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A246165
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Permutation of natural numbers: a(1) = 1, a(n) = A064989(n)-th integer among those positive integers not occurring earlier in the sequence. [A064989(n) shifts the prime factorization of n one step right].
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5
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1, 2, 4, 3, 7, 6, 11, 5, 12, 10, 17, 9, 23, 16, 19, 8, 29, 18, 35, 15, 28, 25, 41, 14, 31, 34, 30, 24, 51, 27, 59, 13, 44, 43, 47, 26, 67, 52, 58, 22, 77, 42, 83, 38, 49, 61, 89, 21, 70, 46, 73, 53, 99, 45, 69, 37, 88, 75, 111, 40, 119, 85, 72, 20, 94, 64, 127, 63, 103, 68, 137, 39, 143, 97, 79, 78, 106, 87, 151, 36
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OFFSET
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1,2
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COMMENTS
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Terms at a(2^n) are: 1, 2, 3, 5, 8, 13, 20, 32, 48, 71, 105, 156, 236, 354, 542, 815, 1228, ...
Fixed points begin as: 1, 2, 6, 10, 18, 42, 92, 26372, ...
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LINKS
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EXAMPLE
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By definition, a(1) = 1.
After that, for n = 2, when its prime factorization is shifted once right, results A064989(2) = 1, so we select the 1st of still unused positive natural numbers, which is 2, thus a(2) = 2.
For n = 3 = p_2 (3 is the second prime), when its prime factorization is shifted once right, results A064989(3) = 2 = p_1, so we select 2nd of still unused numbers, which is 4, thus a(3) = 4.
For n = 4, like for all powers of two, the result of right shifting is 1, so we select the smallest still unused number, which is 3, thus a(4) = 3.
For n = 5 = p_3, A064989(5) = 3 = p_2, so we select the 3rd smallest still unused number from [5, 6, 7, 8, ...] which is 7, thus a(5) = 7.
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PROG
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(defineperm1 (A246165 n) (if (<= n 1) n (let loop ((i 1) (the-n-th-one (- (A064989 n) 1))) (cond ((not-lte? (A246166 i) n) (if (zero? the-n-th-one) i (loop (+ i 1) (- the-n-th-one 1)))) (else (loop (+ i 1) the-n-th-one))))))
;; We consider a > b (i.e. not less than or equivalent to b) also in case a is nil.
;; (Which is needed when using 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
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AUTHOR
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STATUS
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approved
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