cp's OEIS Frontend

A253680 Numbers c(n) whose square are equal to the sum of an odd number M of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c(n)^2, starting at b(n) (A253679).

%I A253680 #51 Sep 08 2022 08:46:10
%S A253680 204,2940,16296,57960,159060,368004,754320,1412496,2465820,4070220,
%T A253680 6418104,9742200,14319396,20474580,28584480,39081504,52457580,
%U A253680 69267996,90135240,115752840,146889204,184391460,229189296,282298800,344826300,417972204,503034840
%N A253680 Numbers c(n) whose square are equal to the sum of an odd number M of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c(n)^2, starting at b(n) (A253679).
%C A253680 Numbers c(n) such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for M being an odd positive integer.
%C A253680 To every odd positive integer M corresponds a sum of M consecutive cubed integers starting at b^3 having at least one nontrivial solution. For n>=1, M(n)=(2n+1) (A005408), b(n) = M^3 -(3M-1)/2 = (2n+1)^3 - (3n+1) (A253679) and c(n) = M*(M^2-1)*(2M^2-1)/2 = 2n*(n+1)*(2n+1)*(8n*(n+1)+1) (A253680).
%C A253680 The trivial solutions with M < 1 and b < 2 are not considered here.
%C A253680 Stroeker stated that all odd values of M yield a solution to  b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2. This was further demonstrated by Pletser.
%H A253680 Vladimir Pletser, <a href="/A253680/b253680.txt">Table of n, a(n) for n = 1..50000</a>
%H A253680 Vladimir Pletser, <a href="/A253680/a253680.txt">File Triplets (M,b,c) for M=(2n+1)</a>
%H A253680 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 A253680 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 A253680 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 A253680 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F A253680 c(n) = 2n(n+1)*(2n+1)*(8n*(n+1)+1).
%F A253680 G.f.: 12*x*(x+1)*(17*x^2+126*x+17) / (x-1)^6. - _Colin Barker_, Jan 09 2015
%e A253680 For n=1, M(n)=3, b(n)=23, c(n)=204.
%e A253680 See "File Triplets (M,b,c) for M=(2n+1)" link.
%p A253680 restart: for n from 1 to 50000 do c:=2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1): print (c); end do:
%t A253680 f[n_] := 2 n (n + 1) (2 n + 1) (8 n (n + 1) + 1); Array[f, 36] (* _Michael De Vlieger_, Jan 10 2015 *)
%o A253680 (PARI) Vec(12*x*(x+1)*(17*x^2+126*x+17)/(x-1)^6 + O(x^100)) \\ _Colin Barker_, Jan 09 2015
%o A253680 (Magma) [2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1): n in [1..30]]; // _Vincenzo Librandi_, Feb 19 2015
%Y A253680 Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253681.
%K A253680 nonn,easy
%O A253680 1,1
%A A253680 _Vladimir Pletser_, Jan 08 2015