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Search: id:A098825
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| A098825 |
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Triangle read by rows: T(n,k) = number of partial derangements, that is, the number of permutations of n distinct, ordered items in which exactly k of the items are in their natural ordered positions, for n >= 0, k = n, n-1, ..., 1, 0. |
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3
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| 1, 1, 0, 1, 0, 1, 1, 0, 3, 2, 1, 0, 6, 8, 9, 1, 0, 10, 20, 45, 44, 1, 0, 15, 40, 135, 264, 265, 1, 0, 21, 70, 315, 924, 1855, 1854, 1, 0, 28, 112, 630, 2464, 7420, 14832, 14833, 1, 0, 36, 168, 1134, 5544, 22260, 66744, 133497, 133496, 1, 0, 45, 240, 1890, 11088, 55650
(list; table; graph; listen)
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OFFSET
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0,9
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COMMENT
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In other words, T(n,k) is the number of permutations of n letters that are at Hammimg distance k from the identity permutation (Cf. Diaconis, p. 112). - N. J. A. Sloane (njas(AT)research.att.com), Mar 02 2007
The sequence d(n) = 1, 0, 1, 2, 9, 44, 265, 1854, 14833, ... (A000166) is the number of derangements, that is, the number of permutations of n distinct, ordered items in which none of the items is in its natural ordered position.
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REFERENCES
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P. Diaconis, Group Representations in Probability and Statistices, IMS, 1988; see p. 112.
Chris D. H. Evans, John Hughes and Julia Houston. Significance-testing the validity of idiographic methods: A little derangement goes a long way, British Journal of Mathematical and Statistical Psychology, 1 November 2002, Vol. 55, pp. 385-390.
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LINKS
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Eric Weisstein's World of Mathematics, Partial Derangement
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FORMULA
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T(0, 0) = 1; d(0) = T(0, 0); for k = n, n-1, ..., 1, T(n, k) = c(n, k) d(n-k) where c(n, k) = n! / [(k!) (n-k)! ]; T(n, 0) = n! - [ T(n, n) + T(n, n-1) + ... + T(n, 1) ]; d(n) = T(n, 0).
T(n,k) = A008290(n,n-k). - Vladeta Jovovic (vladeta(AT)eunet.rs), Sep 04 2006
Assuming a uniform distribution on S_n, the mean Hamming distance is n-1 and the variance is 1 (Diaconis, p. 117). - N. J. A. Sloane (njas(AT)research.att.com), Mar 02 2007
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EXAMPLE
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Assume d(0), d(1), d(2) are given. Then
T(3, 3) = c(3, 3) d(0) = (1) (1) = 1
T(3, 2) = c(3, 2) d(1) = (3) (0) = 0
T(3, 1) = c(3, 1) d(2) = (3) (1) = 3
T(3, 0) = 3! - [ 1 + 0 + 3 ] = 6 - 4 = 2
d(3) = T(3, 0)
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MATHEMATICA
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t[0, 0] = 1; t[n_, k_] := Binomial[n, k]*k!*Sum[(-1)^j/j!, {j, 0, k}]; Flatten[ Table[ t[n, k], {n, 0, 10}, {k, 0, n}]] (from Robert G. Wilson v Nov 04 2004)
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PROGRAM
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This program generates the number of partial derangements T(n, k) for n = 0, 1, ..., 8 and k = n, n-1, ..., 0. Let T(0, 0) = 1 Let d(0) = T(0, 0) For n = 1 to 8 Let PartialSum = 0 For k = n to 1 Step -1 Let c(n, k) = Factorial(n) / (Factorial(k) * Factorial(n-k)) Let T(n, k) = c(n, k) * d(n-k) Let PartialSum = PartialSum + T(n, k) Next k Let T(n, 0) = Factorial(n) - PartialSum Let d(n) = T(n, 0) Next n
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CROSSREFS
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Cf. A000166.
Sequence in context: A128317 A084269 A051427 this_sequence A111460 A035327 A004444
Adjacent sequences: A098822 A098823 A098824 this_sequence A098826 A098827 A098828
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KEYWORD
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nonn,tabl
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AUTHOR
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Gerald P. Del Fiacco (gerrydelfiacco(AT)hotmail.com), Nov 02 2004
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EXTENSIONS
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More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Nov 04 2004
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