A332202 Largest k >= 0 such that 3^k divides 2^(2^n-1) + 1.

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

Offset: 0

Author

M. F. Hasler, Mar 05 2020

Keywords

• nonn

Comments

Behaves like a mixture of 2-adic and 3-adic ruler function, cf. formula.

Examples

			a(0) = 0 since 2^(2^0-1) + 1 = 2^0 + 1 = 2 is not divisible by 3.
a(1) = 1 since 2^(2^1-1) + 1 = 2^1 + 1 = 3 is divisible just once by 3.
a(2) = 2 since 2^(2^2-1) + 1 = 2^3 + 1 = 9 is divisible by 3^2.
a(3) = 1 since 2^(2^4-1) + 1 = 2^15 + 1 = 32769 is divisible only once by 3.
		

Crossrefs

Cf. A007949, A051064, A001511 (2-adic ruler)

Programs

PARI

apply( {A332202(n)=if(bittest(n,0), 1, n, valuation(n\2,3)+2)}, [0..99])

Formula

For all n > 0, a(2n-1) = 1; a(2n) = 2 + A007949(n) = 1 + A051064(n).