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A356257
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Irregular triangle: row n consists of the frequencies of positive distances between permutations P and reverse(P), as P ranges through the permutations of (1, 2, ..., n); see Comments.
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0
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1, 2, 4, 2, 8, 16, 24, 16, 32, 32, 16, 48, 192, 192, 288, 192, 144, 576, 576, 576, 576, 960, 576, 576, 288, 384, 2304, 4608, 7680, 9216, 6912, 9216, 1920, 1536, 9216, 9216, 16128, 18432, 29184, 26112, 36864, 32256, 41472, 23040, 39168, 32256, 18432, 18432
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OFFSET
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1,2
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COMMENTS
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For n >= 1, let P = (p(1),p(2),...,p(n)) and Q = (q(1),q(2),...,q(n)) be permutations of (1,2,...,n). The distance between P and Q is defined by |p(1)-q(1)| + |p(2)-q(2)| + ... + |p(n)-q(n)|. For fixed n >= 1, let m be the least distance that occurs and let M be the greatest. If n is odd, let S = (m, m+2, m+4, ..., M); if n > 2 is even, let S = (m, m+4, m+8, ..., M). Then S gives all the positive distances that occur, and the frequencies in row n of the array account for the distances in S. Four open questions about the numbers in row n follow. (1) How many are there? (2) What are the first and last? (3) What are the least and greatest? (4) What is the greatest common divisor?
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LINKS
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EXAMPLE
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First 8 rows:
1
2
4 2
8 16
24 16 32 32 16
48 192 192 288
192 144 576 576 576 576 960 576 576 288
384 2304 4608 7680 9216 6912 9216
For n=3, the 6 permutations and their reverses are represented by
123 132 213 231 212 321
321 231 312 132 213 123,
so the 6 distances are 4,2,2,2,2,4, whence row 3 accounts for four 2's and two 4's.
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MATHEMATICA
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p[n_] := p[n] = Permutations[Range[n]];
f[n_, k_] := f[n, k] = Abs[p[n][[k]] - Reverse[p[n][[k]]]]
c[n_, k_] := c[n, k] = Total[f[n, k]]
t[n_] := t[n] = Table[c[n, k], {k, 1, n!}]
z = 6; Table[t[n], {n, 1, z}]
u = Table[Count[t[n], k], {n, 1, z}, {k, Min[t[n]], Max[t[n]], 2}]
v[n_] := Select[u[[n]], # > 0 &]
w = Table[v[n], {n, 1, z}]
TableForm[w] (* 356257 array *)
Flatten[w] (* 356257 sequence *)
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CROSSREFS
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KEYWORD
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nonn,tabf,more
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AUTHOR
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STATUS
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approved
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