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%I #14 Jun 14 2017 11:00:26
%S 10,23,65,94,168,213,319,380,518,595,765,858,1060,1169,1403,1528,1794,
%T 1935,2233,2390,2720,2893,3255,3444,3838,4043,4469,4690,5148,5385,
%U 5875,6128,6650,6919,7473,7758,8344,8645,9263,9580,10230,10563,11245,11594
%N Integers of the form (k+1)*(2k+1)/12.
%C This can also be defined as integer averages of the first k halved squares, 1^2/2, 2^2/2, 3^2/2,... , 3^k/2, because sum_{j=1..k} j^2/2 = k*(k+1)*(2k+1)/12. The generating k are in A168489.
%H Colin Barker, <a href="/A164578/b164578.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).
%F a(n) = +a(n-1) +2*a(n-2) -2*a(n-3) -a(n-4) +a(n-5). G.f. x*(-10-13*x-22*x^2-3*x^3) / ((1+x)^2*(x-1)^3). - _R. J. Mathar_, Jan 25 2011
%F From _Colin Barker_, Jan 26 2016: (Start)
%F a(n) = (24*n^2+6*n-(-1)^n*(8*n+1)+1)/4.
%F a(n) = (12*n^2-n)/2 for n even.
%F a(n) = (12*n^2+7*n+1)/2 for n odd.
%F (End)
%t s=0;lst={};Do[a=(s+=(n^2)/2)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n,2*6!}];lst
%t Select[Table[((n+1)(2n+1))/12,{n,300}],IntegerQ] (* or *) LinearRecurrence[ {1,2,-2,-1,1},{10,23,65,94,168},60] (* _Harvey P. Dale_, Jun 14 2017 *)
%o (PARI) Vec(x*(10+13*x+22*x^2+3*x^3)/((1-x)^3*(1+x)^2) + O(x^100)) \\ _Colin Barker_, Jan 26 2016
%Y Cf. A078617, A078618, A154293, A164576, A164577.
%K nonn,easy
%O 1,1
%A _Vladimir Joseph Stephan Orlovsky_, Aug 16 2009