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A101387
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Quet transform of A002260.
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3
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1, 1, 2, 1, 3, 1, 5, 1, 1, 7, 1, 2, 1, 9, 1, 3, 1, 12, 1, 1, 4, 1, 15, 1, 2, 1, 5, 1, 18, 1, 3, 1, 7, 1, 1, 21, 1, 4, 1, 9, 1, 2, 1, 24, 1, 5, 1, 11, 1, 3, 1, 28, 1, 1, 6, 1, 13, 1, 4, 1, 32, 1, 2, 1, 7, 1, 15, 1, 5, 1, 36, 1, 3, 1, 9, 1, 1, 17, 1, 6, 1, 40, 1, 4, 1, 11, 1, 2, 1, 19, 1, 7, 1, 44, 1, 5, 1
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OFFSET
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1,3
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COMMENTS
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The Quet transform converts any sequence of positive integers containing an infinite number of 1's into another sequence of positive integers containing an infinite number of 1's.
Start with a sequence, {a(k)}, of only positive integers and an infinite number of 1's. Example: 1,1,2,1,2,3,1,2,3,4,1,... (A002260).
Form the sequence {b(k)} (which is a permutation of the positive integers), given by b(k) = the a(k)th smallest positive integer not yet in the sequence b, with b(1)=a(1).
In the example b is 1,2,4,3,6,8,5,9,11,13,7,12,15,... (A065562).
Let {c(k)} be the inverse of {b(k)}. In the example c = 1,2,4,3,7,5,11,6,8,16,9,12... (A065579).
Form the final sequence {d(k)}, where each d(k) is such that c(k) = the d(k)th smallest positive integer not yet in the sequence c, with d(1)=c(1).
In the example d is 1,1,2,1,3,1,5,1,1,7,1,2,1,9,1,3,1,12,1,1,4,1,15,... (the current sequence).
A more formal description of the Quet transform is as follows.
Let N denote the positive integers. For any permutation p: N -> N, let T(p): N -> N be given by T(p)(n) = # of elements in {m in N | m >= n AND p(m) <= p(n)}. Observe that T is a bijection from the set of permutations N -> N onto the set of sequences N -> N that contain infinitely many 1's.
Now suppose f: N -> N contains infinitely many 1's; then its Quet transform Q(f): N -> N is T^(-1)[(T(f))^(-1)], which also contains infinitely many 1's. Q is self-inverse; f and Q(f) correspond via T to a permutation and its inverse.
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LINKS
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PROG
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(PARI)
\\ PARI code to compute the Quet transform. Put the first n terms of the sequence
\\ into a vector v; then Q(v) returns the transformed sequence. The output is a
\\ vector, containing as many terms as can be computed from the given data.
TInverse(v) = local(l, w, used, start, x); l = length(v); w = vector(l); used = vector(l); start = 1; for (i = 1, l, while (start <= l && used[start], start++); x = start; for (j = 2, v[i], x++; while (x <= l && used[x], x++)); if (x > l, return (vector(i - 1, k, w[k])), w[i] = x; used[x] = 1)); w;
PInverse(v) = local(l, w); l = length(v); w = vector(l); for (i = 1, l, if (v[i] <= l, w[v[i]] = i)); w;
T(v) = local(l, w, c); l = length(v); w = vector(l); for (n = 1, l, if (v[n], c = 0; for (m = 1, n - 1, if (v[m] < v[n], c++)); w[n] = v[n] - c, return (vector(n - 1, i, w[i])))); w;
Q(v) = T(PInverse(TInverse(v)));
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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