cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A301311 G.f.: Sum_{n>=0} 2^n * (1-x)^(-n^2) / 3^(n+1).

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%I A301311 #14 Mar 21 2018 08:21:13
%S A301311 1,10,370,22570,1924270,210821290,28223418010,4464779024650,
%T A301311 814901395935550,168556843188104050,38965275697707264970,
%U A301311 9955529477371346769010,2785811940289987110605590,847316256984037311888049090,278329013908504193489288029090,98197864581209379156337136722690,37034491818759647215732974465421990,14868275488492647637389364332301206490
%N A301311 G.f.: Sum_{n>=0} 2^n * (1-x)^(-n^2) / 3^(n+1).
%C A301311 Is there a finite expression for the terms of this sequence?
%C A301311 a(n) is divisible by 10 for n>0 (conjecture).
%H A301311 Paul D. Hanna, <a href="/A301311/b301311.txt">Table of n, a(n) for n = 0..100</a>
%F A301311 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/(1-x), a continued fraction due to a partial elliptic theta function identity.
%F A301311 G.f.: Sum_{n>=0} 2^n/3^(n+1) * (1-x)^n * Product_{k=1..n} (3*(1-x)^(4*k-3) - 2) / (3*(1-x)^(4*k-1) - 2), due to a q-series identity.
%F A301311 a(n) = Sum_{k>=0} 2^k * binomial(k^2 + n-1, n) / 3^(k+1).
%F A301311 a(n) ~ 2^(2*n + 1/2 - log(3/2)/8) * 3^(log(3/2)/8 - 1) * n^n / (exp(n) * (log(3/2))^(2*n + 1)). - _Vaclav Kotesovec_, Mar 21 2018
%e A301311 G.f.: A(x) = 1 + 10*x + 370*x^2 + 22570*x^3 + 1924270*x^4 + 210821290*x^5 + 28223418010*x^6 + 4464779024650*x^7 + 814901395935550*x^8 + ...
%e A301311 such that
%e A301311 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 + ...
%o A301311 (PARI) /* Continued fraction expression: */
%o A301311 {a(n) = my(CF=1, q = 1/(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 A301311 for(n=0, 30, print1(a(n), ", "))
%Y A301311 Cf. A104209, A301310.
%K A301311 nonn
%O A301311 0,2
%A A301311 _Paul D. Hanna_, Mar 18 2018