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%I #38 Mar 06 2022 23:09:30
%S 23,118,333,716,1315,2178,3353,4888,6831,9230,12133,15588,19643,24346,
%T 29745,35888,42823,50598,59261,68860,79443,91058,103753,117576,132575,
%U 148798,166293,185108,205291,226890,249953,274528,300663,328406,357805,388908,421763,456418,492921,531320,571663,613998,658373,704836,753435,804218,857233,912528,970151,1030150,1092573,1157468
%N Numbers that begin a run of an odd number of consecutive integers whose cubes sum to a square.
%C Numbers k such that k^3 + (k+1)^3 + ... + (k+M-1)^3 = c^2 has nontrivial solutions over the integers where M is an odd positive integer.
%C To every odd positive integer M corresponds a sum of M consecutive cubes starting at a(n) having at least one nontrivial solution. For n >= 1, M(n) = (2n+1) (A005408), a(n) = M^3 - (3M-1)/2 = (2n+1)^3 - (3n+1) and c(n) = M*(M^2-1)*(2M^2-1)/2 = 2n*(n+1)*(2n+1)*(8n*(n+1)+1) (A253680).
%C The trivial solutions with M < 1 and k < 2 are not considered here.
%C Stroeker stated that all odd values of M yield a solution to k^3 + (k+1)^3 + ... + (k+M-1)^3 = c^2. This was further demonstrated by Pletser.
%H Vladimir Pletser, <a href="/A253679/b253679.txt">Table of n, a(n) for n = 1..50000</a>
%H Vladimir Pletser, <a href="/A253679/a253679.txt">File Triplets (M,a,c) for M=(2n+1)</a>
%H 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 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 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 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F a(n) = (2n+1)^3 - (3n+1).
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - _Colin Barker_, Jan 09 2015
%F G.f.: -x*(x^2-26*x-23) / (x-1)^4. - _Colin Barker_, Jan 09 2015
%e For n=1, M(n)=3, a(n)=23, c(n)=204.
%e See "File Triplets (M,a,c) for M=(2n+1)" link.
%p for n from 1 to 50 do a:=(2*n+1)^3-(3*n+1): print (a); end do:
%t a253679[n_] := (2 # + 1)^3 - (3 # + 1) & /@ Range@ n; a253679[52] (* _Michael De Vlieger_, Jan 10 2015 *)
%o (PARI) Vec(-x*(x^2-26*x-23)/(x-1)^4 + O(x^100)) \\ _Colin Barker_, Jan 09 2015
%Y Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253680, A253681.
%K nonn,easy
%O 1,1
%A _Vladimir Pletser_, Jan 08 2015
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