Talk:Derangement

d(n) also satisfies the recurrence: d(n) = n*d(n-1) + (-1)^n.
see: http://mathworld.wolfram.com/Derangement.html
I've seen the relation proved by inclusion-exclusion.
DonkeyKong the mathematician (in training) 08:19, 17 July 2006 (UTC)

General formula

The page about rencontres numbers gives a general formula for the derangements as the closest integer to $n! \over e$. Is this formula valid for all n? I checked with the first points (ok), and it's obviously valid in the limit. Is there a proof? A corollary of this formula would be a proof that e is irrational. Albmont 13:24, 14 November 2006 (UTC)

It is the closest for all natural numbers except n=0. You can always get the correct integer for d_n by rounding up at even n and rounding down at odd n. JRSpriggs 05:10, 23 December 2006 (UTC)

Probably the formula $!n=\left\lfloor {\frac {n!}{e}}+{\frac {1}{2}}\right\rfloor$ should somehow state this, that it doesn't work for n=0?--Thomasda (talk) 15:06, 28 October 2010 (UTC)

Yeah, of course, I added it to the article. Paul Breeuwsma (talk) 15:25, 5 December 2010 (UTC)

START Zlajos 17 jun 2007

1.PART

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]]

The proof is an application of the inclusion-exclusion principle. I'm a bit surprised that this page doesn't say that. Michael Hardy 16:44, 19 June 2007 (UTC)

The details are on the page for Random Permutation Statistics], Maybe we should link there? -Zahlentheorie 09:45, 20 June 2007 (UTC)

1.table

fixed point: character numbers:	free or "0"	1	2	3	4	5	6	7	8	9	10	11	12
"0"	1
1	0	1
11	1	0	1
111	2	3	0	1
1111	9	8	6	0	1
11111	44	45	20	10	0	1
111111	265	264	135	40	15	0	1
1111111	1854	1855	924	315	70	21	0	1

• If all character twice: example: AABBCC....
• A059056 Penrice Christmas gift numbers, Card-matching numbers (Dinner-Diner matching numbers). [[2]]

COMMENT: Analogous to A008290. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 10 2005

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

fixed point: character numbers:	free or "0"	1	2	3	4	5	6	7	8	9	10	11	12
"0"	1
2	0	0	1
22	1	0	4	0	1
222	10	24	27	16	12	0	1
2222	297	672	736	480	246	64	24	0	1
22222	13756	30480	32365	21760	10300	3568	970	160	40	0	1
222222	925705	2016480	2116836	1418720	677655	243360	67920	14688	2655	320	60	0	1
2222222	85394646	183749160	191384599	128058000	61585776	22558928	6506955	1507392	284550	43848	5901	560	84

If original or classic table: (1.table)

• "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.

column > free or 0 :

• 1, 0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961...
• 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.]]

then:

• 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.

• column > free or 0 :
• 1, 0, 1, 10, 297, 13756, 925705, 85394646,...
• A059072 Penrice Christmas gift numbers; card-matching numbers; dinner-diner matching numbers.[[3]]

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) )

• 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

• Question:
  • 2.table
    • column: 2,3,4,5,...
    • where is it :formula or generating function(?)
    • where is it :bibliography?

3.table

fixed point: character numbers:	free or "0"	1	2	3	4	5	6	7	8	9	10	11	12
"0"	1
3	0	0	0	1
33	1	0	9	0	9	0	1
333	56	216	378	435	324	189	54	27	0	1
3333	13833	49464	84510	90944	69039	38448	16476	5184	1431	216	54	0	1
33333	6699824	23123880	38358540	40563765	30573900	17399178	7723640	2729295	776520	180100	33372	5355	540
333333	5691917785	19180338840	31234760055	32659846104	24571261710	14125889160	6433608330	2375679240	722303568	182701480	38712600	6889320	1035330
3333333	7785547001784	25791442770240	etc

If original or classic table: (1.table)

• "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.
  • column > free or 0 :
  • 1, 0, 1, 2, 9, 44, 265, 1854, 14833, 133496, 1334961...
• 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.]]

then:

• 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.
• column > free or 0 :
  • 1, 0, 1, 56, 13833, 6699824, 5691917785, 7785547001784,
• A059073 Card-matching numbers (Dinner-Diner matching numbers).

