A126200 -id:A126200 - cp's OEIS frontend

A218979 Numbers n such that some sum of n consecutive positive cubes is a square.

1, 3, 5, 7, 8, 9, 11, 12, 13, 15, 17, 18, 19, 21, 23, 25, 27, 28, 29, 31, 32, 33, 35, 37, 39, 40, 41, 42, 43, 45, 47, 48, 49, 50, 51, 53, 54, 55, 57, 59, 60, 61, 63, 64, 65, 67, 69, 71, 72, 73, 75, 76, 77, 79, 81, 82, 83, 85, 87, 89, 91, 92, 93, 94, 95, 97, 98, 99

Offset: 1

Michel Marcus, Nov 08 2012

nonn

The trivial solutions with x = 0 and x = 1 are not considered here.
Numbers n such that x^3 + (x+1)^3 + ... + (x+n-1)^3 = y^2 has nontrivial solutions over the integers.
The nontrivial solutions are found by solving Y^2 = X^3 + d(n)*X with d(n) = n^2*(n^2-1)/4 (A006011), Y = n*y and X = n*x + (1/2)*n*(n-1). [Corrected by Derek Orr, Aug 30 2014]
x^3 + (x+1)^3 + ... + (x+n-1)^3 = y^2 can also be written as y^2 = n(x + (n-1)/2)(n(x + (n-1)/2) + x(x-1)). - Vladimir Pletser, Aug 30 2014
There are 892 triples (n,x,y), with n and x less than 10^5 (1 < n,x < 10^5), which are nontrivial solutions of x^3 + (x+1)^3 + ... + (x+n-1)^3 = y^2 (note that (n,x,y) corresponds to (M,a,c) in A253679, A253680, A253681, A253707, A253708, A253709, A253724, A253725). - Vladimir Pletser, Jan 10 2015

			See "Examples of triples" link.

Michel Marcus, Examples of triples up to n=50
Vladimir Pletser, Triplets (n, x, y) with n,x less than 10^5
Vladimir Pletser, Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares, Research Gate, 2015.
Vladimir Pletser, Fundamental solutions of the Pell equation X^2-(sigma^4-delta^4)Y^2=delta^4 for the first 45 solutions of the sums of consecutive cubed integers equalling integer squares, Research Gate, 2015. See Reference 19.
V. Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253680, A253681, A253707, A253708, A253709.

a(n)=for(x=2,10^7, /* note this limit only generates the terms in the data section */ X = n*x + (1/2)*n*(n-1); d=n^2*(n^2-1)/4;if(issquare(X^3+d*X),return(x)))
n=1;while(n<100,if(a(n),print1(n,", "));n++) \\ Derek Orr, Aug 30 2014

Name changed, a(1) = 1 prepended and a(39)-a(68) from Derek Orr, Aug 30 2014

More terms for 50Vladimir Pletser, Jan 10 2015

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.

Original entry on oeis.org

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.

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A218979 Numbers n such that some sum of n consecutive positive cubes is a square.

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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.

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

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