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A266351
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Start with a(1) = 1, then always choose for a(n) the least unused number such that A057889(a(n)*a(n-1)) = A057889(a(n)) * A057889(a(n-1)), where A057889 is a bijective base-2 reverse.
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6
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1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 11, 17, 13, 32, 15, 24, 18, 20, 19, 33, 21, 36, 28, 27, 56, 34, 22, 64, 23, 65, 25, 40, 35, 72, 42, 48, 30, 51, 60, 66, 26, 68, 37, 96, 31, 99, 62, 128, 29, 129, 38, 80, 49, 73, 70, 130, 39, 256, 41, 131, 74, 136, 44, 132, 46, 257, 43, 258, 45, 260, 47, 512, 50, 133, 76, 160, 67, 84, 97, 137, 112, 54
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OFFSET
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1,2
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COMMENTS
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Equally: always choose for a(n) the least unused number such that a(n)*a(n-1) = A057889(A057889(a(n)) * A057889(a(n-1))).
Note that the adjacent terms of permutation A266195 satisfy the same condition, except that permutation is not the lexicographically earliest sequence of this kind (because it has a more restrictive condition). See A266194.
This is a bijection for the same reason that A266195 is. Any high enough 2^k will always save the permutation of being stuck, and will also immediately pick up as its succeeding pair the least term unused so far.
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LINKS
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PROG
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(Scheme, with defineperm1-macro from Antti Karttunen's IntSeq-library)
(defineperm1 (A266351 n) (cond ((= 1 n) n) (else (let ((prev (A266351 (- n 1)))) (let loop ((k 1)) (cond ((and (not-lte? (A266352 k) (- n 1)) (= (A057889 (* k prev)) (* (A057889 k) (A057889 prev)))) k) (else (loop (+ 1 k)))))))))
;; We consider a > b (i.e. not less than b) also in case a is #f.
;; (Because of the stateful caching system used by defineperm1-macro):
(define (not-lte? a b) (cond ((not (number? a)) #t) (else (> a b))))
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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