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A332010
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Irregular triangle of denominators of the average value of the first letter over all derangements of {1, 2, ..., n} with k descents.
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1
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1, 2, 4, 1, 1, 8, 3, 4, 16, 104, 40, 24, 1, 32, 392, 896, 480, 54, 64, 152, 308, 1496, 63, 108, 1, 128, 2276, 30384, 14410, 4315, 3024, 14, 256, 7340, 153400, 235252, 24766, 180416, 4984, 448, 1, 512, 23172, 365520, 1713160, 5794944, 3739512, 881152, 63840, 453
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OFFSET
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2,2
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COMMENTS
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Even-indexed rows have length n - 1; odd-indexed rows have length n - 2.
Conjecture: T(n, 1) = 2^(n-2).
T(2n, 2n-1) = 1, since there is only one derangement of 2n letters with 2n-1 descents.
The analogous sequence for permutations is T'(n, k) = 1. That is, the expected value of the first letter of a permutation with k descents is an integer (namely k + 1).
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LINKS
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EXAMPLE
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Triangle begins:
1;
2;
4, 1, 1;
8, 3, 4;
16, 104, 40, 24, 1;
32, 392, 896, 480, 54;
64, 152, 308, 1496, 63, 108, 1;
128, 2276, 30384, 14410, 4315, 3024, 14.
T(4,1) = 4 because the derangements of four letters with one descent are
[2,3,4,1], [2,4,1,3], [3,4,1,2], and [4,1,2,3], and the expected value of the first letter is (2+2+3+4)/4 = 11/4, which has 4 as its denominator.
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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