A176060 a(n) = n*(n+1)*(3*n^2+5*n+4)/12.
0, 2, 13, 46, 120, 260, 497, 868, 1416, 2190, 3245, 4642, 6448, 8736, 11585, 15080, 19312, 24378, 30381, 37430, 45640, 55132, 66033, 78476, 92600, 108550, 126477, 146538, 168896, 193720, 221185, 251472, 284768, 321266, 361165, 404670, 451992
Offset: 0
Examples
For n=5, a(5)=1*(1*0+6)+2*(2*1+6)+3*(3*2+6)+4*(4*3+6)+5*(5*4+6)=260.
References
- "Supplemento al Periodico di Matematica", Raffaello Giusti Editore (Livorno), May 1908, p. 111 (Problem 923).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..10000
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Crossrefs
Cf. A213821.
Programs
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Magma
[n*(n+1)*(3*n^2+5*n+4)/12: n in [0..40]]; // Vincenzo Librandi, Jul 02 2011
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Mathematica
Table[n(n+1)(3n^2+5n+4)/12,{n,0,40}] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{0,2,13,46,120},40] (* Harvey P. Dale, Jul 14 2011 *)
Formula
G.f.: x*(2+3*x+x^2)/(1-x)^5.
a(n) = sum(k*(k*(k-1)+n+1), k=1..n) with n>0 (summation proposed in the Problem 923, see References).
a(0)=0, a(1)=2, a(2)=13, a(3)=46, a(4)=120, a(n)=5*a(n-1)- 10*a(n-2)+ 10*a(n-3)-5*a(n-4)+a(n-5). - Harvey P. Dale, Jul 14 2011
Comments