A247107 a(0) = 0, a(n) = previous term + repunit of length of previous term for n > 0.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 220, 331, 442, 553, 664, 775, 886, 997, 1108, 2219, 3330, 4441, 5552, 6663, 7774, 8885, 9996, 11107, 22218, 33329, 44440, 55551, 66662, 77773, 88884, 99995, 111106, 222217, 333328, 444439

Offset: 0

Author

Dhilan Lahoti, Nov 30 2014

Keywords

• nonn
• base
• easy

Examples

			98 = 9*10 + 8 -> 10*10 + 9 = 109.
109 = 1*100 + 0*10 + 9*1 -> 2*100 + 1*10 + 10*1 = 220.
a(42) = 44440 + (10^(floor(log_10(44440))+1)-1) / 9 = 44440 + (10^(4+1)-1) / 9 = 44440 + 99999/9 = 44440 + 11111 = 55551.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,10,-20,10).

Crossrefs

Similar to A158699, but with simpler rules.

Cf. A002275, A055642.

Programs

Mathematica

a[0]=0; a[n_]:=FromDigits[IntegerDigits[a[n-1]]+1]; Array[a,50,0] (* Stefano Spezia, Sep 19 2024 *)

Formula

a(0) = 0, a(n) = a(n-1) + A002275(A055642(a(n-1))) for n>0.

From Jon E. Schoenfield, Nov 30 2014: (Start)

For n > 1, a(n) = a(n-1) + (10^(floor(log_10(a(n-1))) + 1) - 1) / 9.

For n > 0, a(n) = ((n-1) mod 9 + 1) * (10^D - 1) / 9 + 1 - D, where D = floor((n-1)/9) + 1. (There are exactly D digits in a(n).) (End)

G.f.: -(10*x^10-10*x^9+1)*x/((10*x^9-1)*(x-1)^2). - Alois P. Heinz, Nov 30 2014