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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.
2
1, 2079, 7876385, 30254180671, 116236127290689, 446579144331338591, 1715756954644453458529, 6591937773063166150358655, 25326223208345427203876398721, 97303342974524967600723097592479, 373839418381901692962342398114034081
OFFSET
1,2
COMMENTS
Colin Barker's linear recurrence conjecture confirmed, see A180920. - Ray Chandler, Jan 12 2024
LINKS
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
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)
a(n) = Sqrt(A240137(A180920(n))). - Ray Chandler, Jan 12 2024
EXAMPLE
a(3) = 2017*(31*(2079/33) + 8*sqrt(15*((2079/33)^2) + 1)).
PROG
(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
CROSSREFS
Sequence in context: A096927 A076425 A249654 * A270537 A076581 A183675
KEYWORD
easy,nonn
AUTHOR
Vladimir Pletser, Sep 24 2010
EXTENSIONS
Name clarified by Jon E. Schoenfield, Mar 11 2022
STATUS
approved