A341537 a(n) is the number of digits from the end of the concatenation of all previous terms where n last appears. If n has not previously appeared then a(n) = n.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 2, 14, 15, 16, 17, 18, 19, 20, 15, 22, 33, 24, 25, 26, 27, 28, 29, 30, 31, 17, 20, 54, 35, 36, 37, 38, 39, 40, 53, 35, 17, 44, 75, 46, 5, 48, 49, 50, 16, 52, 23, 13, 55, 95, 57, 58, 7, 60, 87, 69, 51, 64, 26, 66, 115, 68, 15, 70, 106, 79, 71, 61, 36

Offset: 0

Author

Scott R. Shannon, Feb 14 2021

Keywords

• nonn
• base

Examples

			a(9) = 9 as the concatenation of all previous terms is "012345678" which does not include 9, so a(9) = 9.
a(12) = 13 as the concatenation of all previous terms is "01234567891011" which includes "12" as a substring, starting 13 digits from the end of the concatenation.
a(13) = 2 as the concatenation of all previous terms is "0123456789101113" which includes "13" as a substring, starting 2 digits from the end of the concatenation.
		

Links

• Scott R. Shannon, Image of the terms for n=0..500000.

Crossrefs

Cf. A333921, A337227, A007908, A116700, A131881.