G.f.: Product_{m>=1} (1 + x^m) = 1/Product_{m>=0} (1-x^(2m+1)) = Sum_{k>=0} Product_{i=1..k} x^i/(1-x^i) = Sum_{n>=0} x^(n*(n+1)/2) / Product_{k=1..n} (1-x^k).
G.f.: Sum_{n>=0} x^n*Product_{k=1..n-1} (1+x^k) = 1 + Sum_{n>=1} x^n*Product_{k>=n+1} (1+x^k). -
Joerg Arndt, Jan 29 2011
Product_{k>=1} (1+x^(2k)) = Sum_{k>=0} x^(k*(k+1))/Product_{i=1..k} (1-x^(2i)) - Euler (Hardy and Wright, Theorem 346).
Asymptotics: a(n) ~ exp(Pi l_n / sqrt(3)) / ( 4 3^(1/4) l_n^(3/2) ) where l_n = (n-1/24)^(1/2) (Ayoub).
For n > 1, a(n) = (1/n)*Sum_{k=1..n} b(k)*a(n-k), with a(0)=1, b(n) =
A000593(n) = sum of odd divisors of n; cf.
A000700. -
Vladeta Jovovic, Jan 21 2002
a(n) = t(n, 0), t as defined in
A079211.
Expansion of 1 / chi(-x) = chi(x) / chi(-x^2) = f(-x) / phi(x) = f(x) / phi(-x^2) = psi(x) / f(-x^2) = f(-x^2) / f(-x) = f(-x^4) / psi(-x) in powers of x where phi(), psi(), chi(), f() are Ramanujan theta functions. -
Michael Somos, Mar 12 2011
G.f. is a period 1 Fourier series which satisfies f(-1 / (1152 t)) = 2^(-1/2) / f(t) where q = exp(2 Pi i t). -
Michael Somos, Aug 16 2007
Expansion of q^(-1/24) * eta(q^2) / eta(q) in powers of q.
Expansion of q^(-1/24) 2^(-1/2) f2(t) in powers of q = exp(2 Pi i t) where f2() is a Weber function. -
Michael Somos, Oct 18 2007
Given g.f. A(x), then B(x) = x * A(x^3)^8 satisfies 0 = f(B(x), B(x^2)) where f(u, v) = v - u^2 + 16*u*v^2 . -
Michael Somos, May 31 2005
Given g.f. A(x), then B(x) = x * A(x^8)^3 satisfies 0 = f(B(x), B(x^3)) where f(u, v) = (u^3 - v) * (u + v^3) - 9 * u^3 * v^3. -
Michael Somos, Mar 25 2008
From Evangelos Georgiadis, Andrew V. Sutherland, Kiran S. Kedlaya (egeorg(AT)mit.edu), Mar 03 2009: (Start)
a(0)=1; a(n) = 2*(Sum_{k=1..floor(sqrt(n))} (-1)^(k+1) a(n-k^2)) + sigma(n) where sigma(n) = (-1)^j if (n=(j*(3*j+1))/2 OR n=(j*(3*j-1))/2) otherwise sigma(n)=0 (simpler: sigma =
A010815). (End)
The product g.f. = (1/(1-x))*(1/(1-x^3))*(1/(1-x^5))*...; = (1,1,1,...)*
(1,0,0,1,0,0,1,0,0,1,...)*(1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,...) * ...; =
a*b*c*... where a, a*b, a*b*c, ... converge to
A000009:
1, 1, 1, 2, 2, 2, 3, 3, 3, 4, ... = a*b
1, 1, 1, 2, 2, 3, 4, 4, 5, 6, ... = a*b*c
1, 1, 1, 2, 2, 3, 4, 5, 6, 7, ... = a*b*c*d
1, 1, 1, 2, 2, 3, 4, 5, 6, 8, ... = a*b*c*d*e
1, 1, 1, 2, 2, 3, 4, 5, 6, 8, ... = a*b*c*d*e*f
... (cf. analogous example in
A000041). (End)
a(n) = P(n) - P(n-2) - P(n-4) + P(n-10) + P(n-14) + ... + (-1)^m P(n-2p_m) + ..., where P(n) is the partition function (
A000041) and p_m = m(3m-1)/2 is the m-th generalized pentagonal number (
A001318). -
Jerome Malenfant, Feb 16 2011
G.f.: 1/2 (-1; x)_{inf} where (a; q)_k is the q-Pochhammer symbol. -
Vladimir Reshetnikov, Apr 24 2013
More precise asymptotics: a(n) ~ exp(Pi*sqrt((n-1/24)/3)) / (4*3^(1/4)*(n-1/24)^(3/4)) * (1 + (Pi^2-27)/(24*Pi*sqrt(3*(n-1/24))) + (Pi^4-270*Pi^2-1215)/(3456*Pi^2*(n-1/24))). -
Vaclav Kotesovec, Nov 30 2015
a(n) ~ exp(Pi*sqrt(n/3))/(4*3^(1/4)*n^(3/4)) * (1 + (Pi/(48*sqrt(3)) - (3*sqrt(3))/(8*Pi))/sqrt(n) + (Pi^2/13824 - 5/128 - 45/(128*Pi^2))/n).