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) [[4]]

• 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. [[5]]

• 3.column ( "1" -fixed point)

111 :3

222 :24

333 :216

444 :1824

555 :15150

etc... A000279 Card matching. [[6]] 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. [[7]]

• 5.column ( "3" fixed point)

111 :1

222 :16

333 :435

444 :7136

555 :99350

etc... A000489 Card matching. [[8]]

• 3.table
  • column: 2,3,4,5,...
  • where is it :formula or generating function(?)
  • where is it :bibliography?

continued:

• 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 !!

Zlajos

19. jun. 2007.

• copy:[[9]]

Zlajos 21. jun. 2007. Extension: If all character twice : example: AABBCC, which has 2 A, 2 B's, and 2 C's, is

${6 \choose 2,2,2}={\frac {6!}{2!\,2!\,2!}}=90$

Compare the all distinct anagram for AABBCC to CCBBAA (90) one after the other :template (or schema)

AAAAAA or 6 0 0 equal, (identical): BBBBBB and CCCCCC

AAAAAB or 5 1 0 equal, (identical): BBBBBC and CCCCCA etc.

AAAABB or 4 2 0 equal, (identical): AAAACC and BBBBAA etc.

AAAABC or 4 1 1 equal, (identical): CCCCAB and BBBBAC etc.

AAABBB or 3 3 0 equal, (identical): AAACCC and BBBCCC etc.

AABBCC or 2 2 2

AAABBC or 3 2 1 equal, (identical): BBBCCA and CCCAAB etc.

4.table

fixed point: character numbers:	free or "0"	1	2	3	4	5	6	sum
6 0 0 or AAAAAA	0	0	90	0	0	0	0	90
5 1 0 or AAAAAB	0	30	30	30	0	0	0	90
4 2 0 or AAAABB	6	24	30	24	6	0	0	90
4 1 1 or AAAABC	6	24	36	12	12	0	0	90
3 3 0 or AAABBB	9	18	36	18	9	0	0	90
2 2 2 or AABBCC	10	24	27	16	12	0	1	90
3 2 1 or AAABBC	12	27	33	15	3	0	0	90

Extension: If all character thrice : example: AAABBBCCC, which has 3 A, 3 B's, and 3 C's, is

${9 \choose 3,3,3}={\frac {9!}{3!\,3!\,3!}}=1680$

Compare the all distinct anagram for AAABBBCCC to CCCBBBAAA (1680) one after the other :template (or schema)

AAAAAAAAA or 9 0 0 equal, (identical): BBBBBBBBB and CCCCCCCCC

AAAAAAAAB or 8 1 0 equal, (identical): BBBBBBBBC and CCCCCCCCA etc.

AAAAAAABB or 7 2 0 equal, (identical): AAAAAAACC and BBBBBBBAA etc.

AAAAAAABC or 7 1 1 equal, (identical): CCCCCCCAB and BBBBBBBAC etc.

AAAAAABBB or 6 3 0 equal, (identical): AAAAAACCC and BBBBBBCCC etc.

AAAAAABBC or 6 2 1 equal, (identical): AAAAAACCB and BBBBBBCCA etc. .................... AAABBBCCC or 3 3 3 etc...

5.table

fixed point: character numbers:	free or "0"	1	2	3	4	5	6	7	8	9			sum
9 0 0 or AAAAAAAAA	0	0	0	1680	0	0	0	0	0	0			1680
8 1 0 or AAAAAAAAB	0	0	560	560	560	0	0	0	0	0			1680
7 2 0 or AAAAAAABB	0	140	420	560	420	140	0	0	0	0			1680
7 1 1 or AAAAAAABC	0	140	420	630	280	210	0	0	0	0			1680
6 3 0 or AAAAAABBB	20	180	360	560	360	180	20	0	0	0			1680
6 2 1 or AAAAAABBC	20	180	420	480	380	140	60	0	0	0			1680
5 4 0 or AAAAABBBB	40	160	400	480	400	160	40	0	0	0			1680
5 3 1 or AAAAABBBC	40	190	400	460	360	160	60	10	0	0			1680
5 2 2 or AAAAABBCC	40	200	400	460	320	200	40	20	0	0			1680
4 4 1 or AAAABBBBC	48	192	384	480	320	192	48	16	0	0			1680
4 3 2 or AAAABBBCC	52	208	388	436	340	168	72	12	4	0			1680
3 3 3 or AAABBBCCC	56	216	378	435	324	189	54	27	0	1			1680

