12 Other binomial coefficients

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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 $2n C 6n$ :

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 $4n C 8n$ :

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 $C244n8n$ :

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,

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

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 a $arcsin$ . 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

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)

$\text{[math]}$

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:

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 !

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

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

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

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

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

$\text{[math]}$

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

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

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

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)

$\text{[math]}$ 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)

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

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 !

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