   4n+ 3
(543)

   oo      n(           )
- sum  -(--1)-  54-+ --72-- = pV ~ 3
 n=13nC2n4n  2n   2n- 1
(Guillera)

  • With C2n5n    :

 sum  oo    n (                                )
   (-1)-- --416- + ---44-+ --16-- + ---14--  = 125p
n=02nC25nn  5 n+ 1   5n + 2  5 n+ 3   5n + 4
(Guillera)

         (                                )
 sum  oo  (-1)n--339-   -154--  --54--   --14--
   2nC25nn  5 n+ 1 + 5n + 2 + 5 n+ 3 + 5n + 4 = 625 ln2
n=0
(Guillera)

  • With C2n
  7n    :

As with Bellard's formula, we can find a lot of very interesting sums for the C2n
 7n

 sum  oo  (- 1)n( 3342    2248    480      117     297 )        V~ -
    -C2n-  7n+-1-- 7n-+2-- 7n-+-3- 7n-+-4-  7n-+-5- = 343p  3
n=0   7n
(544)

 sum  oo    n (                                                )
   -(--1)-  386252- 135507 + 21244-+ -8112-- -7029-+ -2484-  = 470596ln(2)
n=04nC2n7n  7n +1    7n + 2   7n+ 3   7n +4   7n + 5  7n + 6
(545)

  oo        (                                               )
 sum   --1--- -59296-  -10326   3200--  -1352--  -792--  -552--
    2nC27nn  7n + 1- 7n + 2-  7n+ 3 - 7n+ 4 - 7n+ 5 + 7n + 6  = 16807p
n=0
(546)

 sum  oo   1   ( 47272    74376    8924    3718     5445     552  )
    2nC2n- 7n-+-1 + 7n-+-2- 7n-+-3-+ 7n+-4-- 7n+-5-- 7n-+6-  = 117649 ln(2)
n=0    7n
(547)

 oo  sum         (                                                )
   ---1--- 1147198 + 299568+ -34772-- -9867--  6633--- -1656-- = 1882384ln(2)
n=025nC27nn   7n+ 1    7n+ 2   7n + 3  7n + 4   7n+ 5   7n+ 6
(548)

We also obtain as often some representations of 0, uniquely for the combination which gives values to some remarquable constants (p  and ln(2)  ).

  oo       (                                               )
 sum  (--1)n   52229-  90984-  11996-  -5577-  -8811-  --828-
    C2n7n    7n+ 1 - 7n+ 2 - 7n+ 3 - 7n + 4- 7n + 5 + 7n + 6 = 0
n=0
(549)

By looking onto the side of the powers of 3, we can also find some happiness

 sum  oo  (- 1)n (116793   53244    3876    1521     3267     828  )         V~ -
   3nC2n-  7n-+1-- 7n-+-2 + 7n-+-3-- 7n+-4-+ 7n+-5-- 7n-+6-  = 16807p  3
n=1    7n
(550)

but to be truthful, that is it!

  • With   3n
C 7n

  oo  sum  --1---(1747810   185625  -98650-   31200-  5720--  -4675-)
   8nC3n   7n +1  - 7n +2 - 7n + 3 + 7n + 4 + 7n+ 5 - 7n+ 6  + 1630279 = 2352980ln(2)
n=1    7n
(551)

We don't know any formula for p  ...

  • With   2n
C 8n    :

 sum  oo   1  (  881485   371072   97275   16709    9856     3315 )
    n--2n- ------- + ------- ------+  ------- ------+ ------ = - 220ln(2)
n=0 4C 8n    8n+ 1   8n + 2  8n + 3   8n+ 5   8n+ 6   8n+ 7
(552)

  oo        (                                                  )
 sum  ---1---  9975455+ 2965376 + 550905-- -47047-- -20608--  4641-- = 224ln(2)
n=064nC2n8n   8n + 1    8n+ 2    8n+ 3   8n + 5  8n + 6   8n + 7
(553)

Note that hear also we can not apparently find similar formula for p,pV ~ 2,p V~ 3,pV ~ 5,p V~ 6  , which is quite suprising!Furthermore, the repartition of the signs of the terms in the brackets does not seem to be random at all... Is it really a primitive formula ? ?

