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A107863 Column 1 of triangle A107862; a(n) = binomial(n*(n+1)/2 + n, n).

%I A107863 #22 Feb 20 2022 02:28:52
%S A107863 1,2,10,84,1001,15504,296010,6724520,177232627,5317936260,
%T A107863 179013799328,6681687099710,273897571557780,12233149001721760,
%U A107863 591315394579074378,30756373941461374800,1712879663609111933495,101696990867999141755140
%N A107863 Column 1 of triangle A107862; a(n) = binomial(n*(n+1)/2 + n, n).
%H A107863 G. C. Greubel, <a href="/A107863/b107863.txt">Table of n, a(n) for n = 0..350</a>
%H A107863 R. Mestrovic, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv:1111.3057 [math.NT], 2011.
%F A107863 a(n) = [x^(n*(n+1)/2)] 1/(1 - x)^(n+1). - _Ilya Gutkovskiy_, Oct 10 2017
%F A107863 From _Peter Bala_, Feb 23 2020: (Start)
%F A107863 Put b(n) = a(n-1). We have the congruences:
%F A107863 b(p) == 1 (mod p^3) for prime p >= 5 (uses Mestrovic, equation 35);
%F A107863 b(2*p) == 2*p (mod p^4) for prime p >= 5 (uses Mestrovic, equation 44 and the von Staudt-Clausen theorem).
%F A107863 Conjectural congruences:
%F A107863 b(3*p) == (81*p*2 - 1)/8 (mod p^3) for prime p >= 3;
%F A107863 3*b(4*p) == -4*p (mod p^3) for all prime p. Cf. A135860 and A135861. (End)
%t A107863 Table[Binomial[n*(n+3)/2, n], {n,0,40}] (* _G. C. Greubel_, Feb 19 2022 *)
%o A107863 (PARI) a(n)=binomial(n*(n+1)/2+n,n)
%o A107863 (Sage) [binomial(n*(n+3)/2, n) for n in (0..40)] # _G. C. Greubel_, Feb 19 2022
%Y A107863 Cf. A014068, A107862, A135860, A135861.
%K A107863 nonn
%O A107863 0,2
%A A107863 _Paul D. Hanna_, Jun 04 2005