cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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).

This page as a plain text file.
%I A253708 #43 Jan 08 2025 10:58:26
%S A253708 323,7497,57618,262430,878445,2399103,5669972,12026988,23457735,
%T A253708 42785765,73877958,121874922,193444433,297057915,443289960,645140888,
%U A253708 918382347,1281925953,1758214970,2373639030,3158971893,4149832247,5387167548,6917760900,8794760975
%N 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).
%C A253708 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).
%C A253708 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).
%C A253708 The trivial solutions with M < 1 and b < 2 are not considered here.
%H A253708 Vladimir Pletser, <a href="/A253708/b253708.txt">Table of n, a(n) for n = 1..50000</a>
%H A253708 Vladimir Pletser, <a href="/A253707/a253707_1.txt">File Triplets (M,b,c) for a=(2n+1)^2</a>
%H A253708 Vladimir Pletser, <a href="http://www.researchgate.net/profile/Vladimir_Pletser/publication/271272786">Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares</a>, Research Gate, 2015.
%H A253708 Vladimir Pletser, <a href="http://arxiv.org/abs/1501.06098">General solutions of sums of consecutive cubed integers equal to squared integers</a>, arXiv:1501.06098 [math.NT], 2015.
%H A253708 R. J. Stroeker, <a href="http://www.numdam.org/item?id=CM_1995__97_1-2_295_0">On the sum of consecutive cubes being a perfect square</a>, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
%H A253708 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).
%F A253708 a(n) = (n(n+1)/2)(4(2n+1)^4-1).
%F A253708 G.f.: -x*(323*x^4+5236*x^3+11922*x^2+5236*x+323) / (x-1)^7. - _Colin Barker_, Jan 14 2015
%e A253708 For n=1, b(n)=9, M(n)=17, a(n)=323.
%e A253708 See "File Triplets (M,b,c) for a=(2n+1)^2" link.
%p A253708 restart: for n from 1 to 50000 do a:= (n*(n+1)/2)(4*(2*n+1)^4-1): print (a); end do:
%t A253708 f[n_] := (n (n + 1)/2) (4 (2 n + 1)^4 - 1); Array[f, 33] (* _Michael De Vlieger_, Jan 10 2015 *)
%o A253708 (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
%o A253708 (Magma) [(n*(n+1)/2)*(4*(2*n+1)^4-1): n in [1..30]]; // _Vincenzo Librandi_, Feb 19 2015
%Y A253708 Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253681, A253707, A253709.
%K A253708 nonn,easy
%O A253708 1,1
%A A253708 _Vladimir Pletser_, Jan 09 2015