...4. table, 5.table sum: 90, 1680, etc.:A006480 De Bruijn's s(3,n): (3n)!/(n!)^3. [[10]]

continued! Zlajos 28. jun. 2007.

Compare the all distinct anagram for AAAAAABBBBBBB to BBBBBBAAAAAA (924) one after the other :template (or schema)

one after the other :template (or schema)

AAAAAAAAAAAA or 12 0

AAAAAAAAAAAB or 11 1

AAAAAAAAAABB or 10 2

....................

....................

BBBBBBBBBBAA or 2 10

....................

BBBBBBBBBBBB or 0 12

analogous or similar: A129352 [[11]]

MAPLE:with(combinat):T:=(n,i)->binomial(i,n)*binomial(12-i,6-n): for n from 0 to 6 do seq(T(n, i), i=0+n..12-6+n) od; #Warning, new definition for Chi

924, 462, 210, 84, 28, 7, 1

462, 504, 378, 224, 105, 36, 7

210, 378, 420, 350, 225, 105, 28

84, 224, 350, 400, 350, 224, 84

28, 105, 225, 350, 420, 378, 210

7, 36, 105, 224, 378, 504, 462

1, 7, 28, 84, 210, 462, 924

If this is table rotated right by Pi/4. then equal 6.table

6.table

fixed point: character numbers:	"0"	1	2	3	4	5	6	7	8	9	10	11	12	sum
12 0 or AAAAAAAAAAAA							924							924
11 1 or AAAAAAAAAAAB						462		462						924
10 2 or AAAAAAAAAABB					210		504		210					924
9 3 or AAAAAAAAABBB				84		378		378		84				924
8 4 or AAAAAAAABBBB			28		224		420		224		28			924
7 5 or AAAAAAABBBBB		7		105		350		350		105		7		924
6 6 or AAAAAABBBBBB	1		36		225		400		225		36		1	924
5 7 or AAAAABBBBBBB		7		105		350		350		105		7		924
4 8 or AAAABBBBBBBB			28		224		420		224		28			924
3 9 or AAABBBBBBBBB				84		378		378		84				924
2 10 or AABBBBBBBBBB					210		504		210					924
1 11 or ABBBBBBBBBBB						462		462						924
0 12 or BBBBBBBBBBBB							924							924

