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A164577 Integer averages of the first perfect cubes up to some n^3. 4
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Integers of the form A000537(k)/k, created by the k>0 listed in A042965. [R. J. Mathar, Aug 20 2009]

Also, integers of the form (1/4)*n*(n+1)^2 for some n. [From Zak Seidov, Aug 17 2009]

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).

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

EXAMPLE

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).

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 *)

CROSSREFS

Cf. A050248, A051456, A078617, A078618, A154293, A164576

Sequence in context: A132754 A250666 A224669 * A195143 A280321 A198274

Adjacent sequences:  A164574 A164575 A164576 * A164578 A164579 A164580

KEYWORD

nonn,easy

AUTHOR

Vladimir Joseph Stephan Orlovsky, Aug 16 2009

EXTENSIONS

Changed comments to examples - R. J. Mathar, Aug 20 2009

STATUS

approved

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Last modified April 6 08:27 EDT 2020. Contains 333268 sequences. (Running on oeis4.)