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A180921 Numbers a(n) whose square equals sums of cubes of b(n) consecutive integers starting from b(n), where b(n)= A180920. 2
1, 2079, 7876385, 30254180671, 116236127290689, 446579144331338591, 1715756954644453458529, 6591937773063166150358655, 25326223208345427203876398721, 97303342974524967600723097592479, 373839418381901692962342398114034081 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

LINKS

Colin Barker, Table of n, a(n) for n = 1..279

V. Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.

FORMULA

a(n) = b(n)*(31*(a(n-1)/b(n-1)) + 8*(15*((a(n-1)/b(n-1))^2) + 1)^(1/2)) where b(n) = A180920.

Conjectures 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)

EXAMPLE

For n=3, a(3)=2017*(31*(2079/33) + 8*(15*((2079/33)^2) + 1)^(1/2)).

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

Cf. A180920.

Sequence in context: A096927 A076425 A249654 * A270537 A076581 A183675

Adjacent sequences:  A180918 A180919 A180920 * A180922 A180923 A180924

KEYWORD

easy,nonn

AUTHOR

Vladimir Pletser, Sep 24 2010

STATUS

approved

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Last modified June 26 04:11 EDT 2019. Contains 324369 sequences. (Running on oeis4.)