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%I #14 Jun 19 2015 11:19:42
%S 1,25333,2050333330,200500333333300,20005000333333333000,
%T 2000050000333333333330000,200000500000333333333333300000,
%U 20000005000000333333333333333000000,2000000050000000333333333333333330000000,200000000500000000333333333333333333300000000
%N Sum of the first 10^n 4th powers.
%H Colin Barker, <a href="/A130615/b130615.txt">Table of n, a(n) for n = 1..50</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (111010,-1111110000,1011100000000,-10000000000000).
%F Sum of the first m fourth powers = m(m+1)(2m+1)(3m^2+3m-1)/30 (see A000538).
%F From _Colin Barker_, Jun 14 2015: (Start)
%F a(n) = A000538(10^n).
%F 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.
%F a(n) = 111010*a(n-1) - 1111110000*a(n-2) + 1011100000000*a(n-3) - 10000000000000*a(n-4).
%F G.f.: x*(29480000000*x^3+349227000*x^2-85677*x+1) / ((10*x-1)*(1000*x-1)*(10000*x-1)*(100000*x-1)).
%F (End)
%o (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","))) }
%o (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
%K nonn,easy
%O 1,2
%A _Cino Hilliard_, Jun 18 2007
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