A098693 G.f.: q*Product_{k>0} (1-q^(12k))(1+q^(12k-1))(1+q^(12k-11))/(1-q^k).
1, 2, 3, 5, 8, 12, 18, 26, 37, 52, 72, 99, 134, 180, 240, 317, 416, 542, 702, 904, 1158, 1476, 1872, 2364, 2973, 3724, 4647, 5778, 7160, 8844, 10890, 13370, 16368, 19984, 24336, 29561, 35822, 43308, 52242, 62884, 75536, 90552, 108342, 129384, 154232
Offset: 1
Keywords
Links
- G. E. Andrews, Three aspects of partitions, Séminaire Lotharingien de Combinatoire, B25f (1990), 1 p.
Crossrefs
Cf. A036018.
Programs
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Mathematica
nmax = 100; Rest[CoefficientList[Series[x*Product[(1-x^(12*k)) * (1+x^(12*k-1)) * (1+x^(12*k-11))/(1-x^k), {k, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Aug 31 2015 *) -
PARI
{a(n)=if(n<0, 0, polcoeff( sum(k=1, sqrtint(n), x^k^2/(1+x^k)* prod(i=1, k, (1+x^i)^2/(1-x^(2*i-1))/(1-x^(2*i)), 1+x*O(x^(n-k^2)))), n))} /* Michael Somos, Sep 19 2006 */
Formula
Euler transform of period 24 sequence [ 2, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 0, 2, 1, 1, 1, 1, 1, 1, 1, 1, 0, 2, 0, ...]. - Michael Somos, Sep 19 2006
Given g.f. A(x), then B(x)=A(x)+A(x)^2 satisfies 0=f(B(x), B(x^2)) where f(u, v)=(1+6*u)*v*(1+2*v)-u^2. - Michael Somos, Sep 19 2006
G.f.: q*{Sum_{k} q^(24k^2+10k) +q^(24k^2+14k+1) }/{Sum_{k} (-1)^k q^((3k^2+k)/2) }. - Michael Somos, Sep 19 2006
G.f.: q*Product_{k>0} (1-q^(12k))(1+q^(12k-1))(1+q^(12k-11))/(1-q^k).
G.f.: Sum_{k>0} Prod[i=1..k, (1+q^i)^2]*(1+q^k)*q^(k^2) /{(1-q)(1-q^2)...(1-q^(2k))}.
a(n) ~ exp(Pi*sqrt(2*n/3)) / (2^(9/4) * 3^(3/4) * n^(3/4)). - Vaclav Kotesovec, Aug 31 2015
Comments