OFFSET
1,2
COMMENTS
Exponents of powers of q in expansion of eta(q^24).
Odd squares not divisible by 3. - Reinhard Zumkeller, Nov 14 2015
From Peter Bala, Jan 03 2025: (Start)
Exponents of q in the expansion of q*Product_{n >= 1} (1 - q^(24*n))^5/(1 - q^(48*n))^2 = q - 5*q^(5^2) + 7*q^(7^2) - 11*q^(11^2) + 13*q^(13^2) - 17*q^(17^2) + 19*q^(19)^2 - + ... (a consequence of the quintuple product identity).
Also, exponents in the expansion of q*Product_{n >= 1} (1 - q^(48*n))^13 / ( (1 - q^(24*n))*(1 - q^(96*n)) )^5 = q + 5*q^(5^2) + 7*q^(7^2) + 11*q^(11^2) - 13*q^(13^2) - 17*q^(17^2) - 19*q^(19^2) - 23*q^(23^2) + + + + - - - - ... (see Oliver, Theorem 1.1). (End)
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Robert J. Lemke Oliver, Eta quotients and theta functions, Advances in Mathematics, Vol. 241, Jul. 2013, pp. 1-17.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1)
FORMULA
A033683(a(n)) = 1.
G.f.: ( -1-24*x-22*x^2-24*x^3-x^4 ) / ( (1+x)^2*(x-1)^3 ). - R. J. Mathar, Feb 20 2011
a(n) = 9*n^2 - 9*n + 5/2 + (-1)^n * (3*n - 3/2). a(n+4) = 2*a(n+2) - a(n) + 72. - Robert Israel, Dec 12 2014
a(n) == 1 (mod 24). - Joerg Arndt, Jan 03 2017
Sum_{n>=1} 1/a(n) = Pi^2/9 (A100044). - Amiram Eldar, Dec 19 2020
EXAMPLE
eta(q^24) = q - q^25 - q^49 + q^121 + q^169 - q^289 - q^361 + ...
MAPLE
seq(9*(n-1/2)^2 + 1/4 + (-1)^n * (3*n - 3/2), n = 1 .. 100); # Robert Israel, Dec 12 2014
MATHEMATICA
Select[Range[130]^2, Mod[#, 6]==1&] (* or *) LinearRecurrence[{1, 2, -2, -1, 1}, {1, 25, 49, 121, 169}, 50] (* Harvey P. Dale, Mar 09 2017 *)
PROG
(PARI) {a(n) = (3*n - 1 - n%2)^2};
(Haskell)
a104777 = (^ 2) . a007310 -- Reinhard Zumkeller, Nov 14 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Michael Somos, Mar 24 2005
STATUS
approved