A277808 a(n) = number of iterations of map k -> A003188(A006068(k)/2) that are required (when starting from k = n) until k is an odious number.

0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 0, 1, 2, 0, 3, 0, 0, 1, 4, 0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 4, 0, 0, 1, 0, 1, 2, 0, 5, 0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 0, 1, 2, 0, 3, 0, 0, 1, 4, 0, 0, 1, 0, 1, 2, 0, 5, 0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 6, 0, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 3, 0, 0, 1, 0, 1, 2, 0, 3, 0, 0, 1, 4

Offset

1,6

Links

Antti Karttunen, Table of n, a(n) for n = 1..16384

Index entries for sequences related to binary expansion of n

Formula

a(n) = A010059(n) * A001511(n).

If A010060(n) = 1 [when n is one of the odious numbers, A000069], then a(n) = 0, otherwise a(n) = 1 + a(A003188(A006068(n)/2)).

Other identities:

For all n >= 0, a(A003945(n)) = n.

Prog

(Scheme, with memoization-macro definec)
(definec (A277808 n) (if (= 1 (A010060 n)) 0 (+ 1 (A277808 (A003188 (/ (A006068 n) 2))))))
;; Other versions:
(define (A277808 n) (if (= 1 (A010060 n)) 0 (A001511 n)))
(define (A277808 n) (* (A010059 n) (A001511 n)))

Crossrefs

One less than A277822.

A left inverse of A003945.

Cf. A277812 (gives the odious number where such an iteration is finished at when starting from k=n).

Cf. A000069, A001969, A001511, A003188, A006068, A010059, A010060, A245710, A268389.

Sequence in context: A359865 A035217 A357237 * A105964 A303051 A324044

Adjacent sequences: A277805 A277806 A277807 * A277809 A277810 A277811

Keyword

nonn,base

Author

Antti Karttunen, Nov 03 2016

Status

approved