OFFSET
0,2
COMMENTS
The number of zero terms is infinite.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..10000
FORMULA
G.f. A(x) satisfies:
A(x^2)^(1/6) = Sum_{n>=0} x^n / Product_{k=0..n} (1 + x^(2*k+1)), by an identity due to Ramanujan.
EXAMPLE
G.f.: A(x) = 1 + 6*x + 15*x^2 + 20*x^3 + 15*x^4 - 29*x^6 - 60*x^7 - 66*x^8 +...
By definition, A(x) = B(x)^6 where
B(x) = 1 + x - x^5 - x^8 + x^16 + x^21 - x^33 - x^40 + x^56 + x^65 - x^85 - x^96 + x^120 + x^133 - x^161 - x^176 + x^208 + x^225 +...
By Ramanujan's identity,
B(x^2) = 1/(1+x) + x/((1+x)*(1+x^3)) + x^2/((1+x)*(1+x^3)*(1+x^5)) + x^3/((1+x)*(1+x^3)*(1+x^5)*(1+x^7)) + x^4/((1+x)*(1+x^3)*(1+x^5)*(1+x^7)*(1+x^9)) +...
Zero-valued coefficients in g.f. A(x) = B(x)^6 are found at positions:
[5, 11, 35, 69, 75, 151, 177, 180, 226, 516, 539, 728, 922, 2544, 3320, 3936, 4796, 5658, 5702, 6968, 7887, 8603, 9472, ...];
higher powers of B(x) have at most a finite number of zero coefficients.
PROG
(PARI) {a(n)=local(A=sum(m=0, n, (-1)^m*x^(m*(3*m+2))*(1+x^(2*m+1))+x*O(x^n))); polcoeff(A^6, n)}
for(n=0, 60, print1(a(n), ", "))
CROSSREFS
KEYWORD
sign
AUTHOR
Paul D. Hanna, Oct 20 2012
STATUS
approved