

A103115


a(n) = 6*n*(n1)1.


3



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
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OFFSET

0,3


COMMENTS

Star numbers A003154 minus 2.
What A163433 does for a triangle, this sequence is doing for a square but giving onehalf 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.: (12*x11*x^2)/(x1)^3.  R. J. Mathar, May 11 2009 (adapted by Vincenzo Librandi, May 16 2017).
a(n) = 3*a(n1)3*a(n2)+a(n3), with a(0)=1, a(1)=1, a(2)=11.  Harvey P. Dale, Nov 14 2011
a(n) = (n2)*(n1 + n + n+1) + (n1)*(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(n1)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*(n1)1: n in [0..50]]; // Vincenzo Librandi, May 16 2017
(PARI) a(n)=6*n*(n1)1 \\ Charles R Greathouse IV, Jun 16 2017


CROSSREFS

Cf. A000914, A003154, A163433.
Sequence in context: A348845 A233546 A092069 * A003777 A222512 A297539
Adjacent sequences: A103112 A103113 A103114 * A103116 A103117 A103118


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



