A257081 a(n) = Number of iterations of A257080 needed, starting from n, before a fixed point is reached.

0, 2, 1, 2, 0, 3, 1, 3, 1, 1, 2, 2, 0, 4, 2, 2, 0, 3, 0, 2, 3, 2, 0, 3, 1, 3, 1, 3, 1, 2, 1, 4, 1, 1, 7, 3, 1, 6, 1, 3, 2, 5, 1, 4, 1, 1, 2, 3, 0, 4, 2, 2, 0, 4, 2, 2, 9, 10, 4, 8, 0, 6, 3, 3, 0, 6, 0, 3, 6, 3, 0, 6, 0, 2, 2, 2, 0, 3, 2, 2, 2, 2, 1, 3, 0, 1, 3, 1, 0, 2, 0, 2, 2, 3, 0, 4, 0, 3, 8, 4, 0, 5, 6, 5, 3, 2, 6, 4, 0, 3, 1, 5, 0, 5, 0, 2, 2, 2, 0, 6, 1

Offset

0,2

Comments

Note: when at some point of iteration we reach some k whose factorial representation (A007623) does not contain any 1's, then at next step A257080(k) = 1*k, and thus a fixed point has been reached.

Links

Antti Karttunen, Table of n, a(n) for n = 0..10080

Eric Angelini, et al, "Multiply by the fantom digit", Discussion on Seqfan-list

Formula

If A257079(n) = 1, a(n) = 0, otherwise, a(n) = 1 + a(A257080(n)).

Example

For n = 5, with factorial representation A007623(5) = "21", the least missing nonzero digit is 3, thus A257080(5) = 3*5 = 15. 15 has factorial representation "211", so again we multiply by 3, resulting 3*15 = 45, with factorial representation "1311", thus the least missing nonzero digit is now 2, and 2*45 = 90, "3300" in factorial base, for which the least missing digit is 1, resulting 1*90 = 90 forever after, thus we have reached a fixed point after three iteration steps (5 -> 15 -> 45 -> 90) and a(5) = 3.

Prog

(Scheme)
(define (A257081 n) (let loop ((oldn n) (n (A257080 n)) (s 1)) (if (= oldn n) s (loop n (A257080 n) (+ 1 s)))))
;; Alternative, recursive version, optionally using the memoizing definec-macro:
(definec (A257081 n) (if (= 1 (A257079 n)) 0 (+ 1 (A257081 (A257080 n)))))

Crossrefs

A255411 gives the positions of zeros.

Cf. A007623, A257079, A257080.

Sequence in context: A339471 A159834 A274576 * A271484 A199920 A177995

Adjacent sequences: A257078 A257079 A257080 * A257082 A257083 A257084

Keyword

nonn

Author

Antti Karttunen, Apr 15 2015

Status

approved