OFFSET
1,2
COMMENTS
From Fermat's Little Theorem all these numbers are congruent to 1 mod 3.
From Peter Bala, Jan 26 2025: (Start)
The sequence terms are the exponents of q in the expansion of q*Product_{n >= 1} (1 - q^(3*n))*(1 - q^(18*n))^2/( (1 - q^(6*n))*(1 - q^(9*n)) ) = q - q^4 - q^16 + q^25 + q^49 - q^64 - q^100 + + - - ....
Also, the exponents of q in the expansion of q*Product_{n >= 1} (1 - q^(6*n))^5/(1 - q^(3*n))^2 = q + 2*q^4 - 4*q^16 - 5*q^25 + 7*q^49 + 8*q^64 -10*q^100 - 11*q^121 + + - - .... See Lemke Oliver, Theorem 1.2. (End)
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
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
G.f.: -x*(1+3*x+10*x^2+3*x^3+x^4) / ((1+x)^2*(x-1)^3). - R. J. Mathar, Feb 16 2011
From Colin Barker, Jan 26 2016: (Start)
a(n) = (18*n^2-6*(-1)^n*n-18*n+3*(-1)^n+5)/8.
a(n) = (9*n^2-12*n+4)/4 for n even.
a(n) = (9*n^2-6*n+1)/4 for n odd. (End)
E.g.f.: (1/8)*( (3 + 6*x)*exp(-x) - 8 + (5 + 18*x^2)*exp(x)). - G. C. Greubel, Oct 19 2016
Sum_{n>=1} 1/a(n) = 4*Pi^2/27 (A214549). - Amiram Eldar, Dec 19 2020
MATHEMATICA
LinearRecurrence[{1, 2, -2, -1, 1}, {1, 4, 16, 25, 49}, 25] (* or *) Table[(18*n^2-6*(-1)^n*n-18*n+3*(-1)^n+5)/8, {n, 1, 25}] (* G. C. Greubel, Oct 19 2016 *)
Flatten[Partition[Range[70], 2, 3, {1, 1}, {}]]^2 (* Harvey P. Dale, Jun 19 2018 *)
PROG
(PARI) isok(n) = issquare(n) && (n % 3 == 1); \\ Michel Marcus, Nov 02 2013
(PARI) Vec(-x*(1+3*x+10*x^2+3*x^3+x^4) / ( (1+x)^2*(x-1)^3 ) + O(x^100)) \\ Colin Barker, Jan 26 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Artur Jasinski, Nov 25 2007
STATUS
approved