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%I #7 Mar 21 2018 16:21:00
%S 1,21,1512,182448,30845052,6706403424,1782361433664,559861341721920,
%T 202922346528231120,83358099246202940880,38271708686845732234752,
%U 19421327571536329073316864,10794249397953336851774993664,6521104275997643157262604783616,4254768377324045826054766465227264,2981719456871640091643441908508931072
%N G.f.: Sum_{n>=0} 3^n * (1+x)^(n^2) / 4^(n+1).
%C a(n) is divisible by 21 for n>0 (conjecture).
%C In general, for s > 1, Sum_{k>=0} binomial(k^2, n) / s^k is asymptotic to 2^(2*n + 1/2) * n^n / (s^(log(s)/8) * exp(n) * (log(s))^(2*n + 1)). - _Vaclav Kotesovec_, Mar 21 2018
%H Paul D. Hanna, <a href="/A301432/b301432.txt">Table of n, a(n) for n = 0..100</a>
%F G.f.: 1/(4 - 3*q/(1 - 3*q*(q^2-1)/(4 - 3*q^5/(1 - 3*q^3*(q^4-1)/(4 - 3*q^9/(1 - 3*q^5*(q^6-1)/(4 - 3*q^13/(1 - 3*q^7*(q^8-1)/(4 - ...))))))))) where q = (1+x), a continued fraction due to a partial elliptic theta function identity.
%F G.f.: Sum_{n>=0} 3^n/4^(n+1) * (1+x)^n * Product_{k=1..n} (4 - 3*(1+x)^(4*k-3)) / (4 - 3*(1+x)^(4*k-1)), due to a q-series identity.
%F a(n) = Sum_{k>=0} 3^k * binomial(k^2, n) / 4^(k+1).
%F a(n) ~ 2^(2*n - 3/2) * n^n / ((4/3)^(log(4/3)/8) * exp(n) * (log(4/3))^(2*n + 1)). - _Vaclav Kotesovec_, Mar 21 2018
%e G.f.: A(x) = 1 + 21*x + 1512*x^2 + 182448*x^3 + 30845052*x^4 + 6706403424*x^5 + 1782361433664*x^6 + 559861341721920*x^7 + ...
%e such that
%e A(x) = 1/4 + 3*(1+x)/4^2 + 3^2*(1+x)^4/4^3 + 3^3*(1+x)^9/4^4 + 3^4*(1+x)^16/4^5 + 3^5*(1+x)^25/4^6 + 3^6*(1+x)^36/4^7 + 3^7*(1+x)^49/4^8 + 3^8*(1+x)^64/4^9 + ...
%t Table[Sum[StirlingS1[n, j] * HurwitzLerchPhi[3/4, -2*j, 0]/4, {j, 0, n}] / n!, {n, 0, 20}] (* _Vaclav Kotesovec_, Mar 21 2018 *)
%o (PARI) /* Continued fraction expression: */
%o {a(n) = my(CF=1, q = 1+x +x*O(x^n)); for(k=0, n, CF = 1/(4 - 3*q^(4*n-4*k+1)/(1 - 3*q^(2*n-2*k+1)*(q^(2*n-2*k+2) - 1)*CF)) ); polcoeff(CF, n)}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A301310, A173217, A301311.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Mar 21 2018
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