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A164579 Integer averages of halves of first cubes of natural numbers (n^3)/2 for some n. 1
6, 56, 81, 198, 480, 578, 950, 1656, 1875, 2646, 3968, 4356, 5670, 7800, 8405, 10406, 13536, 14406, 17238, 21560, 22743, 26550, 32256, 33800, 38726, 46008, 47961, 54150, 63200, 65610, 73206, 84216, 87131, 96278, 109440, 112908, 123750, 139256 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Also, integers of the form (1/8)*n*(n+1)^2 for some n. - Zak Seidov, Aug 17 2009
LINKS
FORMULA
G.f.: ( x*(6+50*x+25*x^2+99*x^3+132*x^4+23*x^5+39*x^6+10*x^7) ) / ( (1+x+x^2)^3*(x-1)^4 ). - R. J. Mathar, Jan 25 2011
EXAMPLE
1/2, 9/4, 6, 25/2, 45/2, 147/4, 56, 81, ...
MATHEMATICA
s=0; lst={}; Do[a=(s+=(n^3)/2)/n; If[Mod[a, 1]==0, AppendTo[lst, a]], {n, 3*5!}]; lst
LinearRecurrence[{1, 0, 3, -3, 0, -3, 3, 0, 1, -1}, {6, 56, 81, 198, 480, 578, 950, 1656, 1875, 2646}, 40] (* Harvey P. Dale, Jul 26 2017 *)
Module[{nn=200, ac}, ac=Accumulate[Range[nn]^3/2]; Select[#[[1]]/#[[2]]&/@ Thread[{ac, Range[nn]}], IntegerQ]] (* Harvey P. Dale, Jan 28 2020 *)
PROG
(PARI) forstep(n=3, 150, [4, 1, 3], print1(n*(n+1)^2>>3, ", ")); \\ Charles R Greathouse IV, Nov 02 2009
CROSSREFS
Sequence in context: A335199 A137033 A045526 * A137034 A177059 A345473
KEYWORD
nonn,easy,changed
AUTHOR
STATUS
approved

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Last modified April 25 03:15 EDT 2024. Contains 371964 sequences. (Running on oeis4.)