A249068 a(n+1) gives the number of occurrences of the last digit of a(n) in octal base so far, up to and including a(n), with a(0)=0.

0, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 4, 4, 5, 4, 6, 4, 7, 4, 8, 5, 5, 6, 5, 7, 5, 8, 6, 6, 7, 6, 8, 7, 7, 8, 8, 9, 18, 10, 11, 9, 22, 9, 24, 10, 13, 9, 28, 9, 30, 10, 14, 11, 13, 10, 15, 9, 38, 12, 11, 14, 13, 11, 15

Offset: 0

Author

Antti Karttunen, Oct 21 2014

Keywords

• nonn
• base

Comments

This is the octal version of Eric Angelini's A248034.

Examples

			For n=16, we see that a(15) = 8, '10' in octal, and '0' has occurred just twice in the octal representations of terms a(0) .. a(15), namely in a(0) = 0 (which is also zero when read in octal base) and a(15), thus a(16) = 2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..32768

Crossrefs

Cf. A248034 (analogous sequence in base-10), A007094 (octal representation of n).