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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
OFFSET
0,4
COMMENTS
Set k = n, take the k-th 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
FORMULA
a(0) = 0, for n >= 1, a(n) = 1 + a(A268395(n)).
PROG
(Scheme, with memoization-macro definec)
(definec (A268708 n) (if (zero? n) 0 (+ 1 (A268708 (A268395 n)))))
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Feb 11 2016
STATUS
approved