A134393 Row sums of triangle A134392.
1, 3, 8, 20, 45, 91, 168, 288, 465, 715, 1056, 1508, 2093, 2835, 3760, 4896, 6273, 7923, 9880, 12180, 14861, 17963, 21528, 25600, 30225, 35451, 41328, 47908, 55245, 63395, 72416, 82368, 93313, 105315, 118440, 132756, 148333, 165243, 183560, 203360, 224721, 247723, 272448
Offset: 1
Examples
a(4) = 20 = (1, 3, 3, 1) dot (1, 2, 3, 4) = (1 + 6 + 9 + 4). a(4) = sum of row 4 terms of triangle A134392: (8 + 7 + 4 + 1).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Programs
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Magma
[Binomial(n+3, 4)-2*Binomial(n+2, 4)+ 3*Binomial(n+1, 4): n in [1..40]]; // Vincenzo Librandi, Feb 04 2013
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Mathematica
Table[(n^4 - 2*n^3 + 5*n^2 + 8*n)/12, {n, 1, 40}] (* Vincenzo Librandi, Feb 04 2013 *) LinearRecurrence[{5,-10,10,-5,1},{1,3,8,20,45},50] (* Harvey P. Dale, May 28 2018 *)
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PARI
x='x+O('x^99); Vec(x*(1-2*x+3*x^2)/(1-x)^5) \\ Altug Alkan, Aug 16 2017
Formula
From R. J. Mathar, Jun 08 2008: (Start)
O.g.f.: x*(1-2*x+3*x^2)/(1-x)^5.
a(n) = A014628(n+1). (End)
a(n) = binomial(n+3,4) - 2*binomial(n+2,4) + 3*binomial(n+1,4). - Johannes W. Meijer, Apr 29 2011, corrected by Eric Rowland, Aug 16 2017
a(n) = n*(n + 1)*(n^2 - 3*n + 8)/12. - Johannes W. Meijer, Apr 29 2011, corrected by Eric Rowland, Aug 16 2017
Comments