OFFSET
0,3
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).
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
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 *)
PROG
(Magma) [6*n*(n-1)-1: n in [0..50]]; // Vincenzo Librandi, May 16 2017
(PARI) a(n)=6*n*(n-1)-1 \\ Charles R Greathouse IV, Jun 16 2017
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Jacob Landon (jacoblandon(AT)aol.com), May 09 2009
EXTENSIONS
Edited and extended by R. J. Mathar, May 11 2009
Entries rechecked by N. J. A. Sloane, Jul 18 2009
STATUS
approved