A164578 Integers of the form (k+1)*(2k+1)/12.

10, 23, 65, 94, 168, 213, 319, 380, 518, 595, 765, 858, 1060, 1169, 1403, 1528, 1794, 1935, 2233, 2390, 2720, 2893, 3255, 3444, 3838, 4043, 4469, 4690, 5148, 5385, 5875, 6128, 6650, 6919, 7473, 7758, 8344, 8645, 9263, 9580, 10230, 10563, 11245, 11594

Offset: 1

Author

Vladimir Joseph Stephan Orlovsky, Aug 16 2009

Keywords

• nonn
• easy

Comments

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.

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Crossrefs

Cf. A078617, A078618, A154293, A164576, A164577.

Programs

Mathematica

s=0;lst={};Do[a=(s+=(n^2)/2)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n,2*6!}];lst
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 *)

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

Formula

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

From Colin Barker, Jan 26 2016: (Start)

a(n) = (24*n^2+6*n-(-1)^n*(8*n+1)+1)/4.

a(n) = (12*n^2-n)/2 for n even.

a(n) = (12*n^2+7*n+1)/2 for n odd.

(End)