OFFSET
0,10
COMMENTS
Also number of partitions of n into distinct parts congruent to 1,2, or 4 (mod 7). Example: a(15)=4 because we have [15],[11,4],[9,4,2] and [8,4,2,1].
REFERENCES
G. E. Andrews, Number Theory, Dover Publications, 1994 (p. 166, Exercise 7).
FORMULA
G.f.=1/product((1-x^(1+14j))(1-x^(9+14j))(1-x^(11+14j)),j=0..infinity). G.f.=product((1+x^(1+7j))(1+x^(2+7j))(1+x^(4+7j)),j=0..infinity).
a(n) ~ exp(sqrt(n/7)*Pi) / (2*7^(1/4)*n^(3/4)). - Vaclav Kotesovec, Mar 07 2016
EXAMPLE
a(15)=4 because we have [15],[11,1,1,1,1],[9,1,1,1,1,1,1] and [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1].
MAPLE
g:=product((1+x^(1+7*j))*(1+x^(2+7*j))*(1+x^(4+7*j)), j=0..15): gser:=series(g, x=0, 95): seq(coeff(gser, x, n), n=0..77);
MATHEMATICA
nmax = 100; CoefficientList[Series[Product[(1+x^(7*k-6))*(1+x^(7*k-5))*(1+x^(7*k-3)), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Mar 07 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Emeric Deutsch, Feb 16 2006
STATUS
approved