7.table

fixed point: character numbers:	"0"	1	2	3	4	5	6	7	8	9	10	11	12	sum
12 0 or AAAAAAAAAAAA							C(0,0)*C(12,6)							924
11 1 or AAAAAAAAAAAB						C(1,0)*C(11,6)		C(1,1)*C(11,5)						924
10 2 or AAAAAAAAAABB					C(2,0)*C(10,6)		C(2,1)*C(10,5		C(2,2)*C(10,4)					924
9 3 or AAAAAAAAABBB				C(3,0)*C(9,6)		C(3,1)*C(9,5)		C(3,2)*C(9,4)		C(3,3)*C(9,3)				924
8 4 or AAAAAAAABBBB			C(4,0)*C(8,6)		C(4,1)*C(8,5)		C(4,2)*C(8,4)		C(4,3)*C(8,3)		C(4,4)*C(8,2)			924
7 5 or AAAAAAABBBBB		C(5,0)*C(7,6)		C(5,1)*C(7,5)		C(5,2)*C(7,4)		C(5,3)*C(7,3)		C(5,4)*C(7,2)		C(5,5)*C(7,1)		924
6 6 or AAAAAABBBBBB	C(6,0)*C(6,6)		C(6,1)*C(6,5)		C(6,2)*C(6,4)		C(6,3)*C(6,3)		C(6,4)*C(6,2)		C(6,5)*C(6,1)		C(6,6)*C(6,0)	924
5 7 or AAAAABBBBBBB		C(7,1)*C(5,5)		C(7,2)*C(5,4)		C(7,3)*C(5,3)		C(7,4)*C(5,2)		C(7,5)*C(5,1)		C(7,6)*C(5,0)		924
4 8 or AAAABBBBBBBB			C(8,2)*C(4,4)		C(8,3)*C(4,3)		C(8,4)*C(4,2)		C(8,5)*C(4,1)		C(8,6)*C(4,0)			924
3 9 or AAABBBBBBBBB				C(9,3)*C(3,3)		C(9,4)*C(3,2)		C(9,5)*C(3,1)		C(9,6)*C(3,0)				924
2 10 or AABBBBBBBBBB					C(10,4)*C(2,2)		C(10,5)*C(2,1)		C(10,6)*C(2,0)					924
1 11 or ABBBBBBBBBBB						C(11,5)*C(1,1)		C(11,6)*C(1,0)						924
0 12 or BBBBBBBBBBBB							C(12,6)*C(0,0)							924

8.table

PASCAL TRIANGLE item, (portion)

fixed point: character numbers:	"0"	1	2	3	4	5	6	7	8	9	10	11	12
............................							C(0,0)*
						C(1,0)*		C(1,1)*
					C(2,0)*		C(2,1)*		C(2,2)*
				C(3,0)*		C(3,1)*		C(3,2)*		C(3,3)*
			C(4,0)*		C(4,1)*		C(4,2)*		C(4,3)*		C(4,4)*
		C(5,0)*		C(5,1)*		C(5,2)*		C(5,3)*		C(5,4)*		C(5,5)*
	C(6,0)*		C(6,1)*		C(6,2)*		C(6,3)* centre		C(6,4)*		C(6,5)*		C(6,6)*	..................

9.table

.............................

*C(6,6)

*C(6,5)

*C(6,4)

*C(6,3) centre

*C(6,2)

*C(6,1)

*C(6,0)

.................

.

*C(5,5)

*C(5,4)

*C(5,3)

*C(5,2)

*C(5,1)

*C(5,0)

.

*C(4,4)

*C(4,3)

*C(4,2)

*C(4,1)

*C(4,0)

.

*C(3,3)

*C(3,2)

*C(3,1)

*'C(3,0)'

.

*C(2,2)

*C(2,1)

*C(2,0)

.

*C(1,1)

*C(1,0)

.

*C(0,0)

continued! Zlajos 04. jul. 2007.

2. PART

Maple list:

for n from 0 to 0 do seq(binomial(i,n)*binomial(2-i,0-n), i=0+n..2-0+n ); od;#

for n from 0 to 1 do seq(binomial(i,n)*binomial(2-i,1-n), i=0+n..1-0+n ); od;#

for n from 0 to 2 do seq(binomial(i,n)*binomial(4-i,2-n), i=0+n..4-2+n ); od;#

for n from 0 to 3 do seq(binomial(i,n)*binomial(6-i,3-n), i=0+n..6-3+n ); od;

for n from 0 to 4 do seq(binomial(i,n)*binomial(8-i,4-n), i=0+n..8-4+n ); od;

for n from 0 to 5 do seq(binomial(i,n)*binomial(10-i,5-n), i=0+n..10-5+n );od

for n from 0 to 6 do seq(binomial(i,n)*binomial(12-i,6-n), i=0+n..12-6+n ); od;#

for n from 0 to 7 do seq(binomial(i,n)*binomial(14-i,7-n), i=0+n..14-7+n ); od;#

for n from 0 to 8 do seq(binomial(i,n)*binomial(16-i,8-n), i=0+n..16-8+n ); od;#

for n from 0 to 9 do seq(binomial(i,n)*binomial(18-i,9-n), i=0+n..18-9+n ); od;#

for n from 0 to 10 do seq(binomial(i,n)*binomial(20-i,10-n), i=0+n..20-10+n ); od;#

To simplify table (simple table): for 1 to 8

• • • 0.

