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%I #34 Jan 14 2024 14:39:35
%S 1,2079,7876385,30254180671,116236127290689,446579144331338591,
%T 1715756954644453458529,6591937773063166150358655,
%U 25326223208345427203876398721,97303342974524967600723097592479,373839418381901692962342398114034081
%N a(n) is the square root of the sum of the cubes of the b(n) consecutive integers starting from b(n), where b(n) = A180920.
%C Colin Barker's linear recurrence conjecture confirmed, see A180920. - _Ray Chandler_, Jan 12 2024
%H Colin Barker, <a href="/A180921/b180921.txt">Table of n, a(n) for n = 1..279</a>
%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 (3904,-238206,3904,-1).
%F a(n) = b(n)*(31*(a(n-1)/b(n-1)) + 8*sqrt(15*((a(n-1)/b(n-1))^2) + 1)) where b(n) = A180920(n).
%F From _Colin Barker_, Feb 19 2015: (Start)
%F a(n) = 3904*a(n-1) - 238206*a(n-2) + 3904*a(n-3) - a(4).
%F G.f.: x*(x+1)*(x^2-1826*x+1) / ((x^2-3842*x+1)*(x^2-62*x+1)). (End)
%F a(n) = Sqrt(A240137(A180920(n))). - _Ray Chandler_, Jan 12 2024
%e a(3) = 2017*(31*(2079/33) + 8*sqrt(15*((2079/33)^2) + 1)).
%o (PARI)
%o default(realprecision, 1000);
%o b=vector(20, n, if(n==1, t=1, t=round(31*t-14+8*((3*t-1)*(5*t-3))^(1/2))));
%o vector(#b, n, if(n==1, t=1, t=round(b[n]*(31*(t/b[n-1])+8*(15*((t/b[n-1])^2)+1)^(1/2))))) \\ _Colin Barker_, Feb 19 2015
%Y Cf. A180920, A240137.
%K easy,nonn
%O 1,2
%A _Vladimir Pletser_, Sep 24 2010
%E Name clarified by _Jon E. Schoenfield_, Mar 11 2022
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