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A256993
a(1) = 0; for n > 1, a(n) = 1 + a(A256992(n)).
9
0, 1, 2, 3, 2, 3, 4, 3, 4, 4, 5, 3, 4, 5, 4, 5, 4, 5, 6, 5, 5, 4, 5, 6, 6, 5, 4, 5, 6, 5, 6, 5, 6, 6, 7, 5, 6, 6, 6, 7, 5, 6, 6, 6, 5, 7, 7, 6, 6, 5, 7, 7, 6, 7, 6, 6, 7, 5, 6, 7, 6, 7, 6, 7, 6, 7, 8, 7, 7, 6, 7, 8, 7, 7, 6, 7, 7, 8, 6, 7, 7, 7, 8, 6, 7, 6, 7, 8, 8, 7, 7, 6, 8, 7, 7, 8, 6, 8, 7, 7, 8, 7, 6, 8, 7, 8, 8, 7, 7, 8, 8, 6, 7, 7, 7, 8, 7, 8, 8, 7, 6, 7, 8, 7, 8, 7, 8, 7
OFFSET
1,3
COMMENTS
Number of iterations of A256992 needed to reach one when starting from n.
LINKS
FORMULA
a(1) = 0; for n > 1, a(n) = 1 + a(A256992(n)).
Other observations. For all n >= 1 it holds that:
a(n) >= A254110(n).
a(n) >= A256989(n).
a(n) >= A255559(n)-1.
Also it seems that a(n) - A070939(n) = -1, 0 or +1 for all n >= 1. [Compare A256991 and A256992 to see the connection.]
It is also very likely that a(n) <= A071542(n) for all n.
From Antti Karttunen, Dec 10 2016: (Start)
For all n >= 2, a(n) = A070939(A279341(n)) = A070939(A279343(n)).
For all n >= 2, a(n) = A279345(n) + A279346(n) - 1.
(End)
PROG
(Scheme, with memoization macro definec)
(definec (A256993 n) (if (= 1 n) 0 (+ 1 (A256993 (A256992 n)))))
KEYWORD
nonn
AUTHOR
Antti Karttunen, Apr 15 2015
STATUS
approved