A350151 Variation of Van Eck's sequence A181391: a(1) = 0; a(n+1) is the positive offset from the end of the string concatenating a(1) through a(n-1) to the first appearance of a(n). a(n+1) = 0 if a(n) does not appear in the string.

0, 0, 1, 0, 2, 0, 2, 2, 1, 6, 0, 5, 0, 2, 6, 5, 4, 0, 5, 3, 0, 3, 2, 9, 0, 4, 9, 3, 6, 14, 0, 7, 0, 2, 12, 0, 4, 8, 0, 3, 14, 13, 0, 6, 19, 0, 4, 8, 13, 9, 6, 9, 2, 22, 53, 43, 0, 15, 0, 3, 5, 3, 2, 12, 38, 0, 9, 21, 7, 49, 57, 0, 9, 4, 6, 33, 0, 6, 4, 6, 2

Offset: 1

Author

Ya-Ping Lu, Dec 16 2021

Keywords

• nonn
• base

Comments

This sequence and A181391 differ starting at n = 32.

It seems that a(n)/(n*log(n)) <= 1/(3*log(3)) and lim sup a(n)/(n*log(n)) = 0.24 approximately.

Examples

			a(2) = 0 as a(1) = 0, which does not appear before a(1).
a(15) = 6 as a(14) = 2, which first appears 6th digits back from the end of the string, '0010202216050', the decimal concatenation of a(1) through a(13).
a(42) = 13 as a(41) = 14, which first appears 13 digits back from the end of the string, '001020221605026540530329049361407021204803', the decimal concatenation of a(1) through a(40).
		

Crossrefs

Cf. A181391, A333359.

Programs

Mathematica

a[1]=0;a[n_]:=a[n]=Block[{c=""<>ToString/@Most[s=Array[a,n-1]]},Check[Max[StringLength@c-Last@StringPosition[c,ToString@Last@s]]+1,0]];Array[a,81] (* Giorgos Kalogeropoulos, Dec 17 2021 *)

Python

a1 = '0'; print(a1, end =', '); S = a2 = ''
for n in range(2, 100):
    p = S.rfind(a1); a = str(len(S) - p) if p != -1 else '0'
    print(a, end = ", "); S += a1; a2 = a1; a1 = a