  • With C4n8n    :

    sum  oo -(--1)k(64-  -39---  --42--  --60--)
- 8   4nC4n   4n + 4n- 1 + 4n- 2 + 4n -3   = p
   n=1    8n
(Guillera)

Jesús Guillera gave this very nice and simple formula in november 2001 but it is not a real primitive formulae but is more from a formula in   n
C 2n  .

  • With   2n
C 11n    :

         (     8764354488   -1493114756   -2349941760   )
  oo             -11n-+-1--+ --11n-+-2---+ --11n+-3----
 sum  --1---       + 519883000+ 138758080 + 72870896-+       = 2357947691p
n=02nC21n1n          11n +4     11n + 5    11n + 6
            73886592 + -91670800+  -15892240+ 15259166
             11n+ 7     11n +8      11n+ 9     11n + 10
(Guillera)

  • With C4n12n    :

The following formulae are most likely not primitive formulae but so well.... they are still very nice :-)

           (                        )
 oo  sum                1311520n+4190-+ 314279n1+429
   --n1-4n-    +40125n84+04 + 11502n2+957+ 21220n11+7    = 314p
n=016 C12n   + 1852n8+08-+ 1123n09+10-+ 142n9+411-
(554)

You will have noticed that 314p  , that's really nice !

 sum  oo        (     9111209n5+110- 2411024n+8297-    )
   ---1-4n   + 281211n8+545- 10124n17+256+ 11522n59+27   = 23315ln(2)
n=016nC 12n    --59865+  -9163-- --3952-
               12n+8   12n+10  12n+11
(555)

            (                           )
 sum  oo               2311420n45+8120- 4815291n8+7273
   ---n1-4n   + 351712n1+0425-+ 104192n8+6556-- 91692n5+470   = -28315ln(2)
n=0256 C 12n    - 310204n9+58 + 1320n10+710 + 1928n8+011
(556)

the speed is of                (      )
256nC4n   ~   a 2561212-n  ~~  5314300n
     12n n--> oo       4488

  • With C8n
  24n    :

 sum  oo   P(n)
    256nC8n-+ 249571370427585429609 = 243305.7.13.53p
n=0      24n
(557)

where          1806413025061839495168   218419888968315575695  57786660321980351944
       -    6741325446n73325629700+  24969542542n6+8117163140 +366197104243n86+9203405
          +   14312234n5+54387552768+0  22416244n41+9551770020 + 8952639234n0+49755710-
P(n) =      +   1442646n82+9832173570+  5845274n05+4140103700- +9624924471n4+0153160
              +   329241n3+60103445635+ 6594294n1+213544940  +275042564n7+9137600
                + ---24n+1757155+645674244n+13981351+20502254n+20--
                        +--24n+22--+ --24n+23--

 sum  oo --P-(n)----                            14 31
   65 536nC8n24n + 137241086910742281289392 = 2 3 5.7.13.43ln(2)
n=0
(558)

where        - 355863365937214n82380548096+ 124730316421647n+210676678485+ 26191725202422n8+82686956188
          - 19248505621494n0+5449790400-  5658706026148n0+6659307245+ 522798918244n2+5876025310
             +16217632046764162560- 1600130400592066420-  507218088018305300-
P(n) =         + 5163214n6+268496164715 + 1655820749n+81280767400- 17164642464n7+01111840
                  - 55244n4+5216390038555-+ 5821246n12+118407480+ 189347244n57+3176600
                       24n+1-7192082308891242n-+1951971757145024n+20
                              24n+22       24n+23

We can also find a function P(n)  of the same kind as  sum o o  -P-(n)--
  n=0 224nC82n4n = 0  , which represent the fastest way since a long time to calculate 0 ! !

  • With C164n8n    :

