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%I #22 Dec 26 2015 14:23:28
%S 1,11,20,46,63,105,130,188,221,295,336,426,475,581,638,760,825,963,
%T 1036,1190,1271,1441,1530,1716,1813,2015,2120,2338,2451,2685,2806,
%U 3056,3185,3451,3588,3870,4015,4313,4466,4780,4941,5271,5440,5786,5963,6325,6510
%N Integer averages of the set of the first positive squares up to some n^2.
%C Integers of the form A000330(k)/k, k listed in A007310. - _R. J. Mathar_, Aug 20 2009
%H Zak Seidov, <a href="/A164576/b164576.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).
%F a(n) = A000330(A007310(n)) / A007310(n) = A175485(A007310(n)). - _Jaroslav Krizek_, May 28 2010
%F G.f. ( -x*(1+10*x+7*x^2+6*x^3) ) / ( (1+x)^2*(x-1)^3 ). - _R. J. Mathar_, Jan 25 2011
%F a(n) = 1/4*(12*n^2 - 6*n + (-1)^n*(4*n-1) + 1). - _Colin Barker_, Dec 26 2015
%e a(1) = 1^2/1 is integer. The average of the first two squares is (1^2+2^2)/2=5/2, not integer.
%e The average of the first three squares is (1^2+2^2+3^2)/3=14/3, not integer.
%e The average of the first five squares is (1^2+2^2+3^2+4^2+5^2)/ 5=11, integer, and constitutes a(2).
%t s=0;lst={};Do[a=(s+=n^2)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n,6!}];lst
%t Flatten[Table[{(1 + 3 k) (1 + 4 k), (1 + k) (11 + 12 k)}, {k, 0, 499}]] (* _Zak Seidov_, Aug 15 2012 *)
%t Module[{nn=150,sq},sq=Range[nn]^2;Select[Table[Mean[Take[sq,n]],{n,nn}],IntegerQ]] (* or *) LinearRecurrence[{1,2,-2,-1,1},{1,11,20,46,63},50] (* _Harvey P. Dale_, Oct 31 2013 *)
%o (PARI) a(n) = 1/4*(12*n^2 - 6*n + (-1)^n*(4*n-1) + 1) \\ _Colin Barker_, Dec 26 2015
%Y Cf. A050248, A078617, A154293.
%K nonn,easy
%O 1,2
%A _Vladimir Joseph Stephan Orlovsky_, Aug 16 2009
%E Edited by _R. J. Mathar_, Aug 20 2009
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