A098752 a(1) = 1 and a(n+1) is the least number > a(n) that begins with the last digit of a(n) and doesn't end with 0.

1, 11, 12, 21, 101, 102, 201, 1001, 1002, 2001, 10001, 10002, 20001, 100001, 100002, 200001, 1000001, 1000002, 2000001, 10000001, 10000002, 20000001, 100000001, 100000002, 200000001, 1000000001, 1000000002, 2000000001, 10000000001

Offset: 1

Author

Eric Angelini, Oct 01 2004

Keywords

• base
• easy
• nonn

Comments

a(n) must be chosen with nonzero rightmost digit.

Crossrefs

Cf. A076654.

Cf. A101233.

Formula

For n == 0 (mod 3), a(n) = 10^(n/3) + 2; for n == 1 (mod 3), n>1, a(n) = 2*10^((n-1)/3) + 1; for n == 2 (mod 3), a(n) = 10^((n+1)/3) + 1. - Sam Alexander, Jan 04 2005

From Chai Wah Wu, Jun 02 2016: (Start)

a(n) = 11*a(n-3) - 10*a(n-6) for n > 7.

G.f.: x*(1 + x + 2*x^2)*(1 + 10*x - 10*x^3 - 10*x^4)/((1 - x)*(1 - 10*x^3)*(1 + x + x^2)). (End)

a(n) = (1-sign((n-1) mod 3))*10^floor(n/3)+10^floor((n+1)/3)-sign(n mod 3)+2, for n > 1. - Wesley Ivan Hurt, Mar 06 2022

Extensions

More terms from Sam Alexander, Jan 04 2005

More terms from David Wasserman, Feb 26 2008