A164577 Integer averages of the first perfect cubes up to some n^3.
1, 12, 25, 45, 112, 162, 225, 396, 507, 637, 960, 1156, 1377, 1900, 2205, 2541, 3312, 3750, 4225, 5292, 5887, 6525, 7936, 8712, 9537, 11340, 12321, 13357, 15600, 16810, 18081, 20812, 22275, 23805, 27072, 28812, 30625, 34476, 36517, 38637, 43120
Offset: 1
Examples
The average of the first cube is 1^3/1=1=a(1). The average of the first two cubes is (1^3+2^3)/2=9/2, not integer, and does not contribute to the sequence. The average of the first three cubes is (1^3+2^3+3^3)/3=12, integer, and defines a(2).
Links
- Harvey P. Dale, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (1,0,3,-3,0,-3,3,0,1,-1).
Programs
-
Mathematica
Timing[s=0;lst={};Do[a=(s+=n^3)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n, 5!}];lst] With[{nn=80},Select[#[[1]]/#[[2]]&/@Thread[{Accumulate[Range[ nn]^3],Range[ nn]}],IntegerQ]] (* or *) LinearRecurrence[{1,0,3,-3,0,-3,3,0,1,-1},{1,12,25,45,112,162,225,396,507,637},50] (* Harvey P. Dale, Mar 14 2020 *)
Formula
G.f.: ( x*(1+11*x+13*x^2+17*x^3+34*x^4+11*x^5+6*x^6+3*x^7) ) / ( (1+x+x^2)^3*(x-1)^4 ). - R. J. Mathar, Jan 25 2011
Extensions
Changed comments to examples - R. J. Mathar, Aug 20 2009
Comments