A253708 Numbers c(n) whose squares are equal to the sums of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, for a first term b(n) being an odd squared integer (A016754).

323, 7497, 57618, 262430, 878445, 2399103, 5669972, 12026988, 23457735, 42785765, 73877958, 121874922, 193444433, 297057915, 443289960, 645140888, 918382347, 1281925953, 1758214970, 2373639030, 3158971893, 4149832247, 5387167548, 6917760900, 8794760975

Offset: 1

Author

Vladimir Pletser, Jan 09 2015

Keywords

• nonn
• easy

Comments

Numbers c(n) such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for b being an odd squared integer (A016754).

To every odd squared integer b corresponds a sum of M consecutive cubed integers starting at b^3 equaling a squared integer and having at least one nontrivial solution. For n>=1, b(n) = (2n+1)^2 (A016754), M(n) = (sqrt(b)-1)(2b-1)/2 = n(8n(n+1)+1) (A253707), and c(n)= (b-1)(4b^2-1)/8 = (n(n+1)/2)(4(2n+1)^4-1) (this sequence).

The trivial solutions with M < 1 and b < 2 are not considered here.

Examples

			For n=1, b(n)=9, M(n)=17, a(n)=323.
See "File Triplets (M,b,c) for a=(2n+1)^2" link.
		

Links

• Vladimir Pletser, Table of n, a(n) for n = 1..50000
• Vladimir Pletser, File Triplets (M,b,c) for a=(2n+1)^2
• 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, 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.
• Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Crossrefs

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

Programs

Magma

[(n*(n+1)/2)*(4*(2*n+1)^4-1): n in [1..30]]; // Vincenzo Librandi, Feb 19 2015

Maple

restart: for n from 1 to 50000 do a:= (n*(n+1)/2)(4*(2*n+1)^4-1): print (a); end do:

Mathematica

f[n_] := (n (n + 1)/2) (4 (2 n + 1)^4 - 1); Array[f, 33] (* Michael De Vlieger, Jan 10 2015 *)

PARI

Vec(-x*(323*x^4+5236*x^3+11922*x^2+5236*x+323)/(x-1)^7 + O(x^100)) \\ Colin Barker, Jan 14 2015

Formula

a(n) = (n(n+1)/2)(4(2n+1)^4-1).

G.f.: -x*(323*x^4+5236*x^3+11922*x^2+5236*x+323) / (x-1)^7. - Colin Barker, Jan 14 2015