A222739 Partial sums of the first 10^n terms in A181482.

76, 57256, 55722556, 55572225556, 55557222255556, 55555722222555556, 55555572222225555556, 55555557222222255555556, 55555555722222222555555556, 55555555572222222225555555556, 55555555557222222222255555555556, 55555555555722222222222555555555556

Offset

1,1

Comments

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.

Let x = 10^n, y = floor(x/3), and B(n) = Sum_{k<=10^n} binomial(floor(k/3),2).

6*B(n) differs from a(n) by (x*(x+1)*(1+(2*x+1)/3))/4-3*y*(3*y+1).

Links

R. J. Cano, Table of n, a(n) for n = 1..49

R. J. Cano, Demonstrative program and additional information.

Index entries for linear recurrences with constant coefficients, signature (1111,-112110,1111000,-1000000).

Formula

a(n) = Sum_{k<=10^n} A181482(k).

From Colin Barker, Oct 31 2015: (Start)

a(n) = 1111*a(n-1)-112110*a(n-2)+1111000*a(n-3)-1000000*a(n-4) for n>4.

G.f.: -4*x*(250000*x^3-157875*x^2+6795*x-19) / ((x-1)*(10*x-1)*(100*x-1)*(1000*x-1)).

(End)

Example

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.

Prog

(PARI) repdigit(n, k)=(n!=0)*floor((10/9)*n*10^(k-1));
palindrome(n)=repdigit(5, n)*10^(2*n-1)+repdigit(2, n-1)*10^n+repdigit(5, n);
a(n)=palindrome(n)+(1+2*10^(2*n-1));
(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

Crossrefs

Cf. A181482, A213203.

Sequence in context: A271242 A241878 A033521 * A060716 A116255 A136609

Adjacent sequences: A222736 A222737 A222738 * A222740 A222741 A222742

Keyword

nonn,easy

Author

R. J. Cano, Mar 07 2013

Status

approved