A180921 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.
1, 2079, 7876385, 30254180671, 116236127290689, 446579144331338591, 1715756954644453458529, 6591937773063166150358655, 25326223208345427203876398721, 97303342974524967600723097592479, 373839418381901692962342398114034081
Offset: 1
Examples
a(3) = 2017*(31*(2079/33) + 8*sqrt(15*((2079/33)^2) + 1)).
Links
- Colin Barker, Table of n, a(n) for n = 1..279
- Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
- R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
- Index entries for linear recurrences with constant coefficients, signature (3904,-238206,3904,-1).
Programs
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Mathematica
LinearRecurrence[{3904,-238206,3904,-1},{1,2079,7876385,30254180671},20] (* Harvey P. Dale, Apr 27 2025 *) -
PARI
default(realprecision, 1000); b=vector(20, n, if(n==1, t=1, t=round(31*t-14+8*((3*t-1)*(5*t-3))^(1/2)))); 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
Formula
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).
From Colin Barker, Feb 19 2015: (Start)
a(n) = 3904*a(n-1) - 238206*a(n-2) + 3904*a(n-3) - a(4).
G.f.: x*(x+1)*(x^2-1826*x+1) / ((x^2-3842*x+1)*(x^2-62*x+1)). (End)
Extensions
Name clarified by Jon E. Schoenfield, Mar 11 2022
Comments