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For integers $n>1$, let $$ s_n=\prod_{k=0}^{n}\binom{n}{k}=\prod_{k=1}^{n-1}\binom{n}{k} $$ $s_n$ denoting the product of a row of the Pascal Triangle.

Also let $$ t_n=\frac{s_n}{n^{n-1}} =\prod_{k=1}^{n-1}\frac{\binom{n}{k}}{n} $$

If $n$ is prime, by this result, each term in the product for $t_n$ is an integer, and thus $t_n$ is an integer.

My question is, for what composite numbers $n$ is $t_n$ an integer?

Note:

I believe the first composite number to satisfy the above, to be $n = 40$, with $t_n$ equalling:$$ 36990392961319419974642642130188695887669034562410788297544527529420038637863756835263205790947107511723092990405151720013340165976289167766496428365384796272978063487332374591282929938651354547637805475295730509532867521205942260299808874191672263200800 $$

Note 2: It turns out the above is incorrect: $n=36$ is the first valid solution. Thank you Hagen von Eitzen!

Garam Lee
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1 Answers1

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The sequence of such numbers up to $1000$ is:

36, 40, 63, 80, 84, 90, 105, 108, 132, 144, 150, 154, 160, 165, 168, 175, 176, 180, 182, 195, 198, 200, 208, 210, 220, 260, 264, 270, 273, 275, 280, 286, 288, 297, 300, 306, 308, 312, 315, 320, 324, 330, 340, 351, 357, 360, 364, 374, 378, 380, 385, 390, 392, 396, 399, 400, 408, 416, 418, 420, 425, 429, 432, 440, 441, 442, 450, 455, 456, 462, 468, 476, 490, 513, 520, 525, 528, 532, 539, 540, 544, 546, 550, 552, 560, 567, 570, 572, 575, 576, 585, 588, 595, 598, 608, 612, 616, 621, 624, 627, 630, 640, 644, 646, 650, 660, 665, 672, 675, 680, 684, 690, 693, 700, 704, 714, 728, 735, 736, 741, 748, 750, 756, 759, 760, 765, 770, 780, 782, 784, 792, 798, 800, 805, 810, 816, 819, 825, 828, 833, 840, 855, 858, 864, 870, 874, 875, 880, 882, 884, 891, 897, 900, 910, 912, 918, 920, 924, 928, 931, 935, 936, 945, 950, 952, 966, 969, 972, 975, 986, 988, 990

In particular, $$t_{36}= 29{,}320{,}324{,}926{,}758{,}361{,}405{,} \\581{,}338{,}584{,}327{,}970{,}520{,}842{,}477{,}135{,}693{,} \\769{,}332{,}565{,}195{,}583{,}960{,}633{,}771{,}857{,}532{,} \\829{,}449{,}603{,}761{,}587{,}137{,}899{,}623{,}812{,}510{,}\\923{,}539{,}078{,}324{,}218{,}025{,}713{,}952{,}323{,}436{,}\\243{,}765{,}489{,}948{,}359{,}999{,}055{,}207{,}080{,}000{,}\\000{,}000{,}000{,}000{,}000{,}000{,}000{,}000{,}000{,}000 $$

Of course, this is already in http://oeis.org/A276710. Fro the comments there, note that it looks as if $n^{n-1}$ can be replaced with $n^n$.

Hagen von Eitzen
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  • Seems I have to upload a new b-file to OEIS later today :) – Hagen von Eitzen Apr 15 '22 at 10:12
  • Thank you! Of course it's on OEIS. Should have done a better search of it... – Garam Lee Apr 15 '22 at 10:22
  • @GaramLee Well, I didn't find it before computing the terms and looking them up ... – Hagen von Eitzen Apr 15 '22 at 10:23
  • Haha May I enquire for the code? - mine is riddled with problems with decimal inaccuracies, leading me to miss 36. – Garam Lee Apr 15 '22 at 10:27
  • @GaramLee Initially, I used PARI/GP with t(n)=prod(k=1,n-1,binomial(n,k)/n) and then for(n=1, 1000, if (!isprime(n) && denominator(t(n))==1, print(n)))) But for finding results for larger $n$ fast, I wrote a more complex C program that checks the $p$-adic valuation of the product for all $p\mid n$. With that optimization, it doesn't take long to find that the millionth such number is 3741190 – Hagen von Eitzen Apr 15 '22 at 19:58