OFFSET
1,3
COMMENTS
Compare with the expansions Sum_{n >= 0} q^(2*n*(n+1))*Product_{k >= 2*n+2} 1 - q^k = 1 - q^2 - q^3 + q^7 + q^17 - q^25 - q^28 + + - - ... (see A268539) and Sum_{n >= 0} q^(n*(n+1))*Product_{k >= 2*n+1} 1 - q^k = 1 - q - q^8 + q^13 + q^17 - q^24 - q^45 + + - - .... (see A204221).
Conjectures:
1) apart from the coefficient of q, the coefficients of the series expansion (see below) belong to {-1, 0, 1}.
2) starting at q^3, the signs of the nonzero coefficients follow the pattern + + - - + + - - ....
It appears that the sequence terms are the exponents in the expansion of Sum_{n >= 0} x^(3*n)/(Product_{k = 1..2*n} 1 + x^k) = 1 + x^3 - x^4 + x^6 - x^8 + x^11 - x^13 + - .... - Peter Bala, Jan 21 2025
FORMULA
The following are conjectural:
a(n) is quasi-polynomial in n:
a(8*n+1) = 12*n^2 + 5*n + 1 = A244806(n+1) for n >= 1;
a(8*n+2) = 12*n^2 + 7*n + 1 = A033577(n); a(8*n+3) = 12*n^2 + 11*n + 3;
a(8*n+4) = 12*n^2 + 13*n + 4; a(8*n+5) = 12*n^2 + 17*n + 6 = A033578(n+1);
a(8*n+6) = 12*n^2 + 19*n + 8; a(8*n+7) = 12*n^2 + 23*n + 11;
a(8*n+8) = 12*n^2 + 25*n + 13.
G.f.: x^2*(x^8 - x^7 - 2*x^6 + 3*x^5 + 2*x^4 - 2*x^3 - x^2 + 2*x + 1)/((1 + x)^2*(1 - x)^3*(1 + x^4)) = x^2 + 3*x^3 + 4*x^4 + 6*x^5 + 8*x^6 + 11*x^7 + ....
EXAMPLE
(1 - q)*Sum_{n >= 0} q^(2*n*(n+1))*Product_{k >= 2*n+1} 1 - q^k = 1 - 2*q + q^3 + q^4 - q^6 - q^8 + q^11 + q^13 - q^18 - q^20 + q^26 + q^29 - q^35 - q^39 + q^46 + q^50 - q^59 - q^63 + + - - ....
MAPLE
series(add((1 - q)*q^(2*n*(n+1))*mul(1 - q^k, k = 2*n+1..1000), n = 0..21), q, 1001);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Dec 16 2024
STATUS
approved