A152577 a(n) = 10^(2*n - 1) + 1.

11, 1001, 100001, 10000001, 1000000001, 100000000001, 10000000000001, 1000000000000001, 100000000000000001, 10000000000000000001, 1000000000000000000001, 100000000000000000000001

Offset

1,1

Comments

These numbers are all divisible by 11. This follows from the identity a^n - b^n = (a+b)*(a^(n-1) - a^(n-2)*b + ... + b^(n-1)) for odd values of n. In this example a=10 and b=1 so a+b = 11. The sum of digits rule for divisibility by 11 also applies.

Bisection of A000533. Also, bisection of A062397. a(n) is also A084508(n+1) written in base 2. a(n) is also A087289(n-1) written in base 2. a(n) is also the concatenation of "1", 2(n-1) digits "0" and "1". - Omar E. Pol, Dec 13 2008

Links

Table of n, a(n) for n=1..12.

Index entries for linear recurrences with constant coefficients, signature (101,-100).

Formula

a(n) = 100*a(n-1) - 99 (with a(1)=11). - Vincenzo Librandi, Dec 14 2010

G.f.: -11*x*(-1+10*x) / ( (100*x-1)*(x-1) ). - R. J. Mathar, Sep 01 2011

a(n) = 11*A095372(n-1). - R. J. Mathar, Sep 01 2011

a(n) = 101*a(n-1)-100*a(n-2). - Wesley Ivan Hurt, Apr 24 2021

Example

From Omar E. Pol, Dec 14 2008: (Start)
n ....... a(n)
1 ....... 11
2 ...... 1001
3 ..... 100001
4 .... 10000001
5 ... 1000000001
(End)

Mathematica

LinearRecurrence[{101, -100}, {11, 1001}, 20] (* Harvey P. Dale, Nov 05 2015 *)

Prog

(PARI) g(n)=forstep(x=1, n, 2, y=(10^x+1); print1(y", "))

Crossrefs

Cf. A000533, A062397, A084508, A087289. - Omar E. Pol, Dec 13 2008

Sequence in context: A342036 A004656 A143016 * A278757 A281172 A281218

Adjacent sequences: A152574 A152575 A152576 * A152578 A152579 A152580

Keyword

nonn,easy

Author

Cino Hilliard, Dec 08 2008

Status

approved