A103115 a(n) = 6*n*(n-1) - 1.

-1, -1, 11, 35, 71, 119, 179, 251, 335, 431, 539, 659, 791, 935, 1091, 1259, 1439, 1631, 1835, 2051, 2279, 2519, 2771, 3035, 3311, 3599, 3899, 4211, 4535, 4871, 5219, 5579, 5951, 6335, 6731, 7139, 7559, 7991, 8435, 8891, 9359, 9839, 10331, 10835, 11351, 11879, 12419

Offset: 0

Author

Jacob Landon (jacoblandon(AT)aol.com), May 09 2009

Keywords

• easy
• sign

Comments

Star numbers A003154 minus 2.

What A163433 does for a triangle, this sequence is doing for a square but giving one-half the results. Take a square with vertices n, n+1, n+2, and n+3 and find the sum of the four products of each four vertices times the sum of the other three; at n you have n((n+1)+(n+2)+(n+3)) and so on for the other three vertices. The result of all four is 12*n^2 + 36*n + 22; half this is 6*n^2 + 18*n + 11 and gives the numbers in this sequence starting with n = 0. - J. M. Bergot, May 23 2012

Multiplying a(n) by 16 gives the sum of the convolution with itself of each of the 24 permutations of four consecutive numbers. - J. M. Bergot, May 15 2017

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A000914, A003154, A163433.

Programs

Magma

[6*n*(n-1)-1: n in [0..50]]; // Vincenzo Librandi, May 16 2017

Mathematica

Table[6n(n-1)-1,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{-1,-1,11},50] (* Harvey P. Dale, Nov 14 2011 *)
CoefficientList[Series[(1 - 2 x - 11 x^2) / (x - 1)^3, {x, 0, 50}], x] (* Vincenzo Librandi, May 16 2017 *)

PARI

a(n)=6*n*(n-1)-1 \\ Charles R Greathouse IV, Jun 16 2017

Formula

a(n) = A003154(n) - 2.

G.f.: (1-2*x-11*x^2)/(x-1)^3. - R. J. Mathar, May 11 2009 (adapted by Vincenzo Librandi, May 16 2017).

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=-1, a(1)=-1, a(2)=11. - Harvey P. Dale, Nov 14 2011

a(n) = (n-2)*(n-1 + n + n+1) + (n-1)*(n + n+1) + n*(n+1), which is applying A000914 to four consecutive numbers. - J. M. Bergot, May 15 2017

Sum_{n>=1} 1/a(n) = tan(sqrt(5/3)*Pi/2)*Pi/(2*sqrt(15)). Amiram Eldar, Aug 20 2022

E.g.f.: exp(x)*(6*x^2 - 1). - Elmo R. Oliveira, Jan 16 2025

Extensions

Edited and extended by R. J. Mathar, May 11 2009

Entries rechecked by N. J. A. Sloane, Jul 18 2009