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.

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

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