Talk:Rencontres numbers

START Zlajos 17 jun 2007

Extension: If all character once : example: ABCDE......
 * A008290 Triangle T(n,k) of rencontres numbers (number of *permutations of n elements with k fixed points).[]

1.table
COMMENT: Analogous to A008290. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 10 2005
 * If all character twice: example: AABBCC....
 * A059056 Penrice Christmas gift numbers, Card-matching numbers (Dinner-Diner matching numbers). []

1, 0, 0, 1, 1, 0, 4, 0, 1, 10, 24, 27, 16, 12, 0, 1, 297, 672, 736, 480, 246, 64, 24, 0, 1, 13756, 30480, 32365, 21760, 10300, 3568, 970, 160, 40, 0, 1, 925705, 2016480, 2116836, 1418720, 677655, 243360, 67920, 14688, 2655, 320, 60, 0, 1 =2.table=

If original or classic table: (1.table)

=column > free or 0 := =1, 0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961...= then: =column > free or 0 := =1, 0, 1, 10, 297, 13756, 925705, 85394646,...= FORMULA: MAPLE p := (x, k)->k!^2*sum(x^j/((k-j)!^2*j!), j=0..k); R := (x, n, k)->p(x, k)^n; f := (t, n, k)->sum(coeff(R(x, n, k), x, j)*(t-1)^j*(n*k-j)!, j=0..n*k);seq(f(0, n, 2)/2!^n, n=0..18); (AUTHOR Barbara Haas Margolius (margolius(AT)math.csuohio.edu) )
 * "0" (table sign: "0")then 1 derangements,
 * "A" (table sign: 1)then 0 derangements,
 * "AB" (table sign: 11)then 1 derangements,
 * "ABC" (table sign: 111)then 2 derangements,
 * "ABCD" (table sign: 1111)then 9 derangements, etc.
 * 00166 Subfactorial or rencontres numbers, or derangements: number of permutations of n elements with no fixed points.00166 	 	 Subfactorial or rencontres numbers, or derangements: number of permutations of n elements with no fixed points.
 * analogous (2.table)
 * "0"      (table sign: "0")then 1 derangements,
 * AA       (table sign: 2)then 0 derangements,
 * AABB     (table sign: 22)then 1 derangements,
 * AABBCC   (table sign: 222)then 10 derangements,
 * AABBCCDD (table sign: 2222)then 297 derangements, etc.
 * A059072 Penrice Christmas gift numbers; card-matching numbers; dinner-diner matching numbers.[]


 * COMMENT Number of fixed-point-free permutations of n distinct letters (ABCD...), each of which appears twice. If there is only one letter of each type we get A000166. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 15 2006

where is it :bibliography?
=3.table=

If original or classic table: (1.table)

=column > free or 0 := =1, 0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961...= then: =column > free or 0 := =1, 0, 1, 56, 13833, 6699824, 5691917785, 7785547001784,= FORMULA: MAPLE p := (x, k)->k!^2*sum(x^j/((k-j)!^2*j!), j=0..k); R := (x, n, k)->p(x, k)^n; f := (t, n, k)->sum(coeff(R(x, n, k), x, j)*(t-1)^j*(n*k-j)!, j=0..n*k); seq(f(0, n, 3)/3!^n, n=0..18); (AUTHOR Barbara Haas Margolius (margolius(AT)math.csuohio.edu) []
 * "0" (table sign: "0")then 1 derangements,
 * "A" (table sign: 1)then 0 derangements,
 * "AB" (table sign: 11)then 1 derangements,
 * "ABC" (table sign: 111)then 2 derangements,
 * "ABCD" (table sign: 1111)then 9 derangements, etc.
 * 00166 Subfactorial or rencontres numbers, or derangements: number of permutations of n elements with no fixed points.00166 	 	 Subfactorial or rencontres numbers, or derangements: number of permutations of n elements with no fixed points.
 * analogous (3.table)
 * "0"      (table sign: "0")then 1 derangements,
 * AAA       (table sign: 3)then 0 derangements,
 * AAABBB     (table sign: 33)then 1 derangements,
 * AAABBBCCC   (table sign: 333)then 56 derangements,
 * AAABBBCCCDDD (table sign: 3333)then 13833 derangements, etc.
 * A059073 Card-matching numbers (Dinner-Diner matching numbers).


