A264969 Cycle length of n under bijection A264965 / A264966.

1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 5, 1, 5, 3, 2, 1, 5, 1, 5, 1, 2, 3, 5, 1, 5, 3, 1, 1, 5, 1, 22, 2, 3, 9, 7, 1, 22, 9, 3, 1, 22, 2, 7, 4, 2, 9, 1, 1, 7, 9, 4, 4, 7, 1, 5, 1, 3, 9, 5, 1, 1, 3, 2, 2, 5, 2, 22, 1, 2, 9, 18, 1, 7, 3, 4, 4, 18, 4, 7, 1, 2, 3, 18, 1, 3, 10, 3, 10, 18, 1, 3, 8, 6, 3, 22, 2, 22, 9, 4, 4, 18, 3, 17, 10, 4, 10, 6, 1, 3, 3, 6, 1, 18, 4, 17, 8, 3, 3, 10, 1, 7

Offset: 0

Author

Antti Karttunen, Dec 06 2015

Keywords

• nonn

Comments

By convention: a(n) = 0 if the cycle where n resides can be proven to be infinite. a(n) = -1 is the placeholder value if the cycle where n resides is only suspected to be infinite.

Is n=262 the first representative of runaway cycle? See A264972 and A264973.

Examples

			A264965(5) = 7 and A264965(7) = 5, thus a(5) = a(7) = 2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..261

Crossrefs

Cf. A264965, A264966, A264972, A264973.

Programs

Scheme

;; Very naive program which succumbs to the first infinite cycle:
(define (A264969 n) (let loop ((s 1) (next (A264965 n))) (if (equal? next n) s  (loop (+ 1 s) (A264965 next)))))