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
Keywords
Examples
312^2 = 97344 = 14^3 + 15^3 + ... + 25^3.
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.
Programs
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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 *) -
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
Comments