A366347 a(n) has as many prime factors as the ternary expansion of n has runs of nonzero digits; if the k-th run corresponds to A032924(e) and appears after m-1 0's then the p-adic valuation of a(n) is e (where p corresponds to the m-th prime number).

1, 2, 4, 3, 8, 16, 9, 32, 64, 5, 6, 12, 27, 128, 256, 81, 512, 1024, 25, 18, 36, 243, 2048, 4096, 729, 8192, 16384, 7, 10, 20, 15, 24, 48, 45, 96, 192, 125, 54, 108, 2187, 32768, 65536, 6561, 131072, 262144, 625, 162, 324, 19683, 524288, 1048576, 59049

Offset: 0

Author

Rémy Sigrist, Oct 07 2023

Keywords

• nonn
• base

Comments

This sequence is a variant of the Doudna sequence (A005940); here we consider runs of nonzero digits in ternary expansions, there in binary expansions.

This sequence is a bijection from the nonnegative integers to the positive integers with inverse A366348.

We can devise a similar sequence for any fixed base b >= 2:

- the case b = 2 corresponds (up to the offset) to the Doudna sequence (A005940),

- the case b = 3 corresponds to the present sequence,

- the case b = 10 corresponds to A290389.

Examples

			For n = 46: the ternary expansion of 46 is "1201; we have two runs of nonzero digits: "12" (= 5 = A032924(4)) after 2-1 0's and "1" (= 1 = A032924(1)) after 1-1 0's; so a(46) = prime(2)^4 * prime(1)^1 = 3^4 * 2^1 = 162.
		

Links

• Rémy Sigrist, PARI program
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A003961, A005940, A032924, A060140, A290389, A365278, A366348 (inverse).

Programs

PARI

See Links section.

Formula

a(3*n) = A003961(a(n)).

a(3^k) = prime(1 + k) for any k >= 0.

a(2 * 3^k) = prime(1 + k)^2 for any k >= 0.

a(n) is squarefree iff n belongs to A060140.