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%I #20 Jan 06 2022 15:32:05
%S 1,1,2,10,106,1930,53612,2108560,111482424,7625997280,655331699940,
%T 69110082376388,8775534280695310,1320693932817784342,
%U 232459627389638257316,47311901973588298051380,11025565860152700884475938,2916827988004938784779055448
%N Sum of squared coefficients of q in the q-factorials.
%C Two n-permutations are randomly selected from S_n with replacement. a(n)/(n!)^2 is the probability that they will have the same number of inversions. - _Geoffrey Critzer_, May 15 2010
%H Vaclav Kotesovec, <a href="/A127728/b127728.txt">Table of n, a(n) for n = 0..100</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/q-Factorial.html">q-Factorial</a>.
%F a(n) ~ 6 * sqrt(Pi) * n^(2*n - 1/2) / exp(2*n). - _Vaclav Kotesovec_, Oct 22 2020
%F a(n) = Sum_{k>=0} A008302(n,k)^2. - _R. J. Mathar_, Jan 06 2022
%e Definition of q-factorial of n:
%e faq(n) = Product_{k=1..n} (1-q^k)/(1-q) for n>0, with faq(0)=1;
%e faq(4) = 1*(1 + q)*(1 + q + q^2)*(1 + q + q^2 + q^3) = 1 + 3*q + 5*q^2 + 6*q^3 + 5*q^4 + 3*q^5 + q^6;
%e then a(n) is the sum of squared coefficients of q:
%e a(4) = 1^2 + 3^2 + 5^2 + 6^2 + 5^2 + 3^2 + 1^2 = 106.
%t Table[Total[ CoefficientList[Expand[Product[Sum[x^i, {i, 0, m}], {m, 1, n - 1}]], x]^2], {n, 0, 15}] (* _Geoffrey Critzer_, May 15 2010 *)
%o (PARI) {a(n)=local(faq_n=if(n==0,1,prod(k=1,n,(1-q^k)/(1-q)))); sum(k=0,n*(n-1)/2,polcoeff(faq_n,k,q)^2)}
%Y Cf. A008302.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 25 2007
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