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%I #45 Dec 09 2016 03:45:07
%S 76,57256,55722556,55572225556,55557222255556,55555722222555556,
%T 55555572222225555556,55555557222222255555556,
%U 55555555722222222555555556,55555555572222222225555555556,55555555557222222222255555555556,55555555555722222222222555555555556
%N Partial sums of the first 10^n terms in A181482.
%C Indeed: a(n) is the sum of 2*10^(2n-1)+1 and the palindrome built by repetition of the digits 2 and 5 such that it recalls the number 525.
%C Let x = 10^n, y = floor(x/3), and B(n) = Sum_{k<=10^n} binomial(floor(k/3),2).
%C 6*B(n) differs from a(n) by (x*(x+1)*(1+(2*x+1)/3))/4-3*y*(3*y+1).
%H R. J. Cano, <a href="/A222739/b222739.txt">Table of n, a(n) for n = 1..49</a>
%H R. J. Cano, <a href="/A222739/a222739_1.txt">Demonstrative program and additional information.</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1111,-112110,1111000,-1000000).
%F a(n) = Sum_{k<=10^n} A181482(k).
%F From _Colin Barker_, Oct 31 2015: (Start)
%F a(n) = 1111*a(n-1)-112110*a(n-2)+1111000*a(n-3)-1000000*a(n-4) for n>4.
%F G.f.: -4*x*(250000*x^3-157875*x^2+6795*x-19) / ((x-1)*(10*x-1)*(100*x-1)*(1000*x-1)).
%F (End)
%e When n=1, 10^n is 10. By looking at A181482 for its first 10 terms we have the sum: 1+3+0+4+9+3+10+18+9+19, then a(1)=76.
%o (PARI) repdigit(n,k)=(n!=0)*floor((10/9)*n*10^(k-1));
%o palindrome(n)=repdigit(5,n)*10^(2*n-1)+repdigit(2,n-1)*10^n+repdigit(5,n);
%o a(n)=palindrome(n)+(1+2*10^(2*n-1));
%o (PARI) Vec(-4*x*(250000*x^3-157875*x^2+6795*x-19)/((x-1)*(10*x-1)*(100*x-1)*(1000*x-1)) + O(x^100)) \\ _Colin Barker_, Oct 31 2015
%Y Cf. A181482, A213203.
%K nonn,easy
%O 1,1
%A _R. J. Cano_, Mar 07 2013
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