%I #15 Apr 09 2016 10:33:19
%S 1,1,5,45,781,23981,1371885,145101805,29560055405,11546945197165,
%T 8881721878376045,13338290506465706605,39879639563413780322925,
%U 234862804790553590007179885,2768979430068663216466330446445,64586918396493458414460474344516205,3024204274887062319005574660727125346925
%N Self-convolution square-root of A243950, which is the sums of the squares of the q-binomial coefficients for q=2 in rows of triangle A022166.
%H Paul D. Hanna, <a href="/A243951/b243951.txt">Table of n, a(n) for n = 0..80</a>
%F a(n) ~ c * 2^(n^2/2-1), where c = 18.0796893855819714431... if n is even and c = 18.02126069886312898683... if n is odd (constants same as for A243950). - _Vaclav Kotesovec_, Jun 23 2014
%e G.f.: A(x) = 1 + x + 5*x^2 + 45*x^3 + 781*x^4 + 23981*x^5 + 1371885*x^6 +...
%e where
%e A(x)^2 = 1 + 2*x + 11*x^2 + 100*x^3 + 1677*x^4 + 49974*x^5 + 2801567*x^6 + 293257480*x^7 + 59426801521*x^8 +...+ A243950(n)*x^n +...
%e The terms in this sequence appear to be divisible by 5 everywhere except
%e a(n) == 1 (mod 5) when n = {0,1,4,5,20,21,24,25,100,101,104,105,120,121,124, 125,500,501,...}.
%t a[n_] := SeriesCoefficient[Sqrt[Sum[x^m Sum[QBinomial[m, k, 2]^2, {k, 0, m}], {m, 0, n}]], {x, 0, n}]; Table[a[n], {n, 0, 16}] (* _Jean-François Alcover_, Apr 09 2016 *)
%o (PARI) {A022166(n, k)=polcoeff(x^k/prod(j=0, k, 1-2^j*x+x*O(x^n)), n)}
%o {a(n)=polcoeff(sqrt(sum(m=0,n,x^m*sum(k=0,m,A022166(m, k)^2) +x*O(x^n))),n)}
%o for(n=0,20,print1(a(n),", "))
%Y Cf. A243950, A022166.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jun 21 2014