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A176060 a(n) = n*(n+1)*(3*n^2+5*n+4)/12. 3

%I

%S 0,2,13,46,120,260,497,868,1416,2190,3245,4642,6448,8736,11585,15080,

%T 19312,24378,30381,37430,45640,55132,66033,78476,92600,108550,126477,

%U 146538,168896,193720,221185,251472,284768,321266,361165,404670,451992

%N a(n) = n*(n+1)*(3*n^2+5*n+4)/12.

%C Antidiagonal sums of the convolution array A213821. [_Clark Kimberling_, Jul 04 2012]

%D "Supplemento al Periodico di Matematica", Raffaello Giusti Editore (Livorno), May 1908, p. 111 (Problem 923).

%H Vincenzo Librandi, <a href="/A176060/b176060.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F G.f.: x*(2+3*x+x^2)/(1-x)^5.

%F a(n) = sum(k*(k*(k-1)+n+1), k=1..n) with n>0 (summation proposed in the Problem 923, see References).

%F 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

%F a(-n) = A132117(n-1) with A132117(-1)=A132117(0)=0. - _Bruno Berselli_, Aug 22 2011

%e 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.

%t 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 *)

%o (MAGMA) [n*(n+1)*(3*n^2+5*n+4)/12: n in [0..40]]; // _Vincenzo Librandi_, Jul 02 2011

%Y Cf. A213821.

%K nonn,easy

%O 0,2

%A _Bruno Berselli_, Dec 06 2010

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Last modified April 22 08:25 EDT 2019. Contains 322329 sequences. (Running on oeis4.)