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 A238099 The stonemason's problem: numbers n such that n^2 is the sum of more than three consecutive cubes, the cube 1 being disallowed. 2
 312, 315, 323, 504, 588, 720, 2079, 2170, 2940, 4472, 4914, 5187, 5880, 5984, 6630, 7497, 8721, 8778, 9360, 10296, 10695, 11024, 13104, 14160, 16296, 16380, 18333, 18810, 22022, 22330, 23247, 31248, 36729, 42021, 43065, 43309, 49665 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS A subsequence of both A126200 and A163393. LINKS Vincenzo Librandi, Chai Wah Wu, and Charles R Greathouse IV, Table of n, a(n) for n = 1..1000 (1..57 from Librandi, 58..246 from Wu) H. E. Dudeney, Amusements in Mathematics, Nelson, London, 1917, Problem 135. EXAMPLE 312^2 = 97344 = 14^3 + 15^3 + ... + 25^3. MATHEMATICA 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 *) 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 *) PROG (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 CROSSREFS Cf. A126200, A163393. Sequence in context: A139638 A308002 A112542 * A259720 A011774 A011251 Adjacent sequences: A238096 A238097 A238098 * A238100 A238101 A238102 KEYWORD nonn AUTHOR N. J. A. Sloane, Feb 25 2014 STATUS approved

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Last modified March 3 00:46 EST 2024. Contains 370499 sequences. (Running on oeis4.)