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A031138
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Numbers k such that 1^5 + 2^5 + ... + k^5 is a square.
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11
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1, 13, 133, 1321, 13081, 129493, 1281853, 12689041, 125608561, 1243396573, 12308357173, 121840175161, 1206093394441, 11939093769253, 118184844298093, 1169909349211681, 11580908647818721, 114639177128975533, 1134810862641936613, 11233469449290390601
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OFFSET
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1,2
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COMMENTS
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This is a 6th-degree Diophantine equation 12*m^2 = n^2*(n+1)^2*(2*n^2 + 2*n - 1) which reduces to the generalized Pell equation 6*q^2 = (2*n + 1)^2 - 3 where q = 3*m/(n*(n+1)), so there is no surprise that the solutions satisfy a linear recurrent equation. - Charles R Greathouse IV, Max Alekseyev, Oct 22 2012
Also k such that k^2 + (k+1)^2 is equal to the sum of three consecutive squares, for example 13^2 + 14^2 = 10^2 + 11^2 + 12^2. - Colin Barker, Sep 06 2015
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LINKS
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FORMULA
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a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = -1/2 + ((3 - sqrt(6))/4)*(5 + 2*sqrt(6))^n + ((3 + sqrt(6))/4)*(5 - 2*sqrt(6))^n.
a(n)^2 + (a(n) + 1)^2 = (b(n) - 1)^2 + b(n)^2 + (b(n) + 1)^2 = c(n) = 3*d(n) + 2; where b(n) is A054320, c(n) is A007667 and d(n) is A006061.
a(n) = 10*a(n-1) - a(n-2) + 4; a(0) = a(1) = 1. Also sum of first a(n) fifth powers is a square m^2, where m has factors A000217{a(n)} and A054320(n). - Lekraj Beedassy, Jul 08 2002
G.f.: x*(1+x)^2/((1-x)*(x^2-10*x+1)). - R. J. Mathar, Oct 26 2009
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EXAMPLE
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a(2) = 13 because 1^5+2^5+...13^5 = 1001^2; a(1) = 1 because 1^5 = 1^2.
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MATHEMATICA
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LinearRecurrence[{11, -11, 1}, {1, 13, 133}, 20 ] (* Harvey P. Dale, Oct 23 2012 *)
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PROG
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(PARI) isok(n) = issquare(sum(i=1, n, i^5)); \\ Michel Marcus, Dec 28 2013
(PARI) Vec(x*(1+x)^2/((1-x)*(x^2-10*x+1)) + O(x^40)) \\ Colin Barker, Sep 06 2015
(Magma) [Round(-1/2 + ((3 - Sqrt(6))/4)*(5 + 2*Sqrt(6))^n + ((3 + Sqrt(6) )/4)*(5 - 2*Sqrt(6))^n): n in [0..50]]; // G. C. Greubel, Nov 04 2017
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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