               ( 6355284697691500746427743822689241562572914688  1247765746954168283601712063903296488776216424 )
                   33011744322082215394870n82504638950531572299094+0  38511288486239761948811n5+01768160497305977517600
                  + 14264333966331821549482n5+02092119575576663395250+ 20919712264130242154338n1+3464843587638893080328
                  + ---------------48n+5---------------+ --------------48n+7--------------
                   + 81762055362572179784568n1+684803581674796560400 + 12805790943577678046884n7+8313060036565557507280
                     +5114334327206169964989n64+712159862867874359910+ 826411873463214358418n21+410393110653412727480
                     + 333913091163753891489n74+413440118030547637500+ 549483732443874324084n52+31762928680293835488
                      + 2236393234029626341842n0+51717563617661168900+ 372441164571775004687n78+9109231109083404100
                       + 152338243904226845845n1+42480313872086330000+ 25584803104931302486n2+29292892511257703944
 sum o o n=0 655316nC16n        +10502716832369394288n5+42573463216001826900+ 17750057985560144285n9+9623572596951740300
            48n           + 73065196804183540881n7+822368694863915944-+ 1240837130711064484n1+7229896329332538000
                          + 5118716098513968968017041993051100+ 872653523109331662374029939407900
                           +36061373153746381n+42289800678743296288+ 61672251805694781n09+03216759767867500
                           + 2552183444148085n5+5639227924504332680+ 4376247736732488n1+333148303949011210
                            + 181325034634488n8+6834597161611745680+ 3116954747941781n3+18307304376640400
                               129348550894982n59+03487902628326328  2239327478249487n5+4740895445517950
                             +  93992848404488n8+06431186608045600-+ 184522963090488n9+5942377437828140-
                              +         48n+441078185110096+82190614859364248n+46
                                          +         48n+47

   63                     1266 2
+3  7.74606281.2385 191 = 2  3 5 7.13.29.31.1753p

If this is not beautiful.... admire the perfectly regular decrease in the integer coefficients....

Concerning the speed, we have                   (        48 )n
65536nC1648nn   ~  a  655361468163232
           n-->o o    : 1.21 × 1018n  i.e. 18 new decimals at each itteration, great! But obviously, more terms are needed to be calculated one in the other...

12.6 So, what does it all mmean?

This time again, a few things to note :

  • The big coefficients in front of the constant can be factorised into product of fairly simple prime numbers generally, but this does not bring any structal information on the serie...
  • The integer coefficients on the --1-
an+b  can be factorise but not always with the powers of 2 for example. This does not give much indication if the serie does or not follow from an easy cut of a serie with inferior order.
  • One inconveniant of those formulae is to have to use 3a  termes in -1--
an+b  when we are dealing with a coefficient     -1an-
    C3an  in the serie. For the polynomial series, we had more a  termes in that case...
  • What about greater orders ? That is the constants G  , z(3)  etc... ? the series of kind [link towards my cubic formula] uses central binomial coefficients, but with other binomial coefficients? well funily enough we don't really find any BBP formula in the usual sens, with loads of terms ---1--
(an+b)k  . No explenation to this phenomene if it's not the complexity of the integral coming into play, which does not give any remarquable constant, but that's not really satisfying...We need to study in detail the dividors of the polynomial in the denominators of the formulae. A few elements later on.
  • We can ask ourself seing the serie that are a bit equivalent to polynomials and BBP (and we can even find some for the values m  of the coefficinets   pn
C mn  ) if there really exist some "canonic" formulae, some basic formulae, the simplest possible in the sense of factorial writting. In fact, we can notice that the BBP series can be put under polynomial form if we really want it
           sum  oo  (- 1)n (64   39      42       60 )         oo  sum  (-1)n(4n- 1)!(4n -4)!
p = -8   -n--4n-  --+  ------+ ------+ ------ = - 64   ------n-------------P(n)
      n=14 C 8n   4n   4n- 1   4n- 2   4n- 3        n=1     4 (8n- 1)!


    with            3      2
P (n) = 820n - 927n +296n - 24  . But of course, we can see there that we can not put the factorials under combiatorial form. It is not purely due to notation, the combination have a real sense and a coherance in the development of power series as Newton's binomial formula shows.
    In fact, the integral representation shows that often the BBP form is more natural and simple that the polynomial form, but that only an intuitive remark, there is still missing some clarification to the evaluation of the complexity of one of those series compare to the others.

12.7 Typical Proof

Jesús Guillera and myself offer here a method for proof for those formulae (which does not involve  ---1-- -1-
qnCpnmn mn ), this is done to show the existance of this kind of proof for each formulae, followed by an application of the first formula of the section. This method was inspired from the one [7], by atempting to go just a bit further in the detail...

12.7.1 La méthode

The serie is presented under the form

     sum  oo -1---( --b1---  --b2---      ---bm--1---)
T =    qnCpmnn   mn + 1 + mn + 2 + ...+ mn + (m - 1)
    n=0
(559)

Great...

