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%I #26 Nov 14 2016 02:43:12
%S 312,315,323,504,588,720,2079,2170,2940,4472,4914,5187,5880,5984,6630,
%T 7497,8721,8778,9360,10296,10695,11024,13104,14160,16296,16380,18333,
%U 18810,22022,22330,23247,31248,36729,42021,43065,43309,49665
%N The stonemason's problem: numbers n such that n^2 is the sum of more than three consecutive cubes, the cube 1 being disallowed.
%C A subsequence of both A126200 and A163393.
%H Vincenzo Librandi, Chai Wah Wu, and Charles R Greathouse IV, <a href="/A238099/b238099.txt">Table of n, a(n) for n = 1..1000</a> (1..57 from Librandi, 58..246 from Wu)
%H H. E. Dudeney, <a href="https://archive.org/stream/amusementsinmath00dude">Amusements in Mathematics</a>, Nelson, London, 1917, Problem <a href="https://archive.org/stream/amusementsinmath00dude#page/24/mode/2up/">135</a>.
%e 312^2 = 97344 = 14^3 + 15^3 + ... + 25^3.
%t nn = 500; t = Table[n^3, {n, 2, nn}]; t2 = Table[Total[Take[t, {i, j}]], {i, nn - 1}, {j, i + 3, nn - 1}]; t3 = Select[Union[Flatten[t2]], # <= nn^3 &]; Select[t3, IntegerQ[#^(1/2)] &]^(1/2) (* _T. D. Noe_, Feb 25 2014 *)
%t nn=1000;With[{c=Range[2,nn]^3},Sort[Select[Sqrt[#]&/@ Flatten[ Table[ Total/@ Partition[c,n,1],{n,4,nn}]],IntegerQ]]] (* _Harvey P. Dale_, Apr 28 2014 *)
%o (PARI) list(lim)=my(v=List(),L2=(lim\=1)^2,s,t); for(n=25,sqrtnint(lim^2\3,3)+1, s=3*n^3 - 9*n^2 + 15*n - 9; forstep(k=n-3,2,-1, s+=k^3; if(s>L2, break); if(issquare(s,&t), listput(v,t)))); Set(v) \\ _Charles R Greathouse IV_, Nov 13 2016
%Y Cf. A126200, A163393.
%K nonn
%O 1,1
%A _N. J. A. Sloane_, Feb 25 2014
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