a(n) ~ exp(Pi*sqrt(n/3) + (Pi/(48*sqrt(3)) - 3*sqrt(3)/(8*Pi))/sqrt(n) - (1/32 + 9/(16*Pi^2))/n) / (4*3^(1/4)*n^(3/4)).
(End)
a(n) ~ Pi*BesselI(1, Pi*sqrt((n+1/24)/3)) / sqrt(24*n+1). -
Vaclav Kotesovec, Nov 08 2016
G.f.: (1 + x)*Sum_{n >= 0} x^(n*(n+3)/2)/Product_{k = 1..n} (1 - x^k) =
(1 + x)*(1 + x^2)*Sum_{n >= 0} x^(n*(n+5)/2)/Product_{k = 1..n} (1 - x^k) = (1 + x)*(1 + x^2)*(1 + x^3)*Sum_{n >= 0} x^(n*(n+7)/2)/Product_{k = 1..n} (1 - x^k) = ....
G.f.: (1/2)*Sum_{n >= 0} x^(n*(n-1)/2)/Product_{k = 1..n} (1 - x^k) =
(1/2)*(1/(1 + x))*Sum_{n >= 0} x^((n-1)*(n-2)/2)/Product_{k = 1..n} (1 - x^k) = (1/2)*(1/((1 + x)*(1 + x^2)))*Sum_{n >= 0} x^((n-2)*(n-3)/2)/Product_{k = 1..n} (1 - x^k) = ....
G.f.: Sum_{n >= 0} x^n/Product_{k = 1..n} (1 - x^(2*k)) = (1/(1 - x)) * Sum_{n >= 0} x^(3*n)/Product_{k = 1..n} (1 - x^(2*k)) = (1/((1 - x)*(1 - x^3))) * Sum_{n >= 0} x^(5*n)/Product_{k = 1..n} (1 - x^(2*k)) = (1/((1 - x)*(1 - x^3)*(1 - x^5))) * Sum_{n >= 0} x^(7*n)/Product_{k = 1..n} (1 - x^(2*k)) = .... (End)
G.f.: A(x) = Sum_{n >= 0} x^(n*(2*n-1))/Product_{k = 1..2*n} (1 - x^k). (Set z = x and q = x^2 in Mc Laughlin et al., Section 1.3, Entry 7.)
Similarly, A(x) = Sum_{n >= 0} x^(n*(2*n+1))/Product_{k = 1..2*n+1} (1 - x^k). (End)
G.f.: A(x) = exp ( Sum_{n >= 1} x^n/(n*(1 - x^(2*n))) ) = exp ( Sum_{n >= 1} (-1)^(n+1)*x^n/(n*(1 - x^n)) ). -
Peter Bala, Dec 23 2021
Sum_{n>=0} a(n)/exp(Pi*n) = exp(Pi/24)/2^(1/8) =
A292820. -
Simon Plouffe, May 12 2023 [Proof: Sum_{n>=0} a(n)/exp(Pi*n) = phi(exp(-2*Pi)) / phi(exp(-Pi)), where phi(q) is the Euler modular function. We have phi(exp(-2*Pi)) = exp(Pi/12) * Gamma(1/4) / (2 * Pi^(3/4)) and phi(exp(-Pi)) = exp(Pi/24) * Gamma(1/4) / (2^(7/8) * Pi^(3/4)), see formulas (14) and (13) in I. Mező, 2013. -
Vaclav Kotesovec, May 12 2023]
a(2*n) = Sum_{j=1..n} p(n+j, 2*j) and a(2*n+1) = Sum_{j=1..n+1} p(n+j,2*j-1), where p(n, s) is the number of partitions of n having exactly s parts. -
Gregory L. Simay, Aug 30 2023
