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%I #14 Jun 13 2015 00:52:22
%S 1,385,338350,333833500,333383335000,333338333350000,
%T 333333833333500000,333333383333335000000,333333338333333350000000,
%U 333333333833333333500000000,333333333383333333335000000000
%N Sum of the first 10^n squares.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1110, -111000, 1000000).
%F Sum of the first m squares = m(2m^2+3m+1)/6.
%F From _Robert Israel_, Jan 02 2015: (Start)
%F a(n) = (2*10^(3*n)+3*10^(2*n)+10^n)/6.
%F a(n+3) = 10^6*a(n) - 111000*a(n+1) + 1110*a(n+2)
%F G.f.: 1/(3 - 3000*x) + 1/(2 - 200*x) + 1/(6 - 60*x).
%F E.g.f.: exp(1000*x)/3 + exp(100*x)/2 + exp(10*x)/6.
%F (End)
%p seq((2*10^(3*n)+3*10^(2*n)+10^n)/6, n = 0 .. 30); # _Robert Israel_, Jan 02 2015
%t Table[(2^(x-1)*5^x*(1+2^(x+1)*5^x)(1+10^x))/3,{x,0,20}] (* or *) Join[{1},LinearRecurrence[{1110,-111000,1000000},{385,338350,333833500},20]] (* or *) Join[{1},Table[FromDigits[Join[PadRight[{},n,3],{8},PadRight[{},n-1,3],{5},PadRight[{},n-1,0]]],{n,20}]] (* _Harvey P. Dale_, Jan 02 2015 *)
%o (PARI) sumsq(n) = for(x=0,n,y=10^x;z=y*(y+1)*(2*y+1)/6;(print1(z","))) \Trust but verify, brute force sum g1(n) = for(x=0,n,y=sum(j=1,10^x,j^2);(print1(y",")))
%K nonn
%O 0,2
%A _Cino Hilliard_, Jun 18 2007
%E Offset corrected by _Robert Israel_, Jan 02 2015
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