A104777 Integer squares congruent to 1 mod 6.

1, 25, 49, 121, 169, 289, 361, 529, 625, 841, 961, 1225, 1369, 1681, 1849, 2209, 2401, 2809, 3025, 3481, 3721, 4225, 4489, 5041, 5329, 5929, 6241, 6889, 7225, 7921, 8281, 9025, 9409, 10201, 10609, 11449, 11881, 12769, 13225, 14161, 14641, 15625, 16129

Offset: 1

Author

Michael Somos, Mar 24 2005

Keywords

• nonn
• easy

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)

Examples

			eta(q^24) = q - q^25 - q^49 + q^121 + q^169 - q^289 - q^361 + ...
		

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)

Crossrefs

Disjoint union of A016922 and A016970.

Cf. A007310, A001318, A033683, A100044.

Programs

Haskell

a104777 = (^ 2) . a007310  -- Reinhard Zumkeller, Nov 14 2015

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 *)

PARI

{a(n) = (3*n - 1 - n%2)^2};

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) = A007310(n)^2 = 1 + 24*A001318(n-1).

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