We use the integral representration of the function Beta so to directly obtain that

 1             integral  1                            integral  1
-pn-= (mn + 1)   xpn(1- x)(m-p)ndx = (mn  + 1)    x(m -p)n(1- x)pndx
Cmn            0                              0
(560)

which gives us the sum

pict

from which we can deal with by decomposition of the denominator into simple elements (careful, I never said it was easy ! :-) ).

Good, but what do we do for the   1     1
-n-pn--------
q Cmn mn + k  , k > 1   ? ?

For a fixed k  , the decomposition into simple elements of (pn+1)(pn+2)...(pn+k-1)
(mn+1)(mn+2)...(mn+k)  seing it's a rational fraction in n gives a sequence (ak,j)j=1,..,k  such that

                               (                             )
 integral  1 k-1 pn     (m -p)n     --1--  -ak,1--- --ak,2--      --ak,k--
 0 x   x  (1- x)      dx = (mn ) mn + 1 + mn + 2 + ...+ mn +k
                            pn
(562)

with a     (-  Q
 k,j  . For all values k  from 1  to m - 1  , we then have some sequences (a  )
  k,j j=1,..,k  that we can linearly combine so to form the coefficients b
 k  of the sum T  . This is the same as writing down and solving the system

( a1,1  a2,1  ...  am-1,1 )      (   b1  )
   0   a2,2  ...  am-1,2            b2
   ...   0   ...    ...    .B =     ...
   0    0   ...  am-1,k           bm -1
(563)

The vector B contains the coefficients of the linear combination of the integrals containing xk- 1  , which gives the polynomial            (   1   )
         t     X
P (X) = B  .    ...
             Xm -2 of degree m  -2  such that

pict

There we go, it's not that hard !

We can also, if we worry about the construction of formulae and not of proofs, choose a polynomial P (x)  which voluntary simplify certain divisors of the denominator of the equivalent integral. We then found ourself in a situation which I spend time on in 12.7.4.

12.7.2 Application : Proof of the first formula

We take yet again the first formula by Guillera

  sum  oo      (              )
1   --1--- ---8--+  --2--- = p
3n=02nCn3n  3n + 1   3n+ 2
(565)

We use the integral formula

             integral 
--1----1--=   1x2n(1- x)ndx
3n+ 1 Cn3n    0
(566)

which is proved by recurrance for example for the pure mathematician! We deduce that

pict

For  1    1
Cn--3n+-2-
 3n  , it's a bit more complicated, butwe can found our way around anyway by following the method of the above section, know that

                     integral  1
---(2n+-1)-----1n-=    x2n+1(1 - x)ndx
(3n + 1)(3n + 2)C3n    0
(568)

We have

pict

hence

  1    1      integral  1                  1    1
-------n--= 3  x2n+1(1- x)ndx - --------n--
3n+ 2 C3n     0                 (3n + 1)C3n
(571)

which is equivalent to considering the famous polynomial we are looking for is P (x) = 3x- 1  because we have

--1---1--   integral  1       2n      n
3n + 2Cn3n =  0 (3x- 1)x  (1 - x) dx
(572)

since the polynomial is simple, we can explicitely determine it, but this will not always be the case (see proof below).

The calculation is now direct

pict

hence finaly by recopying the method of 6.2by Gery Huvent , we have

 sum  oo   1  (   a       b   )   (3    12 )        (1     7  )
   -n--n-  ------+ ------ =   -a - --b  ln(2)+  - a+ -- b p
n=02 C 3n  3n+ 1   3n+ 2      5     5           5    10
(574)

  We now choose a = 4b = 4  so to cancel down the participation of our dear ln(2)  and obtain  Guillera. We can also let a = - 72b = - 7  so to obtain 541.

12.7.3 Proof of the formula in C27nn

Jesus Guillera and myself now offer you a quick proof of one of the best formula in the packet,  546.

We remind ourself that the principal is to find a polynomial Pk(x)  such that

--1----1--   integral  1     5n      2n
7n+ k C27nn =  0 Pk(x) x (1 -x) dx

While the solution, for each k, is not unique, we know that the degree of Pk(x)  is k - 1  . By the method of linear system, we get

|------|----------------------------------------------|
|k-----|--------------------Pk(x)---------------------|
|k-=-1-|----------------------1-----------------------|
|k = 2 |                    7x- 2                     |
|------|----------------49--3---3--1------------------|
|k = 3 |                -- x2- 7x- -                  |
|------|------------343-34-294-2---414----2------------|
|k-=-4-|------------13-x-+--13 x---13x---13-----------|
|k = 5 |      - 2401x4 + 2744x3 - 49x2- 56x- 1        |
|------|---------99------99------11-----99---9--------|
|k = 6 |- 16807x5 + 12005x4 - 1715x3- 245x2 - 35x- 2- |
|------|---276------138-------69-----138-----92----23-

