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%I #24 Apr 11 2024 10:10:59
%S 1,12,25,45,112,162,225,396,507,637,960,1156,1377,1900,2205,2541,3312,
%T 3750,4225,5292,5887,6525,7936,8712,9537,11340,12321,13357,15600,
%U 16810,18081,20812,22275,23805,27072,28812,30625,34476,36517,38637,43120
%N Integer averages of the first perfect cubes up to some n^3.
%C Integers of the form A000537(k)/k, created by the k>0 listed in A042965. - _R. J. Mathar_, Aug 20 2009
%C Integers of the form (1/4)*n*(n+1)^2 for some n. - _Zak Seidov_, Aug 17 2009
%H Harvey P. Dale, <a href="/A164577/b164577.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,3,-3,0,-3,3,0,1,-1).
%F 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
%e The average of the first cube is 1^3/1=1=a(1).
%e The average of the first two cubes is (1^3+2^3)/2=9/2, not integer, and does not contribute to the sequence.
%e The average of the first three cubes is (1^3+2^3+3^3)/3=12, integer, and defines a(2).
%t Timing[s=0;lst={};Do[a=(s+=n^3)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n, 5!}];lst]
%t 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 *)
%Y Cf. A000537, A050248, A051456, A078617, A078618, A154293, A164576
%K nonn,easy,changed
%O 1,2
%A _Vladimir Joseph Stephan Orlovsky_, Aug 16 2009
%E Changed comments to examples - _R. J. Mathar_, Aug 20 2009
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