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
Keywords
Examples
See "Examples of triples" link.
Links
- 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.
Crossrefs
Programs
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PARI
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
Extensions
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
Comments