OFFSET
1,2
LINKS
Colin Barker, Table of n, a(n) for n = 1..50
Index entries for linear recurrences with constant coefficients, signature (111010,-1111110000,1011100000000,-10000000000000).
FORMULA
Sum of the first m fourth powers = m(m+1)(2m+1)(3m^2+3m-1)/30 (see A000538).
From Colin Barker, Jun 14 2015: (Start)
a(n) = A000538(10^n).
a(n) = 2^(-5+n) * 5^(-6+n) * (-5000 + 4^(1+n)*5^(3+2*n) + 3*5^(2+3*n)*8^n+3*10^(4*n))/3.
a(n) = 111010*a(n-1) - 1111110000*a(n-2) + 1011100000000*a(n-3) - 10000000000000*a(n-4).
G.f.: x*(29480000000*x^3+349227000*x^2-85677*x+1) / ((10*x-1)*(1000*x-1)*(10000*x-1)*(100000*x-1)).
(End)
PROG
(PARI) sumquartic(n) = { for(x=0, n, m=10^x; z=m*(m+1)*(2*m+1)*(3*m^2+3*m-1)/30; (print1(z", "))) }
(PARI) Vec(x*(29480000000*x^3+349227000*x^2-85677*x+1) / ((10*x-1)*(1000*x-1)*(10000*x-1)*(100000*x-1)) + O(x^15)) \\ Colin Barker, Jun 14 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Cino Hilliard, Jun 18 2007
STATUS
approved