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A332202
Largest k >= 0 such that 3^k divides 2^(2^n-1) + 1.
2
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
0,3
COMMENTS
Behaves like a mixture of 2-adic and 3-adic ruler function, cf. formula.
FORMULA
For all n > 0, a(2n-1) = 1; a(2n) = 2 + A007949(n) = 1 + A051064(n).
EXAMPLE
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.
PROG
(PARI) apply( {A332202(n)=if(bittest(n, 0), 1, n, valuation(n\2, 3)+2)}, [0..99])
CROSSREFS
Cf. A007949, A051064, A001511 (2-adic ruler)
Sequence in context: A257993 A376928 A055881 * A204917 A232098 A055874
KEYWORD
nonn
AUTHOR
M. F. Hasler, Mar 05 2020
STATUS
approved