

A268708


Number of iterations of A268395 needed to reach zero: a(0) = 0, for n >= 1, a(n) = 1 + a(A268395(n)).


5



0, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 18
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OFFSET

0,4


COMMENTS

Set k = n, take the kth Xfactorial A048631(k) and find what is the maximum exponent h so that polynomial (X+1)^h still divides it (over GF(2), this can be computed with A268389, maybe also more directly). Set this h as a new value of k, and repeat. a(n) tells how many steps are needed before zero is reached.
Note that so far it is just an empirical observation that A268672(n) > 0 for all n > 0.


LINKS

Antti Karttunen, Table of n, a(n) for n = 0..65537


FORMULA

a(0) = 0, for n >= 1, a(n) = 1 + a(A268395(n)).


PROG

(Scheme, with memoizationmacro definec)
(definec (A268708 n) (if (zero? n) 0 (+ 1 (A268708 (A268395 n)))))


CROSSREFS

Cf. A048631, A268389, A268395, A268672.
Cf. also A268709, A268710.
Sequence in context: A283297 A091194 A156079 * A061555 A146323 A071626
Adjacent sequences: A268705 A268706 A268707 * A268709 A268710 A268711


KEYWORD

nonn


AUTHOR

Antti Karttunen, Feb 11 2016


STATUS

approved



