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A057063
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Let P(n) of a sequence s(1),s(2),s(3),... be obtained by leaving s(1),...,s(n) fixed and reverse-cyclically permuting every n consecutive terms thereafter; apply P(2) to 1,2,3,... to get PS(2), then apply P(3) to PS(2) to get PS(3), then apply P(4) to PS(3), etc. The limit of PS(n) is A057063.
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6
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1, 2, 4, 6, 3, 10, 12, 7, 16, 18, 11, 22, 13, 5, 28, 30, 19, 20, 36, 23, 40, 42, 9, 46, 29, 31, 52, 37, 35, 58, 60, 15, 44, 66, 43, 70, 72, 25, 54, 78, 8, 82, 61, 55, 88, 65, 59, 68, 96, 27, 100, 102, 38, 106, 108, 71, 112, 85, 33, 89, 94, 79, 48, 126, 83, 130
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OFFSET
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1,2
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COMMENTS
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It appears that this is a permutation of the integers. - Michel Marcus, Feb 19 2016
The fact that this is a permutation is proved at the MathOverflow link below. Also from that link: a(n)+1 is prime if and only if a(n) = 2*(n-1). - Ilya I. Bogdanov, Feb 15 2022
Let {b(n)} be the inverse permutation of this sequence, then each number n >= 3 is moved for b(n)-2 times during the process.
Proof: suppose that this number is k, which is well-defined since PS(1), PS(2), ... has a limit. Suppose that PS(i+1)(c_i) = n for each i >= 0, that is, c_i is the index of n after i steps. In the (i+1)-th step, each group of PS(i+1) contains i+2 elements, and every element is moved if and only if it has index at least i+3. We obtain that k = #{i >= 0 : c_i >= i+3}.
Note that if c_i <= i+2 for some i, then from the (i+1)-th step on, each group contains at least i+2 numbers so the first i+2 numbers in the sequence remain fixed, which means that n = PS(i+1)(c_i) = PS(i+2)(c_i) = ... = a(c_i), so c_i = c_{i+1} = ... = b(n). This shows that {i >= 0 : c_i >= i+3} = {0, 1, ..., k-1}, which implies that k is the smallest number such that c_k <= k+2. On one hand, since c_k <= k+2, we have c_k = b(n), so b(n) <= k+2. On the other hand, from the (b(n)-1)-th step on, each group contains at least b(n) numbers so the first b(n) numbers in the sequence remain fixed, which means that PS(b(n)-1)(b(n)) = PS(b(n))(b(n)) = ... = a(b(n)) = n, so c_{b(n)-2} = b(n), and k <= b(n)-2. In conclusion, we have k = b(n)-2.
By the MathOverflow link, we have a(n) <= 2*n-2 for all n, where the equality holds if and only if a(n)+1 is prime. On the other hand, it is hard to get a lower bound for {a(n)}, so it is infeasible to calculate the inverse permutation of this sequence. (End)
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LINKS
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FORMULA
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EXAMPLE
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PS(2) begins with 1,2,4,3,6,5,8;
PS(3) begins with 1,2,4,6,5,3,7;
PS(4) begins with 1,2,4,6,3,7,10.
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PROG
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(PARI) get(v, iv) = if (iv > #v, 0, v[iv]);
rcp(nbn, nbp, startv, v) = {w = vector(nbn); for (k=1, nbn, if (k % nbp, jv = startv+k, jv = startv+k-nbp); w[k] = get(v, jv); ); w; }
lista(nn) = {v = vector(nn, n, n); print1(v[1], ", ", v[2], ", "); startv = 3; for (n=3, nn, w = rcp(nn-n+1, n-1, startv, v); startv = 2; if (w[1] == 0, break); print1(w[1], ", "); v = w; ); } \\ Michel Marcus, Feb 19 2016]
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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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