Now, we put it together....

 sum  oo       sum 6         integral  1 sum 6      sum  oo                integral  1    sum  oo  5n     2n
   -n1-n-   -ak---=      akPk(x)   x5n(1- x)2n dx =   P (x)   x--(1-nx)--dx =
n=02 C 2n k=1 7n+ k    0 k=1       n=0                0     n=0     2

 integral  1   --------2------
 0 P(x)x7 -2x6 + x5- 2dx

With the ak  of the formula 546, we have         sum 6
P (x) =    akPk(x) = 4- 4x2 + 4x4 - 2x5
       k=1

and

 integral  1                          integral  1     2    4    5       integral  1
   Q(x)-7-----6-2--5---dx = -2   4--7-4x-+64x--5-2x- dx = 4  ---1-2 dx = p
 0     x - 2x  + x - 2        0  x  - 2x + x - 2        0 1 + x
(575)

Incredible, no ? ? That this quite big sum is in fact equivalent to a very simple integral, it's great!

12.7.4 Predictions of formulae

Looking at the proof, we might ask ourself if there exist a way to predict for which combinations there exists a formula. I have not for the moment a complete answer, but a sufficient condition is already quite simple to show.

We have seen in fact with the formula in C12n7n-  that the integral was equivalent to  integral  1  1
   -----2dx
 0 1 + x  . This comes from the fact that  1 + x2  divides x7- 2x6 + x5 - 2  and the polynomial in the numerator (hence the coefficients) is only used to simplify the other factors. But for non alternating formulae, the polynomial in the denominator comes from q- x(m-p)(1 - x)p  which we will call a generating polynomial. Starting from that, a sufficient existance condition for a formula for p  is that

1+ X2 || K(X) = q- X(m -p)(1 - X)p
      |

Hence we need that i  and - i  are roots of K  . In particular, we have for p = 2  , q = 3  ,       2        2
i(1- i) = i(1 + i) = 2  . This proof the existance for a formula in  2n
C3n  for q = 2  . But since i  and -i  are 4th root of 1  , we can multiply        2
x(1- x)  by  4k
x  without changing the existance of the roots i  and -i  . This then proof the existance of a formula in   2n
C (3+4k)n  .

Let us try to specify the sufficient condition :

{i et - i sont racines de K, p < m}   {       (m-p)                }
<==>  q - (± i)    (1± i)p = 0,p < m

<==>  {(1 ± i)p = (- 1)m+pq(± i)m,p < m}

<==>  {(1 + i)p = (- 1)m+pqim, p < m} by conjugate

   {                                       }
<==>   ( V~ 2)p = q et pp  =_  (m + p)p+ m p [2p],p < m
                4               2 by using modulus and argument

   { p                       }   {        k                       }
<==>   22 = q, p < m et - p  =_  2m [8] <==> p = 2k, 2 = q, p < m et 7p  =_  2m[8]

Things are clear. So that we can obtain a non alternating factorial BBP formula thanks to the integral  integral  1
    --1--dx
 0  1+ x2  , it is necessary and sufficient that q  is a power of 2  , p  even and smaller than m  and most of all the great relation of congruence 7p  =_  2m [8]  . We can easily check that the conditions p = 2,q = 2,m = (3+ 4k)  satisfy those conditions. We also find that the condition is satisfied for  p = 4,q = 4,m = 6+ 4k  , formulae that I missed in my experimentations! Just to show, that sometimes the theory is faster than the pratical... :-)

By applying the same method to alternating series, to the roots -1  (which allows us to find ln(2)  thanks to  integral 
  1 -1--dx
 0  1+ x  ), to the polynomial  2
x - x + 1  (so to find          integral 
  V~ -   9  1----1----
p  3 = 2 0 1- x + x2dx  ), we find the following conditions, and hence the following formulae :

For non-alternating series   oo        (                                  )
 sum  ---1--  --a1---+ --a2---+ ...+ -----a2-----
n=0qnCpnmn  mn + 1   mn + 2       mn + (m - 1) =  integral 1----qP(x)----dx
   0 q- x(m-p)(1-x)p






Pole Constant Condition with roots Condition of existance of pole Type of existing formula





1+ x2  p    (m -p)
(i)     (1 - i)p = q  p = 2k, 2k = q,7p  =_  2m [8]    oo  sum      1       sum  oo    1        sum  oo    1
    2nC2n----...,   4nC4n-----...,   8nC6n-----
n=0    (3+4k)n  n=0    (6+4k)n   n=0    (9+4k)n





1+ x  ln(2)      m-p p
(- 1)   2 = q                 p
m = p+ 2k, q = 2  Quite a few seing the relation !





