

A098825


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, n1, ..., 1, 0.


5



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;
refs;
listen;
history;
text;
internal format)



OFFSET

0,9


COMMENTS

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, 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.


REFERENCES

P. Diaconis, Group Representations in Probability and Statistices, IMS, 1988; see p. 112.
Chris D. H. Evans, John Hughes and Julia Houston. Significancetesting 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. 385390.


LINKS

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Eric Weisstein's World of Mathematics, Partial Derangement


FORMULA

T(0, 0) = 1; d(0) = T(0, 0); for k = n, n1, ..., 1, T(n, k) = c(n, k) d(nk) where c(n, k) = n! / [(k!) (nk)! ]; T(n, 0) = n!  [ T(n, n) + T(n, n1) + ... + T(n, 1) ]; d(n) = T(n, 0).
T(n,k) = A008290(n,nk).  Vladeta Jovovic, Sep 04 2006
Assuming a uniform distribution on S_n, the mean Hamming distance is n1 and the variance is 1 (Diaconis, p. 117).  N. J. A. Sloane, Mar 02 2007


EXAMPLE

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)


MATHEMATICA

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}]] (* Robert G. Wilson v, Nov 04 2004 *)


PROG

(Haskell)
a098825 n k = a098825_tabl !! n !! k
a098825_row n = a098825_tabl !! n
a098825_tabl = map (zipWith (*) a000166_list) a007318_tabl
 Reinhard Zumkeller, Dec 16 2013


CROSSREFS

Cf. A000166, A007318.
Sequence in context: A194763 A194741 A194753 * A111460 A035327 A004444
Adjacent sequences: A098822 A098823 A098824 * A098826 A098827 A098828


KEYWORD

nonn,tabl


AUTHOR

Gerald P. Del Fiacco, Nov 02 2004


EXTENSIONS

More terms from Robert G. Wilson v, Nov 04 2004


STATUS

approved



