1, 2, 1, 3, 1, 3, 1, 2, 1, 4, 1, 4, 1, 2, 1, 4, 1, 4, 1, 2, 1, 4, 1, 4, 1, 2, 1, 4, 1, 4, 1, 2, 1, 3, 1, 3, 1, 2, 1, 5, 1, 5, 1, 2, 1, 5, 1, 5, 1, 2, 1, 5, 1, 5, 1, 2, 1, 5, 1, 5, 1, 2, 1, 3, 1, 3, 1, 2, 1, 5, 1, 5, 1, 2, 1, 5, 1, 5, 1, 2, 1, 5, 1, 5, 1, 2, 1, 5, 1, 5, 1, 2, 1, 3, 1, 3, 1, 2, 1, 5, 1, 5, 1, 2, 1, 5
Offset: 0
For n = 2, we divide it with A000040(1) = 2, and it leaves zero remainder, so we have finished on the first round (needing no actual iterations), and thus a(2) = 1. Note that 2 in primorial base (A049345) is written as "10", and indeed the first zero from the right occurs at the position 1.
For n = 5, we first divide 5 with prime(1) = 2, and discarding the remainder, we are left with floor(5/2) = 2. Then we divide that 2 with prime(2) = 3, leaving floor(2/3) = 0 and remainder 2. And finally we divide 0 with prime(3) = 5, and that doesn't leave any remainder, thus we are finished on the third round, and a(5) = 3. Note that 5 in primorial base is written as "21", and allowing here a leading zero, written as "021", we see that it is in this case the least significant zero occurring at position 3 from the right.
For n = 43, we first divide it with prime(1) = 2, leaving a remainder 1 and integer quotient 21. Then we divide 21 with prime(2) = 3, which doesn't leave any remainder, thus we are finished on the second round, and a(43) = 2. Note that 43 is written as "1201" in primorial base, with the least significant zero occurring in the position 2.
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