cp's OEIS Frontend

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

%I A103115 #35 Jan 16 2025 12:11:42
%S A103115 -1,-1,11,35,71,119,179,251,335,431,539,659,791,935,1091,1259,1439,
%T A103115 1631,1835,2051,2279,2519,2771,3035,3311,3599,3899,4211,4535,4871,
%U A103115 5219,5579,5951,6335,6731,7139,7559,7991,8435,8891,9359,9839,10331,10835,11351,11879,12419
%N A103115 a(n) = 6*n*(n-1) - 1.
%C A103115 Star numbers A003154 minus 2.
%C A103115 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
%C A103115 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
%H A103115 Vincenzo Librandi, <a href="/A103115/b103115.txt">Table of n, a(n) for n = 0..1000</a>
%H A103115 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A103115 a(n) = A003154(n) - 2.
%F A103115 G.f.: (1-2*x-11*x^2)/(x-1)^3. - _R. J. Mathar_, May 11 2009 (adapted by _Vincenzo Librandi_, May 16 2017).
%F A103115 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
%F A103115 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
%F A103115 Sum_{n>=1} 1/a(n) = tan(sqrt(5/3)*Pi/2)*Pi/(2*sqrt(15)). _Amiram Eldar_, Aug 20 2022
%F A103115 E.g.f.: exp(x)*(6*x^2 - 1). - _Elmo R. Oliveira_, Jan 16 2025
%t A103115 Table[6n(n-1)-1,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{-1,-1,11},50] (* _Harvey P. Dale_, Nov 14 2011 *)
%t A103115 CoefficientList[Series[(1 - 2 x - 11 x^2) / (x - 1)^3, {x, 0, 50}], x] (* _Vincenzo Librandi_, May 16 2017 *)
%o A103115 (Magma) [6*n*(n-1)-1: n in [0..50]]; // _Vincenzo Librandi_, May 16 2017
%o A103115 (PARI) a(n)=6*n*(n-1)-1 \\ _Charles R Greathouse IV_, Jun 16 2017
%Y A103115 Cf. A000914, A003154, A163433.
%K A103115 easy,sign
%O A103115 0,3
%A A103115 Jacob Landon (jacoblandon(AT)aol.com), May 09 2009
%E A103115 Edited and extended by _R. J. Mathar_, May 11 2009
%E A103115 Entries rechecked by _N. J. A. Sloane_, Jul 18 2009