OFFSET
0,2
COMMENTS
Sequence found by reading the line from 1, in the direction 1, 21, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012
LINKS
Muniru A Asiru, Table of n, a(n) for n = 0..5000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014.
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, Vol. 18 (2015), Article 15.3.2.
Leo Tavares, Illustration: Square Block Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = a(n-1) + 24*n - 4, with n > 0, a(0)=1. - Vincenzo Librandi, Dec 28 2010
G.f.: (1 + 18*x + 5*x^2)/(1 - 3*x + 3*x^2 - x^3). - Colin Barker, Jan 06 2012
a(n) = A289873(6*n+2). - Hugo Pfoertner, Jul 15 2017
From Peter Bala, Jan 22 2018: (Start)
This is the polynomial Qbar(2,n) in Brent. See A160485 for the triangle of coefficients (with signs) of the Qbar polynomials.
a(n) = (1/4^n) * Sum_{k = 0..n} (2*k + 1)^4*binomial(2*n + 1, n - k).
a(n-1) = (2/4^n) * binomial(2*n,n) * ( 1 + 3^4*(n - 1)/(n + 1) + 5^4*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^4*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)
From Amiram Eldar, Feb 27 2022: (Start)
Sum_{n>=0} 1/a(n) = (sqrt(3)*Pi + 3*log(3))/8.
Sum_{n>=0} (-1)^n/a(n) = Pi/8 + sqrt(3)*log(2+sqrt(3))/4. (End)
E.g.f.: exp(x)*(1 + 20*x + 12*x^2). - Stefano Spezia, Apr 16 2022
MAPLE
MATHEMATICA
Table[(2n + 1)(6n + 1), {n, 0, 49}] (* Harvey P. Dale, Mar 24 2011 *)
PROG
(Magma) [ (2*n+1)*(6*n+1) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 08 2014
(PARI) a(n)=(2*n+1)*(6*n+1) \\ Charles R Greathouse IV, Jun 17 2017
(GAP) List([0..50], n->(2*n+1)*(6*n+1)); # Muniru A Asiru, Feb 05 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Mohammad K. Azarian, Dec 11 1999
EXTENSIONS
More terms from Patrick De Geest
Better description from N. J. A. Sloane
STATUS
approved