A255565 a(0) = 0; for n >= 1: if n = A255411(k) for some k, then a(n) = 2*a(k), otherwise, n = A256450(h) for some h, and a(n) = 1 + 2*a(h).

0, 1, 3, 7, 2, 15, 5, 31, 11, 63, 23, 127, 6, 47, 255, 13, 14, 95, 4, 511, 27, 29, 30, 191, 9, 1023, 55, 59, 61, 383, 19, 2047, 111, 119, 123, 767, 39, 4095, 223, 239, 247, 1535, 79, 8191, 447, 479, 495, 3071, 10, 159, 16383, 895, 62, 959, 991, 6143, 21, 319, 32767, 1791, 22, 125, 1919, 1983, 126, 12287, 46, 43, 639, 65535, 254, 3583, 12

Offset

0,3

Comments

Because all terms of A255411 are even it means that even terms can only occur in even positions (together with some odd terms, for each one of which there is a separate infinite cycle), while terms in odd positions are all odd.

Links

Antti Karttunen, Table of n, a(n) for n = 0..5040

Index entries for sequences related to factorial base representation

Index entries for sequences that are permutations of the natural numbers

Formula

a(0) = 0; for n >= 1: if A257680(n) = 0 [i.e., n is one of the terms of A255411], then a(n) = 2*a(A257685(n)), otherwise [when n is one of the terms of A256450], a(n) = 1 + 2*a(A273662(n)).

Other identities:

For all n >= 1, A001511(a(n)) = A257679(n).

For all n >= 1, a(A001563(n)) = A000079(n-1) = 2^(n-1).

For all n >= 1, a(A000142(n)) = A083318(n-1).

Prog

(Scheme, with memoization-macro definec)
(definec (A255565 n) (cond ((zero? n) n) ((zero? (A257680 n)) (* 2 (A255565 (A257685 n)))) (else (+ 1 (* 2 (A255565 (A273662 n)))))))

Crossrefs

Inverse: A255566.

Cf. A000079, A000142, A001511, A001563, A083318, A255411, A256450, A257679, A257680, A257682, A257685.

Cf. also arrays A257503, A257505.

Related or similar permutations: A273665, A273668.

Sequence in context: A266417 A260433 A243343 * A227351 A246377 A260421

Adjacent sequences: A255562 A255563 A255564 * A255566 A255567 A255568

Keyword

nonn,base

Author

Antti Karttunen, May 05 2015

Extensions

Formula changed because of the changed starting offset of A256450 - Antti Karttunen, May 30 2016

Status

approved