A265330 Zero-based row index to A265345; 2-adic valuation of bijective base-3 reversal of n: a(n) = A007814(A263273(n)).

0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 6, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 6, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 3

Offset: 1

Author

Antti Karttunen, Dec 18 2015

Keywords

• nonn
• base

Examples

			For n = 32, in base-3 "1012" [= A007089(32)], when we reverse it, we get "2101" [= A007089(64)], and 2-adic valuation of 64 [= "1000000" = A007088(64)] is 6, thus a(32) = 6.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..6561

Crossrefs

One less than A265331.

Cf. A007088, A007089, A263273, A265345, A265352.

Cf. A265910 (corresponding other index).

Cf. also A265336, A265337, A265340.

Differs from A007814 for the first time at n=32, where a(32) = 6, while A007814(32) = 5.

Formula

a(n) = A007814(A263273(n)).

a(2n+1) = 0, a(2n) = 1 + a(A265352(n)).