A264972 Trajectory of 262 under repeated application of the permutation A264965: a(0) = 262; for n >= 1, a(n) = A264965(a(n-1))

262, 670, 1450, 1690, 2158, 4246, 19522, 18410, 34678, 36926, 118882, 146998, 290566, 377966, 240038, 381466, 407054, 178898, 147146, 149714, 159458, 149378, 129242, 117958, 157822, 350014, 489878, 180770, 155930, 395686, 510386, 292426, 514294, 503114, 264490, 435670, 482882, 311674, 452774, 353570, 323374, 369638, 321926, 293726

Offset

0,1

Comments

The trajectory is probably infinite, not periodic.

Links

Antti Karttunen, Table of n, a(n) for n = 0..1000

Formula

a(0) = 262; for n >= 1, a(n) = A264965(a(n-1)).

Example

a(0) = 262 by definition. Its binary representation is A007088(262) = 100000110. When we reverse the significant prefix (i.e., leave the trailing zeros where they are), we get 386 (A007088(386) = 110000010). 386's ternary representation is A007089(386) = 112022. Reversing the significant prefix (now the whole expansion because no trailing zeros present), we get 220211 (= A007089(670)), thus a(1) = 670.

Prog

(Scheme, with memoizing macro definec)
(definec (A264972 n) (if (zero? n) 262 (A264965 (A264972 (- n 1)))))

Crossrefs

Cf. A007088, A007089, A264965, A264969, A264973.

Sequence in context: A045179 A264973 A250767 * A241877 A270752 A224556

Adjacent sequences: A264969 A264970 A264971 * A264973 A264974 A264975

Keyword

nonn,look,base

Author

Antti Karttunen, Dec 06 2015

Status

approved