OFFSET
0,19
COMMENTS
Also number of partitions of n such that 2k and 2k+1 occur with the same multiplicities. Example: a(26)=3 because we have [11,10,3,2], [9,8,5,4] and [7,7,6,6]. It is easy to find a bijection between these partitions and those described in the definition.
FORMULA
G.f.: 1/product(1-x^(4i+1), i=1..infinity).
a(n) ~ exp(sqrt(n/6)*Pi) * Pi^(1/4) * Gamma(1/4) / (2^(31/8) * 3^(5/8) * n^(9/8)). - Vaclav Kotesovec, Mar 07 2016
EXAMPLE
a(26)=3 because we have [21,5],[17,9] and [13,13].
MAPLE
g:=1/product(1-x^(4*i+1), i=1..50): gser:=series(g, x=0, 93): seq(coeff(gser, x, n), n=0..88);
MATHEMATICA
nmax = 100; CoefficientList[Series[Product[1/(1-x^(4*k+1)), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Mar 07 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Emeric Deutsch, Apr 05 2006
STATUS
approved