A135860 - OEIS

%I #42 Feb 20 2022 02:11:56

%S 1,2,15,220,4845,142506,5245786,231917400,11969016345,706252528630,

%T 46897636623981,3461014728350400,281014969393251275,

%U 24894763097057357700,2389461906843449885700,247012484980695576597296,27361230617617949782033713,3233032526324680287912449550

%N a(n) = binomial(n*(n+1), n).

%H Seiichi Manyama, <a href="/A135860/b135860.txt">Table of n, a(n) for n = 0..337</a>

%H 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 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n^2,k). - _Paul D. Hanna_, Nov 18 2015

%F a(n) is divisible by (n+1): a(n)/(n+1) = A135861(n).

%F a(n) is divisible by (n^2+1): a(n)/(n^2+1) = A135862(n).

%F a(n) = binomial(2*A000217(n),n). - _Arkadiusz Wesolowski_, Jul 18 2012

%F a(n) = [x^n] 1/(1 - x)^(n^2+1). - _Ilya Gutkovskiy_, Oct 03 2017

%F a(n) ~ exp(n + 1/2) * n^(n - 1/2) / sqrt(2*Pi). - _Vaclav Kotesovec_, Feb 08 2019

%F a(p) == p + 1 ( mod p^4 ) for prime p >= 5 and a(2*p) == (4*p + 1)*(2*p + 1) ( mod p^4 ) for all prime p. Apply Mestrovic, equation 37. - _Peter Bala_, Feb 27 2020

%F a(n) = ((n^2 + n)!)/((n^2)! * n!). - _Peter Luschny_, Feb 27 2020

%t Table[Binomial[n^2 + n, n], {n, 0, 16}] (* _Arkadiusz Wesolowski_, Jul 18 2012 *)

%o (PARI) a(n)=binomial(n*(n+1),n)

%o for(n=0,15,print1(a(n),", "))

%o (PARI) a(n)=sum(k=0,n,binomial(n,k)*binomial(n^2,k))

%o for(n=0,15,print1(a(n),", "))

%o (Magma) [Binomial(n*(n+1), n): n in [0..30]]; // _G. C. Greubel_, Feb 20 2022

%o (Sage) [binomial(n*(n+1), n) for n in (0..30)] # _G. C. Greubel_, Feb 20 2022

%Y Cf. A014062, A014068, A054688, A107863, A135861, A135862.

%K easy,nonn

%O 0,2

%A _Paul D. Hanna_, Dec 02 2007