1- x + x2  p V~ 3       (    V~ -)p+m
(- 1)p 1+i-3     = q
        2  q = 1, m = 2p+ 6k    oo  sum        sum   oo       sum   oo 
    -1-...,  --1-...,   -1-...
n=0 Cn2n  n=0C2n4n  n=0 Cn8n





2+ x    (3)
ln 2     (m- p)   p
(- 2)    (3) = q                 2k p
m = p+ 2k, q = 2 3    oo  sum  --1----  sum  oo --1---   sum  oo --1---
    12nCn3n...,   48nCn5n ...,   36nC2n4n ...
n=0         n=0         n=0





z- 1 + x    (z+1)
ln  z      (m-p)   p
(1 - z)    (z) = q                      2kp
m = p+ 2k, q = (1- z) z  ...





1+ (z- 1)x  ln(z)  (    )(m -p)(       )p
 -z1-1        1+ z1-1   = q                    p
m = p+ 2k, q = (z-z1)p+2k  ...







Note that when we need a polynomial with complex roots that divides our generating plynomial, we only need one of the complex root that is root of this polynomial, by conjugates (due to the fact that the generating polynomial is real). This table indicates for example why we find so often some formulae for ln(2)  but not always for p  . While we are dealing here with a sufficient condition and not a necessary one, the conditions on the logarithm shows that it is probable that we can not obtain any other intigers logarithm than ln(2)  with some integer powers q  .

For alternating series  oo  sum      n (                                  )
   (-n1)pn-  --a1--+ ---a2--+ ...+ -----a2-----
n=0q Cmn   mn + 1  mn  + 2      mn + (m - 1)    integral 
=  01q+x(mq-P(xp))(1-x)pdx






Pole Constant Condition with roots Condition of existance of pole p < m  Type of existing formula





    2
1+ x  p    (m -p)     p
(i)     (1 - i)  = -q          k
p = 2k, 2 = q,7p  =_  2m + 4[8]    oo  sum  --(--1)n---  sum  oo -(--1)n---   sum  oo ---1-----
    2nC2n    ...,   4nC4n     ...,   8nC6n
n=0    (5+4k)n  n=0    (8+4k)n   n=0    (7+4k)n





1+ x  ln(2)  (- 1)m-p+12p = q  m = p+ 2k + 1, q = 2p  Quite a few seing the relation  !





1- x + x2    V~ -
p  3         (   V~ -)p+m
(- 1)p+1  1+i2-3     = q  q = 1, m = 2p+ 3+ 6k    oo  sum     n    sum  oo    n
    (--1n)-...,   (-1n)-...
n=0 C 5n   n=0 C11n





2+ x    (3)
ln 2     (m- p)   p
(- 2)    (3) = - q                    2k p
m = p+ 2k + 1, q = 2 3    oo            oo            oo 
  sum  (-1)n--  sum  -(-1)n-   sum  -(-1)n-
    12nCn4n...,   48nCn6n ...,   36nC2n5n ...
n=0         n=0         n=0





z- 1 + x  ln(z+1)
    z (1 - z)(m-p)(z)p = - q  m = p+ 2k + 1, q = (1- z)2kzp  ...





1+ (z- 1)x  ln(z)  (    )(m -p)(       )p
 -z1-1        1+ z1-1   = q                       p
m = p+ 2k + 1, q = (z-z1)p+2k  ...