 * Number of fixed-point-free permutations of n distinct letters (ABCD...), each of which appears thrice. If there is only one letter of each type we get A000166. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 15 2006


 * 2.column (free or "0" -fixed point

" "    :1   111     :2

222    :10

333    :56

444    :346

555    :2252

etc... A000172 Franel number a(n) = Sum C(n,k)^3, k=0..n. []


 * 3.column ( "1" -fixed point)

111    :3

222    :24

333    :216

444    :1824

555    :15150

etc... A000279 Card matching. [] COMMENT

Number of permutations of 3 distinct letters (ABC) each with n copies such that one (1) fixed points. E.g. if AAAAABBBBBCCCCC n=3*5 letters permutations then one fixed points n5=15150 - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Feb 02 2006


 * 4.column ( "2" fixed point)

111    :0

222    :27

333    :378

444    :4536

555    :48600

etc... A000535 Card matching. []


 * 5.column ( "3" fixed point)

111    :1

222    :16

333    :435

444    :7136

555    :99350

etc... A000489 Card matching. []

Number parallelogram based on Pascal's triangle (and special mirror of central and multiply of diagonal)

 * OEIS


 * 1) A113899 >>
 * 2) A129352 >>
 * 3) A129536 >>
 * 4) Demo>>...mirror of central and multiply of diagonal... (Pascal háromszög tükrözése és szorzás. Minta.)

continued:
Zlajos 19. jun. 2007. Zlajos 28. jun. 2007. 16. apr. 2009.
 * charcters:quadruple, example:AAAA, AAAABBBB, AAAABBBBCCCC, AAAABBBBCCCCDDDD, etc...
 * table 1.column :4, 44, 444, 4444, 44444, etc...
 * charcters:quintuple, example:AAAAA, AAAAABBBBB, AAAAABBBBBCCCCC, etc...
 * table 1.column :5, 55, 555, 5555, 55555, etc...
 * a great number of connexion of interesting !!
 * copy:[]

Justification that expectation is 1
To show that for n ≥ 1, the expected number of fixed points is 1 : We'll number the permutations p = 1 to n! Now let X[p,m]=1 if in the p_th permutation, element m is fixed, when it is not fixed, X[p,m]=0 Now the expected number of fixed points is E_n[F] = sum_p_from_1_to_n! { sum_m_from_1_to_n { X[p,m] } } / n! => E_n[F] = sum_m_from_1_to_n  { sum_p_from_1_to_n! { X[p,m] } } / n! => E_n[F] = sum_m_from_1_to_n { (n-1)! } / n! => E_n[F] = n * (n-1)! / n! => E_n[F] = 1

(with thanks to FD) Pnelnik (talk) 08:42, 19 July 2009 (UTC)


 * Simpler argument: Let
 * $$ X_i = \begin{cases} 1 & \text{if }i\text{ is a fixed point}, \\

0 & \text{otherwise}. \end{cases} $$
 * Then the number of fixed points is
 * $$ X_1 + \cdots + X_n, \, $$
 * so the expected number of fixed points is
 * $$ E(X_1) + \cdots + E(X_n) = \frac1n + \cdots + \frac1n = 1. $$
 * Michael Hardy (talk) 15:01, 19 July 2009 (UTC)


 * Well, I certainly prefer your equation mark-up.
 * I'd argue that the two proofs are almost almost identical.
 * The same arguement presented slightly differently.
 * (I'm now going to make the effort to write the equations properly)
 * You use the fact that
 * $$E(\sum_{i=1} ^ {n}X_i) = \sum_{i=1}^ {n} E(X_i)$$
 * (eqn1)
 * Here the probability distribution is discrete, so taking the expectation is a sumation
 * So to prove eqn1, we just need to swap the order of summation.
 * In my version, I've just shown that explicitly
 * For any wikipedians wondering, I should point out that the work or particularly the results
 * are not original work. The results have probably been known for a couple of centuries.
 * Pnelnik (talk) 20:52, 19 July 2009 (UTC)


 * My version doesn't require knowing how many permutations a set has, given its cardinality. In that sense it is simpler. Michael Hardy (talk) 01:07, 20 July 2009 (UTC)

Table format
On the article page the first table is not very pretty. There is no horizontal bar under number 2,3,4,5,6,7 and the verticle bar goes down too far:

Perhaps one solution would be to put in blanks in those extra cells.

It is not ideal, but I think it looks a bit better. Pnelnik (talk) 23:23, 19 July 2009 (UTC)


 * I've now just noticed that the tables look fine using Opera and IE browsers.
 * The only problem is when they are viewed in Firefox (version 3.5.1)
 * Pnelnik (talk) 23:54, 19 July 2009 (UTC)

Formulae need clarification
- This is the coefficient operator. Definition can be found at http://en.wikipedia.org/wiki/Formal_power_series#Extracting_coefficients Heycarnut (talk) 08:37, 10 August 2013 (UTC)
 * Explain meaning of symbol z.


 * Replace approximation by lim specifying range of validity. — Preceding unsigned comment added by 80.219.149.220 (talk) 22:18, 5 March 2012 (UTC)