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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).
6
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.
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. also arrays A257503, A257505.
Related or similar permutations: A273665, A273668.
Sequence in context: A266417 A260433 A243343 * A227351 A246377 A260421
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