1, 1, 1

• • • 1.

2, 1

1, 2

• • • 2.

6, 3, 1

3, 4, 3

1, 3, 6

• • • 3.

20, 10, 4, 1

10, 12, 9, 4

4, 9, 12, 10

1, 4, 10, 20

• • • 4.

70, 35, 15, 5, 1

35, 40, 30, 16, 5

15, 30, 36, 30, 15

5, 16, 30, 40, 35

1, 5, 15, 35, 70

• • • 5.

252, 126, 56, 21, 6, 1

126, 140, 105, 60, 25, 6

56, 105, 120, 100, 60, 21

21, 60, 100, 120, 105, 56

6, 25, 60, 105, 140, 126

1, 6, 21, 56, 126, 252

• • • 6.

924, 462, 210, 84, 28, 7, 1

462, 504, 378, 224, 105, 36, 7

210, 378, 420, 350, 225, 105, 28

84, 224, 350, 400, 350, 224, 84

28, 105, 225, 350, 420, 378, 210

7, 36, 105, 224, 378, 504, 462

1, 7, 28, 84, 210, 462, 924

• • • 7.

3432, 1716, 792, 330, 120, 36, 8, 1

1716, 1848, 1386, 840, 420, 168, 49, 8

792, 1386, 1512, 1260, 840, 441, 168, 36

330, 840, 1260, 1400, 1225, 840, 420, 120

120, 420, 840, 1225, 1400, 1260, 840, 330

36, 168, 441, 840, 1260, 1512, 1386, 792

8, 49, 168, 420, 840, 1386, 1848, 1716

1, 8, 36, 120, 330, 792, 1716, 3432

• • • 8.

12870, 6435, 3003, 1287, 495, 165, 45, 9, 1

6435, 6864, 5148, 3168, 1650, 720, 252, 64, 9

3003, 5148, 5544, 4620, 3150, 1764, 784, 252, 45

1287, 3168, 4620, 5040, 4410, 3136, 1764, 720, 165

495, 1650, 3150, 4410, 4900, 4410, 3150, 1650, 495

165, 720, 1764, 3136, 4410, 5040, 4620, 3168, 1287

45, 252, 784, 1764, 3150, 4620, 5544, 5148, 3003

9, 64, 252, 720, 1650, 3168, 5148, 6864, 6435

1, 9, 45, 165, 495, 1287, 3003, 6435, 12870

etc...

3. PART

all 1.rows 1. numbers (and mirror)

1, 2, 6, 20, 70, 252, 924, 3432, 12870, etc...

Central binomial coefficients: C(2n,n) = (2n)!/(n!)^2.

A000984[[12]]

all 1.rows 2. numbers (and mirror)

1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, etc...

C(2n+1, n+1)

A001700 [[13]]

all 1.rows 3. numbers (and mirror)

1, 4, 15, 56, 210, 792, 3003, 11440, 43758, 167960, etc...

• Binomial coefficients C(2n,n-1).

A001791 [[14]]

all 1.rows 4. numbers (and mirror)

1, 5, 21, 84, 330, 1287, 5005, 19448, 75582, etc...

• Binomial coefficient binomial(2n+1,n-1).

A002054 [[15]]

all 1.rows 5. numbers (and mirror)

1, 7, 36, 165, 715, 3003, 12376, 50388, 203490, 817190, etc...

• Binomial coefficients C(2n+1,n-2).

A003516 [[16]]

all 1.rows 6. numbers (and mirror)

1, 7, 36, 165, 715, 3003, 12376, 50388, 203490, etc...

• Binomial coefficients C(2n+1,n-2).

A003516[[17]]

all 1.rows 7. numbers (and mirror)

1, 8, 45, 220, 1001, 4368, 18564, 77520, 319770, etc...

• Binomial coefficients C(2n,n-3).

