A253709 Integer squares c^2 that are equal to the sums of M (A253707) consecutive cubed integers equaling a squared integer, b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, for a first term b(n) being an odd squared integer (A016754).

104329, 56205009, 3319833924, 68869504900, 771665618025, 5755695204609, 32148582480784, 144648440352144, 550265331330225, 1830621686635225, 5457952678249764, 14853496612506084, 37420748658691489, 88243404864147225, 196505988636801600, 416206765369428544, 843426135281228409, 1643334148974958209, 3091319880732100900, 5634162244739340900

Offset: 1

Author

Vladimir Pletser, Jan 09 2015

Keywords

• nonn
• easy

Comments

Numbers a(n)=c^2 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) and M (A253707).

To every odd squared integer b (A016754) corresponds a sum of M (A253707) consecutive cubed integers starting at b^3 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), c(n)= (b-1)(4b^2-1)/8 = (n(n+1)/2)(4(2n+1)^4-1) (A253708) and a(n)=c(n)^2 (this sequence).

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

Examples

			For n=1, b(1)=9, M(1)=17, c(1)=323, a(1)= 104329 (see File 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, 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 (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Crossrefs

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

Programs

Magma

[((n*(n+1)/2)*(4*(2*n+1)^4-1))^2: n in [1..20]]; // Vincenzo Librandi, Jan 15 2015

Maple

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

Mathematica

f[n_] := ((n (n + 1)/2) (4 (2 n + 1)^4 - 1))^2; Array[f, 20] (* Michael De Vlieger, Jan 10 2015 *)

PARI

Vec(-x*(104329*x^10 [phone]*x^9 [phone]*x^8 [phone]*x^7 [phone]*x^6 [phone]*x^5 [phone]*x^4 [phone]*x^3 [phone]*x^2 [phone]*x +104329) / (x -1)^13 + O(x^100)) \\ Colin Barker, Jan 10 2015

Formula

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

G.f.: -x*(104329*x^10 [phone]*x^9 [phone]*x^8 [phone]*x^7 [phone]*x^6 [phone]*x^5 [phone]*x^4 [phone]*x^3 [phone]*x^2 [phone]*x +104329) / (x -1)^13. - Colin Barker, Jan 10 2015