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G.f.: Sum_{n>=0} 2^n * (1+x)^(n^2) / 3^(n+1).
6

%I #22 Mar 21 2018 12:13:41

%S 1,10,360,21840,1857660,203258160,27188330400,4298562686880,

%T 784233322674120,162161079972261480,37477229047577953920,

%U 9573364920705562944000,2678416661190852872256960,814535089079749159186189440,267528376262254011309768677760,94377360018309519999410315205120,35590366640535756970223476489499280,14287353028920891078189826021459809120

%N G.f.: Sum_{n>=0} 2^n * (1+x)^(n^2) / 3^(n+1).

%C Is there a finite expression for the terms of this sequence?

%C a(n) is divisible by 10 for n>0 (conjecture).

%H Paul D. Hanna, <a href="/A301310/b301310.txt">Table of n, a(n) for n = 0..100</a>

%F G.f.: 1/(3 - 2*q/(1 - 2*q*(q^2-1)/(3 - 2*q^5/(1 - 2*q^3*(q^4-1)/(3 - 2*q^9/(1 - 2*q^5*(q^6-1)/(3 - 2*q^13/(1 - 2*q^7*(q^8-1)/(3 - ...))))))))) where q = (1+x), a continued fraction due to a partial elliptic theta function identity.

%F G.f.: Sum_{n>=0} 2^n/3^(n+1) * (1+x)^n * Product_{k=1..n} (3 - 2*(1+x)^(4*k-3)) / (3 - 2*(1+x)^(4*k-1)), due to a q-series identity.

%F a(n) = Sum_{k>=0} 2^k * binomial(k^2, n) / 3^(k+1).

%F a(n) ~ 2^(2*n + 1/2 + log(3/2)/8) * n^n / (3^(1 + log(3/2)/8) * exp(n) * (log(3/2))^(2*n + 1)). - _Vaclav Kotesovec_, Mar 21 2018

%e G.f.: A(x) = 1 + 10*x + 360*x^2 + 21840*x^3 + 1857660*x^4 + 203258160*x^5 + 27188330400*x^6 + 4298562686880*x^7 + 784233322674120*x^8 + ...

%e such that

%e A(x) = 1/3 + 2*(1+x)/3^2 + 2^2*(1+x)^4/3^3 + 2^3*(1+x)^9/3^4 + 2^4*(1+x)^16/3^5 + 2^5*(1+x)^25/3^6 + 2^6*(1+x)^36/3^7 + 2^7*(1+x)^49/3^8 + 2^8*(1+x)^64/3^9 + ...

%t Table[Sum[StirlingS1[n, j] * HurwitzLerchPhi[2/3, -2*j, 0]/3, {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/(3 - 2*q^(4*n-4*k+1)/(1 - 2*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. A173217, A301311.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Mar 18 2018