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A234743
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Multiplicative permutation of integers: a(n) = A235199(A234840(n)).
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10
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0, 1, 3, 2, 9, 19, 6, 61, 27, 4, 57, 17, 18, 433, 183, 38, 81, 101, 12, 7, 171, 122, 51, 173, 54, 361, 1299, 8, 549, 43, 114, 31, 243, 34, 303, 1159, 36, 1811, 21, 866, 513, 733, 366, 157, 153, 76, 519, 613, 162, 3721, 1083, 202, 3897, 1193, 24, 323, 1647, 14, 129, 59, 342, 5
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OFFSET
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0,3
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COMMENTS
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Consider two self-inverse and multiplicative permutations, b and c defined as follows:
b(0)=0, b(1)=1, b(2)=3, b(3)=2, b(p_i) = p_{b(i+1)-1} for primes with index i > 2, and b(u*v) = b(u)*b(v) for u, v > 0.
c(n)=n if n < 4, c(5)=7 and c(7)=5, c(p_i) = p_{c(i)} for primes with index i > 4, and c(u*v) = c(u)*c(v) for u, v > 0.
This permutation is defined as their composition: a(n) = c(b(n)) = A235199(A234840(n)).
It is also multiplicative: a(u*v) = c(b(u*v)) = c(b(u)*b(v)) = c(b(u))*c(b(v)) = a(u)*a(v). For primes p_i with index i, a(p_i) = c(b(p_i)) = c(p_{b(i+1)-1}) = p_{c(b(i+1)-1)} = A000040(A235047(i)), except for cases i=8 and i=18, use 7 and 5, instead of 5 and 7.
Because 22 = 2*11, and 2 is in a two-cycle and 11 is in a three-cycle, 22 is in a cycle whose length is lcm(2,3) = 6: a(22)=51 (= a(2)*a(11) = 3*17), a(51)=202, a(202)=33, a(33)=34, a(34)=303, a(303)=22.
Among primes, there are at least fixed points (31), two-cycles (2 <-> 3), (37 <-> 1811), three-cycles: (11, 17, 101), (29, 43, 157), four-cycles: (5, 19, 7, 61), (41, 733, 359, 1091), eight-cycles: (47, 613, 2593, 1163, 1733, 409, 73, 131).
How long is the cycle beginning from 13, a(13)=433, a(433)=20693, a(20693)=? or from 23? (23, 173, 24043, ...)
Question: Are there any infinite cycles? If there are, what is the ratio of terms (primes) in finite cycles vs. infinite cycles?
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LINKS
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FORMULA
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CROSSREFS
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KEYWORD
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nonn,mult,nice
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AUTHOR
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STATUS
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approved
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