A163393 -id:A163393 - cp's OEIS frontend

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

N. J. A. Sloane, Feb 25 2014

nonn

A subsequence of both A126200 and A163393.

			312^2 = 97344 = 14^3 + 15^3 + ... + 25^3.

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.

Cf. A126200, A163393.

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 *)

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

A343409 Numbers whose square is the sum of one or more consecutive nonnegative cubes.

Original entry on oeis.org

0, 1, 3, 6, 8, 10, 15, 21, 27, 28, 36, 45, 55, 64, 66, 78, 91, 105, 120, 125, 136, 153, 171, 190, 204, 210, 216, 231, 253, 276, 300, 312, 315, 323, 325, 343, 351, 378, 406, 435, 465, 496, 504, 512, 528, 561, 588, 595, 630, 666, 703, 720, 729, 741, 780, 820

Offset: 1

Lamine Ngom, Apr 14 2021

Roots of square terms of A217843. Sequence contains (but is not limited to) cubes (A000578) and triangular numbers (A000217).

			8 is a term because 8^2 = 64 = 4^3.
10 is a term because 10^2 = 100 = 1^3 + 2^3 + 3^3 + 4^3.

Cf. A000217, A000578, A126200, A163393, A217843.

N:= 1000: # for terms <= N
M:= floor(N^(2/3)):
S:= [seq(n^2*(n+1)^2/4, n=0..M)]:
SD:= {0,seq(seq(S[i]-S[j],j=1..i-1),i=1..M+1)}:
Q:= select(t -> t <= N^2 and issqr(t),SD):
sort(convert(map(sqrt,Q),list)); # Robert Israel, Sep 11 2023

Union of A000217 and A126200.

cp's OEIS Frontend

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.

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