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A253681 Integer squares c^2 that are equal to the sum of an odd number M of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 starting at b(n) (A253679).

%I A253681 #41 Sep 08 2022 08:46:10
%S A253681 41616,8643600,265559616,3359361600,25300083600,135426944016,
%T A253681 568998662400,1995144950016,6080268272400,16566690848400,
%U A253681 41192058954816,94910460840000,205045101804816,419208426176400,817072496870400,1527363954902016,2751797699456400,4798055269856016
%N A253681 Integer squares c^2 that are equal to the sum of an odd number M of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 starting at b(n) (A253679).
%C A253681 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 A253681 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), c(n) = M*(M^2-1)*(2M^2-1)/2 = 2n*(n+1)*(2n+1)*(8n*(n+1)+1) (A253680) and this sequence a(n) = c(n)^2.
%C A253681 The trivial solutions with M < 1 and b < 2 are not considered here.
%C A253681 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 A253681 Vladimir Pletser, <a href="/A253681/b253681.txt">Table of n, a(n) for n = 1..50000</a>
%H A253681 Vladimir Pletser, <a href="/A253680/a253680.txt">File Triplets (M,b,c) for M=(2n+1)</a>
%H A253681 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 A253681 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 A253681 <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).
%F A253681 a(n) = (2n(n+1)*(2n+1)*(8n*(n+1)+1))^2.
%F A253681 G.f.: -144*x*(289*x^8 + 56846*x^7 + 1199784*x^6 + 6296786*x^5 + 10697390*x^4 + 6296786*x^3 + 1199784*x^2 + 56846*x + 289) / (x -1)^11. - _Colin Barker_, Jan 09 2015
%e A253681 For n=1, M(1)=3, b(1)=23, c(1)=204, a(1)=c^2=41616.
%e A253681 See "File Triplets (M,b,c) for M=(2n+1)" link, [where in this File, M is the number of term, a the first term and c the square root of the sum].
%p A253681 restart: for n from 1 to 50000 do a:=(2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1))^2: print (a); end do:
%t A253681 f[n_] := (2 n (n + 1) (2 n + 1) (8 n (n + 1) + 1))^2; Array[f, 21] (* _Michael De Vlieger_, Jan 10 2015 *)
%o A253681 (PARI) Vec(-144*x*(289*x^8 +56846*x^7 +1199784*x^6 +6296786*x^5 +10697390*x^4 +6296786*x^3 +1199784*x^2 +56846*x +289) / (x -1)^11 + O(x^100)) \\ _Colin Barker_, Jan 09 2015
%o A253681 (Magma) [(2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1))^2: n in [1..20]]; // _Vincenzo Librandi_, Feb 19 2015
%Y A253681 Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253680.
%K A253681 nonn,easy
%O A253681 1,1
%A A253681 _Vladimir Pletser_, Jan 08 2015