A002696 [[18]]

all 2.rows 1. numbers (and mirror)equal all 1.rows 2. numbers

all 2.rows 2. numbers (and mirror)

2, 4, 12, 40, 140, 504, 1848, 6864, 25740, 97240, etc...

Twice central binomial coefficients

A028329[[19]]

all 2.rows 3. numbers (and mirror)

3, 9, 30, 105, 378, 1386, 5148, 19305, 72930, 277134, 1058148,etc...

3*C(2*n-1,n).

A003409 [[20]]

all 3.rows 3. numbers (and mirror)

6, 12, 36, 120, 420, 1512, 5544 etc...

A067804 formatted as a square array:3.rows [[21]]

all 4.rows 4. numbers (and mirror)

20, 40, 120, 400, 1400, 5040, etc...

A067804 formatted as a square array:4.rows [[22]]

all 5.rows 5. numbers (and mirror)

70, 140, 420, 1400, 4900,etc...

A067804 formatted as a square array:5.rows [[23]]

etc...

etc...

A067804 formatted as a square array:

1 2 6 20 70 252 924 3432 12870

2 4 12 40 140 504 1848 6864

6 12 36 120 420 1512 5544

20 40 120 400 1400 5040

70 140 420 1400 4900

252 504 1512 5040

924 1848 5544

3432 6864

12870

...................................................

all diagonal left to right and bottom to top

Square the entries of Pascal's triangle.

A008459 [[24]]

all 2.table "center" 1, 4, 36, 400, 4900, 63504, 853776, etc...

• Binomial(2n,n)^2.

A002894 [[25]]

Everything to correlate everything....

I am search: bibliography (internet), proof and etc...

continued! Zlajos 05. jul. 2007.

Not speak English

Sorry, not speak english. All corrections thanks! Zerinvary Lajos, Hungary Zlajos OEIS>>zerinvarylajos or zerinvary >>e-mail — Preceding unsigned comment added by 78.92.185.226 (talk) 06:48, 16 April 2009 (UTC)

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

OEIS

1. A113899 >>[26]
2. A129352 >> [27]
3. A129536 >> [28]
4. Demo>>...mirror of central and multiply of diagonal...[29] (Pascal háromszög tükrözése és szorzás. Minta.)

—Preceding unsigned comment added by Zlajos (talk • contribs) 07:03, 16 April 2009 (UTC)

Applications

Would it be possible to add a section on why this is useful, what are the applications of this math? - Thanks. — Preceding unsigned comment added by 62.60.15.67 (talk) 08:53, 28 August 2012 (UTC)

Dual Dearrangement

Suppose we have a list A of 2n distinct elements {a_1i, a_2i; i=1, 2, ..., n}. A dual dearrangement is a permutation of A in which no element a_1i or a_2i is at any positions 2_i and 2_i-1, and no two elements a_1i and a_2i occupy the positions 2j and 2j-1, for any j, in any order.

The method used to derive derangement problem doesn't seem to work for the dual derangement. Any thought?

Mileszhou (talk) 08:12, 20 March 2014 (UTC) Miles Zhou

Derangement product proof

Does anyone have a proof that the product of all deranged permutations equals the identity?

Darcourse (talk) 15:26, 3 March 2019 (UTC)

Doesn't that follow immediately by symmetry? —David Eppstein (talk) 17:05, 3 March 2019 (UTC)

I think if you assume the product of all permutations is I, and the conjecture is true for all derangements of permutations < k, then by induction on the fixed point permutations of k (all equal I over a particular set of fixed points) then as all perms=fixed point perms + derangements, derangements = I.

I added an answer on my talk. My suggestion that it follows from symmetry assumes that you have a symmetric definition for "the product". In what order are you multiplying them? Also, it's not even true for $n=2$. —David Eppstein (talk) 21:03, 3 March 2019 (UTC)

No matter how you define the product, you will not get the identity when there is an odd number of odd derangements. This appears to happen whenever n = 2 (mod 4); see OEIS:A000387 and note the pattern of odd numbers in it. —David Eppstein (talk) 21:26, 3 March 2019 (UTC)