12.8 Product of combinaisons

A very interesting questiong concern the possible presence of several combinations in a serie giving p  . I have to admit that I thought this was very improbable, until I discovered the following formula in december 2001, thanks to Jesús Guillera

 sum  oo          3n  (                               )
   (-1)n2n--C6n--  -1885--+ --965-+ -363--+ ---51-  = 29p
n=0       C4n8nCn4n  8n+ 1   8n+ 3   8n + 5  8n + 7
(Guillera)

The same with factorials

  oo                   (                               )
 sum      n n(6n)!(4n)!n!  1885--  --965-  -363--  --51--     9
   (- 1) 2  (3n)!(8n)!   8n + 1 + 8n+ 3 + 8n+ 5 + 8n +7   = 2p
n=0
(Guillera)

We also have

 sum  oo        C3n    ( 11855   2295     297      17  )       V~ -
   (-1)n33nC64nnCn--  8n-+-1-- 8n+-3-+ 8n+-5-- 8n-+7-  = 211p  3
n=0         8n  4n
(Gourevitch)

or

 sum  oo                 (                               )       V~ -
   (-1)n-(6n)!(4n)!n!  -11855-  2295--+ -297--- --17-- = 211p  3
n=0     33n(3n)!(8n)! 8n + 1   8n+ 3   8n+ 5   8n+ 7
(Gourevitch)

So, where does this comes from ? Well instead of starting from integers power in the integral, we use rational powers and in particular square roots. In fact, considering the integral

 integral 
  1bnxkn-12(1- x)jndx
 0
(576)

we obtain sums of the form

 sum  oo    n(2kn)!(jn+-kn)!(jn)!
   (-1)  an(2jn + 2kn)!(kn)!
n=0
(577)

or for example for k = 3  and j = 1

 sum  oo 
   (-1)n(6n)!(4n)!(n)!
n=0      an(8n)!(3n)!
(578)

or even for k = 3  and j = 2

 oo  sum     n (6n)!(5n)!(2n)!
   (- 1) an(10n)!(3n)!
n=0
(579)

The BBP sums are then deduced after integration so to give formulae of the type

 sum  oo     (6n)!(4n)!(n)!(   b       b       b       b   )
   (- 1)n--n---------  ---1--+ ---3--+ ---5--+  --7---
n=0      a (8n)!(3n)!  8n +1   8n + 3  8n + 5   8n + 7
(580)

  oo  sum                   (                                            )
   (- 1)n (6n)!(5n)!(2n)!- --b1---+ --b3---+ ---b5---+ ---b7--+ ---b9--
n=0      an(10n)!(3n)!  10n +1   10n + 3  10n + 5  10n + 7  10n + 9
(581)

For the first formula of Guillera, we consider for example the following proof

Proof.  integral  1 V~ -( 2         2-) sum  oo  (--1)nx3n(1--x)n     sum  oo  (-1)n integral  1 V~ -( 2        2)  3n     n
 0  x  x  -2x + 2- x            2n      dx =      2n  0   x  x - 2x+ 2 - x  x  (1 - x)dx
                      n=0                    n=0

and we then use the known values of  (    1)
G n + 2 .  _

An other example contains

pict

where K(k1)  is the elliptical function of first sort in the first singular value, famous constant of elliptical theory !

It is obtained from 

 integral                      (    )  (     )     ( )2
  1 n-14      n- 12    G--n+-34--G-n-+-12-   4G-34--(--1)n -----------(4n)!-----------
 0 x   (1- x)    dx =    G(2n + 54)    =    V~ 2p   8n  (8n + 1)n!(4n - 3)!!!!(8n - 3)!!!!
(583)

and

 integral  1  3         1    G(n + 7)G(n  + 1)   4G(3)2(- 1)n            (4n)!
   xn+4(1- x)n- 2dx =------4(----9)--2-=  - V~ -4---8n- (8n-+-5)n!(4n---3)!!!!(8n---3)!!!!
 0                       G 2n + 4          2p
(584)

and the integral

 integral  1       oo  sum      n
   (x- 2)   (-1)-xn-14(1 -x)n- 12dx = - p
 0       n=0 2n
(585)

All of this open some very interesting perspectives !

Equaly note that the BBP form is is a very general form of series in the sense where even the formulae like the ones by Ramanujan can be put under BBP form, like

-1- sum  oo  (3n)!(2n)!4n-( -2448 -644--  -3552-)   1-
125    (n!)563n27n   2n- 1 + 3n- 1 + 3n- 2  =  p
   n=0
(586)

On the other hand, unfortunatly, we still have not find a method to pove those formulae with the method Bêta...

A lot more details and a few other formulae will be given in the article [13], available sometime soon I hope !


back to home page