A367342 a(n) = n+1 for n = 1 to 8; a(n) = 100 + a(n-8) for n = 9 to 16; thereafter a(8*i+j) = 10^(i+1) + a(8*(i-1)+j) for i >= 2, 1 <= j <= 8.

2, 3, 4, 5, 6, 7, 8, 9, 102, 103, 104, 105, 106, 107, 108, 109, 1102, 1103, 1104, 1105, 1106, 1107, 1108, 1109, 11102, 11103, 11104, 11105, 11106, 11107, 11108, 11109, 111102, 111103, 111104, 111105, 111106, 111107, 111108, 111109, 1111102, 1111103, 1111104, 1111105, 1111106, 1111107, 1111108, 1111109

Offset: 1

Author

Michael S. Branicky and N. J. A. Sloane, Nov 15 2023

Keywords

• nonn
• easy

Comments

Conjectured to equal A367341(n)/9.

Links

• Paolo Xausa, Table of n, a(n) for n = 1..5000
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,11,0,0,0,0,0,0,0,-10).

Crossrefs

Cf. A367341.

Programs

Mathematica

A367342[n_]:=100(10^Floor[(n-1)/8]-1)/9+Mod[n-1,8]+2;Array[A367342,100] (* Paolo Xausa, Nov 17 2023 *)

Formula

a(n) = ((10^i-1)/9)*100 + j, where i = floor((n-1)/8) and j = (n-1) mod 8 + 2.