So, lets take a natural integer *n *different from 0. For example *10*. So far everything is all right !

Now let's considerthe nearest "multiple" bigger than or equal to *n-1*.

In our case, we find *18* because it is a "multiple" of *9=10-1* and bigger than *10*.

Let us repeat this considering the nearest "multiple" bigger or equal to *n-2*, here *24*, then of *n-3* (*28*), of *n-4* (*30*) and so on for *n-k* unto we get to *k=n-1*. We call *f(n)* the result (*f(10)=34*)

Well, funnily enough,

Erdös and Jabotinski proved that more precisely :

The simplest thing is to take K.Brown's example again with the starting number *n=100*.

Here is a table where *x* represents the *n-k* and *y* the multiple bigger or equal to the previous *n-k+1*. It's what we did with *x=9*, *y=18* then *x=8* and *y=24* and so on...

We will write *w=y/x*

x w y x w y x w y x w y
100 1 100 75 26 1950 50 51 2550 25 122 3050
99 2 198 74 27 1998 49 53 2597 24 128 3072
98 3 294 73 28 2044 48 55 2640 23 134 3082
97 4 388 72 29 2088 47 57 2679 22 141 3102
96 5 480 71 30 2130 46 59 2714 21 148 3108
95 6 570 70 31 2170 45 61 2745 20 156 3120
94 7 658 69 32 2208 44 63 2772 19 165 3135
93 8 744 68 33 2244 43 65 2795 18 175 3150
92 9 828 67 34 2278 42 67 2814 17 186 3162
91 10 910 66 35 2310 41 69 2829 16 198 3168
90 11 990 65 36 2340 40 71 2840 15 212 3180
89 12 1068 64 37 2368 39 73 2847 14 228 3192
88 13 1144 63 38 2394 38 75 2850 13 246 3198
87 14 1218 62 39 2418 37 78 2886 12 267 3204
86 15 1290 61 40 2440 36 81 2916 11 292 3212
85 16 1360 60 41 2460 35 84 2940 10 322 3220
84 17 1428 59 42 2478 34 87 2958 9 358 3222
83 18 1494 58 43 2494 33 90 2970 8 403 3224
82 19 1558 57 44 2508 32 93 2976 7 461 3227
81 20 1620 56 45 2520 31 96 2976 6 538 3228
80 21 1680 55 46 2530 30 100 3000 5 646 3230
79 22 1738 54 47 2538 29 104 3016 4 808 3232
78 23 1794 53 48 2544 28 108 3024 3 1078 3234
77 24 1848 52 49 2548 27 112 3024 2 1617 3234
76 25 1900 51 50 2550 26 117 3042 1 3234 3234

You must read the table from *x=100* towards *x=1*. The interval between two following *y* values gets smaller and smaller, but that is normal when you see *y*'s construction. As long as it doesn't reach *0*, *w* gets bigger, by one unit each time. As the interval between two *y* values gets smaller by *2* each time, obviously, this interval reaches *0* for *x=50*.

Up to this stage, we can modelize *y *by the *f*_{1} parabola:

*y=(101-x)x*
Then, naturally, the interval beween two *y* values gets smaller by 4 each time, and so *w* gets bigger by *2* up to *x=38*. The model is then written as the *f*_{2} parabola, whose equation is :

*y=(151-2x)x*
which reaches a maximum, and so its derivative is equal to *0*, so the interval is equal to *0* with the previous one, for and that for the *x *integers

And in fact, from *x=38*, the interval between two following *y* values gets smaller by 6 each time, and w gets bigger by *3*. *y* is then equal to :

*y=(189-3x)x*
which represents the *f*_{3} parabola, and so on.

And so for the *k*th parabola, we can write :

*y=(A*_{k}-k.x)x
with *A*_{k} integer. If we derive to know the maximum of this parabola as we did previously, we get, for *x* :

(1)
To find the value of *A*_{k}, you just need to calculate the intersection point in between the *k*-th parabola and the previous one's maximum, which means :

(2)
We can see it on the following graph, where *f*_{1}, f_{2}, f_{3}... are respectively the parabolas of which we have calculated the equations :

we then replace *y*_{k-1} by the value found in (1) and so we get :

But then, with the *A*_{k} value found in (1), we can write :

for *k>1*, because for *k=1*, we have :

Starting with *x*_{0}=y_{0}=n as we start from the integer considered at the beginning, we get :

When *k* approaches the infinite, *y*_{k} always reaches the *f(n)* value (as for *n=100 *for example, *k* can not go over *100*)

To conclude, when *n* gets towards the infinite, we then get :

according to Wallis !

And that's another nice result.