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a(n) = (1/6)*(2*n - 3)*(n + 2)*(n + 1).
8

%I #35 Feb 17 2021 17:30:08

%S 0,0,-1,-1,2,10,25,49,84,132,195,275,374,494,637,805,1000,1224,1479,

%T 1767,2090,2450,2849,3289,3772,4300,4875,5499,6174,6902,7685,8525,

%U 9424,10384,11407,12495,13650,14874,16169,17537,18980,20500,22099

%N a(n) = (1/6)*(2*n - 3)*(n + 2)*(n + 1).

%C For n>=0, real parts of sum_(k=0)^n (k+i)^2, where i=sqrt(-1). [_Bruno Berselli_, Jan 24 2014]

%H Michael De Vlieger, <a href="/A058373/b058373.txt">Table of n, a(n) for n = -2..10000</a>

%H Francesco Toppan, <a href="https://arxiv.org/abs/2008.11554">Z_2 X Z_2-graded parastatistics in multiparticle quantum Hamiltonians</a>, arXiv:2008.11554 [hep-th], 2020.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(n) = 2*C(n+2,3) - C(n+2,2). - _Zerinvary Lajos_, Nov 25 2006

%p seq((2*n-3)*(n+2)*(n+1)/6, n=-2..40);

%t s=0;lst={0, s};Do[s+=n^2-1;AppendTo[lst, s], {n, 0, 6!, 1}];lst (* _Vladimir Joseph Stephan Orlovsky_, Nov 07 2008 *)

%t Table[((2n-3)(n+2)(n+1))/6,{n,-2,40}] (* _Harvey P. Dale_, Apr 21 2019 *)

%Y Cf. A236377: real part of sum_(k=0)^n (k + i^k)^2, i=sqrt(-1). [_Bruno Berselli_, Jan 25 2014]

%K sign,easy

%O -2,5

%A _N. J. A. Sloane_, Dec 18 2000