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

%I #4 Mar 07 2020 02:11:01

%S 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,

%T 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,

%U 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

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

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

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

%e a(0) = 0 since 2^(2^0-1) + 1 = 2^0 + 1 = 2 is not divisible by 3.

%e a(1) = 1 since 2^(2^1-1) + 1 = 2^1 + 1 = 3 is divisible just once by 3.

%e a(2) = 2 since 2^(2^2-1) + 1 = 2^3 + 1 = 9 is divisible by 3^2.

%e a(3) = 1 since 2^(2^4-1) + 1 = 2^15 + 1 = 32769 is divisible only once by 3.

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

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

%K nonn

%O 0,3

%A _M. F. Hasler_, Mar 05 2020