A173217 G.f.: A(x) = Sum_{n>=0} (1 + x)^(n^2) / 2^(n+1).
1, 3, 36, 744, 21606, 807912, 36948912, 1997801520, 124666314300, 8817945612300, 697162848757056, 60925366551278592, 5831682410241684192, 606763511537812563648, 68184018356901256320192, 8229830886505821175612416, 1061871008421711265790015880, 145851902823090076435152800208, 21247730059665104564252809209792
Offset: 0
Keywords
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..100
- StackExchange, Combinatorial puzzle (2013)
Programs
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Mathematica
Table[Sum[Binomial[k^2, n]/2^(k+1), {k, 0, Infinity}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 21 2018 *) Table[Sum[StirlingS1[n, j] * HurwitzLerchPhi[1/2, -2*j, 0]/2, {j, 0, n}] / n!, {n, 0, 20}] (* Vaclav Kotesovec, Mar 21 2018 *) -
PARI
{a(n)=local(A=sum(m=0,n^2+100,(1+x +O(x^(n+2)))^(m^2)/2^(m+1)));round(polcoeff(A,n))} -
PARI
/* Continued fraction expression: */ {a(n) = my(CF=1, q = 1+x +x*O(x^n)); for(k=0, n, CF = 1/(2 - q^(4*n-4*k+1)/(1 - q^(2*n-2*k+1)*(q^(2*n-2*k+2) - 1)*CF)) ); polcoeff(CF, n)} for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 18 2018
Formula
G.f.: 1/(2 - q/(1 - q*(q^2-1)/(2 - q^5/(1 - q^3*(q^4-1)/(2 - q^9/(1 - q^5*(q^6-1)/(2 - q^13/(1 - q^7*(q^8-1)/(2 - ...))))))))) where q = (1+x), a continued fraction due to a partial elliptic theta function identity. - Paul D. Hanna, Mar 18 2018
G.f.: Sum_{n>=0} 1/2^(n+1) * (1+x)^n * Product_{k=1..n} (2 - (1+x)^(4*k-3)) / (2 - (1+x)^(4*k-1)), due to a q-series identity. - Paul D. Hanna, Mar 18 2018
a(n) ~ 2^(2*n - 1/2 - log(2)/8) * n^n / (exp(n) * log(2)^(2*n + 1)). - Vaclav Kotesovec, Mar 21 2018
Comments