

A031138


1^5+2^5+...+n^5 is a square.


5



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

1,2


COMMENTS

Partial sums of A004291 or convolution of A040000 with A054320. [From R. J. Mathar, Oct 26 2009]
This is a 6thdegree Diophantine equation 12*m^2=n^2*(n+1)^2*(2*n^2+2*n1) which reduces to the generalized Pell equation 6*q^2=(2*n+1)^23 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


LINKS

Table of n, a(n) for n=1..20.
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (11, 11, 1).


FORMULA

a(n) = 11*(a(n1)a(n2)) + a(n3); a(n) = 1/2+((3sqrt(6))/4) *(5+2sqrt(6))^n +((3+sqrt(6))/4)*(52sqrt(6))^n.
a(n)^2+(a(n)+1)^2 = (b(n)1)^2+b(n)^2+(b(n)+1)^2 = c(n) = 3d(n)+2; where b(n) is A054320, c(n) is A007667 and d(n) is A006061.
a(n) = 10*a(n1)  a(n2) + 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
contfrac(sqrt(6)/A054320(n))[4]/2  Thomas Baruchel, Dec 02 2003
G.f.: x*(1+x)^2/((1x)*(x^210*x+1)). [From R. J. Mathar, Oct 26 2009]


EXAMPLE

a(2) = 13 because 1^5+2^5+...13^5 = 1001^2; a(1) = 1 because 1^5 = 1^2.


MATHEMATICA

LinearRecurrence[{11, 11, 1}, {1, 13, 133}, 20 ] (* Harvey P. Dale, Oct 23 2012 *)


PROG

(PARI) isok(n) = issquare(sum(i=1, n, i^5)); \\ Michel Marcus, Dec 28 2013


CROSSREFS

Cf. A000539, A006061, A054320, A007667.
Sequence in context: A081042 A016153 A187732 * A097166 A073556 A154999
Adjacent sequences: A031135 A031136 A031137 * A031139 A031140 A031141


KEYWORD

easy,nonn


AUTHOR

Ignacio Larrosa CaĆ±estro, entry revised Feb 27 2000


STATUS

approved



