A333358 a(1) = 0. For n>1, a(n+1) = the minimum positive m such that the concatenation of the decimal values [a(n-m),a(n-m+1),...,a(n-m+k)] equals a(n) for some k>=0; a(n+1) = a(n) + 1 if no such m exists.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 11, 2, 12, 3, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 34, 1, 38, 39, 40, 41, 42, 43, 44, 45, 43, 3, 33, 14, 33, 2, 39, 13, 38, 16, 37, 38, 3, 11, 48, 49, 50, 51, 52, 53, 54, 55, 56, 65, 66, 67, 67, 1, 36

Offset: 1

Author

Scott R. Shannon, Mar 16 2020

Keywords

• nonn
• base

Examples

			a(n) = n-1 for 2 <= n <= 13 as a(1) = 0 and no choice for a(n) equals the concatenation of one or more previous terms, so a(n) = a(n-1) + 1 = n-1.
a(14) = 11 as a(13) = 12, and the value '12' is the concatenation of a(2) = 1 and a(3) = 2, and a(2) is eleven terms back from a(13).
a(15) = 2 as a(14) = 11, and the value '11' is a(12), which is two terms back from a(14).
		

Crossrefs

Cf. A181391, A007908, A001704.