A047920 -id:A047920 - cp's OEIS frontend

A001565 3rd differences of factorial numbers.

2, 11, 64, 426, 3216, 27240, 256320, 2656080, 30078720, 369774720, 4906137600, 69894316800, 1064341555200, 17255074636800, 296754903244800, 5396772116736000, 103484118786048000, 2086818140639232000, 44150769074700288000, 977904962186600448000

Offset: 0

N. J. A. Sloane

From Emeric Deutsch, Sep 09 2010: (Start)
a(n) is the number of isolated entries in all permutations of [n+2]. An entry j of a permutation p is isolated if it is not preceded by j-1 and not followed by j+1. For example, the permutation 23178564 has 2 isolated entries: 1 and 4. a(1)=11 because in 123, 1'3'2', 2'1'3', 231', 3'12, and 3'2'1' we have a total of 11 isolated entries (they are marked).
a(n) = Sum_{k>=0} k*A180196(n+2,k). (End)

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..100
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
A. van Heemert, Cyclic permutations with sequences and related problems, J. Reine Angew. Math., 198 (1957), 56-72.
Index entries for sequences related to factorial numbers

Cf. A047920.

Cf. A180196.

List([0..20], n-> (n^3+3*n^2+5*n+2)*Factorial(n)); # G. C. Greubel, Apr 29 2019

[(n^3+3*n^2+5*n+2)*Factorial(n): n in [0..20]]; // G. C. Greubel, Apr 29 2019

Table[(n^3 +3*n^2 +5*n +2) n!, {n, 0, 20}] (* T. D. Noe, Aug 09 2012 *)
Differences[Range[0, 25]!, 3] (* Paolo Xausa, May 28 2025 *)

{a(n) = (n^3+3*n^2+5*n+2)*n!}; \\ G. C. Greubel, Apr 29 2019

[(n^3+3*n^2+5*n+2)*factorial(n) for n in (0..20)] # G. C. Greubel, Apr 29 2019

a(n) = (n^3 + 3*n^2 + 5*n + 2)*n!. - Mitch Harris, Jul 10 2008

E.g.f.: (2 + 3*x + x^3)/(1 - x)^4. - Ilya Gutkovskiy, Jan 20 2017

A033815 Number of standard permutations of [ a_1..a_n b_1..b_n ] (b_i is not immediately followed by a_i, for all i).

Original entry on oeis.org

1, 1, 14, 426, 24024, 2170680, 287250480, 52370755920, 12585067447680, 3854801333416320, 1465957162768492800, 677696237345719468800, 374281829360322587827200, 243388909697235614324812800, 184070135024053703140543027200, 160192129141963141211280644352000

Offset: 0

N. J. A. Sloane

Also turns up as the solution to Problem #18, p. 326 of Alan Tucker's Applied Combinatorics, 4th ed, Wiley NY 2002 [Tucker's `n' is the `2n' here]. - John L Leonard, Sep 15 2003
Number of acyclic orientations of the Turán graph T(2n,n). - Alois P. Heinz, Jan 13 2016
n-th term of the n-th forward differences of n!. - Alois P. Heinz, Feb 22 2019

R. P. Stanley, Enumerative Combinatorics I, Chap.2, Exercise 10, p. 89.

Reinhard Zumkeller, Table of n, a(n) for n = 0..200
Leo Chao, Paul DesJarlais and John L Leonard, A binomial identity, via derangements, Math. Gaz. 89 (2005), 268-270.
Ira Gessel, Enumerative applications of symmetric functions, Séminaire Lotharingien de Combinatoire, B17a (1987), 17 pp.

Main diagonal of array in A068106 and of A047920.
Cf. A000142, A002119, A116854, A267383.
Column k=2 of A372326.

a033815 n = a116854 (2 * n + 1) (n + 1)
-- Reinhard Zumkeller, Aug 31 2014

A033815 := proc(n) local i; add(binomial(n, i)*(-1)^i*(2*n - i)!, i = 0 .. n) end;
# second Maple program:
A:= proc(n, k) A(n, k):= `if`(k=0, n!, A(n+1, k-1) -A(n, k-1)) end:
a:= n-> A(n$2):
seq(a(n), n=0..23);  # Alois P. Heinz, Feb 22 2019

a[n_] := (2n)!*Hypergeometric1F1[-n, -2n, -1]; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Jun 13 2012, after Vladimir Reshetnikov *)

a(n) = A002119(n)*n!*(-1)^n.
D-finite with recurrence a(n) = 2n*(2n-1)*a(n-1) + n*(n-1)*a(n-2).
a(n) = Sum_{i=0..n} binomial(n, i)*(-1)^i*(2*n-i)!.
From John L Leonard, Sep 15 2003: (Start)
a(n) = Sum_{i=0..n} C(n, i)*(2n-i)!*Sum_{j=0..2n-i} (-1)^j/j!.
a(n) = n!*Sum_{i=0..n} C(n, i)*n!/(n-i)!*Sum_{j=0..n-i} (-1)^j*C(n-i, j)*(n-j)!/i!. (End)
a(n) = Sum_{k=0..n} binomial(n,k)*A000166(n+k). - Vladeta Jovovic, Sep 04 2006
a(n) = A116854(2*n+1,n+1). - Reinhard Zumkeller, Aug 31 2014
a(n) = A267383(2n,n). - Alois P. Heinz, Jan 13 2016
a(n) ~ sqrt(Pi) * 2^(2*n + 1) * n^(2*n + 1/2) / exp(2*n + 1/2). - Vaclav Kotesovec, Feb 18 2017
a(n) = n!*exp(-1/2)*((-1)^n * BesselI(n+1/2,1/2)*Pi^(1/2) + BesselK(n+1/2,1/2)/Pi^(1/2) ). - Mark van Hoeij, Jul 15 2022

A155521 Smallest fixed point summed over all non-derangement permutations of {1,2,...,n}.

Original entry on oeis.org

0, 1, 1, 7, 31, 191, 1331, 10655, 95887, 958879, 10547659, 126571919, 1645434935, 23036089103, 345541336531, 5528661384511, 93987243536671, 1691770383660095, 32143637289541787, 642872745790835759, 13500327661607550919

Offset: 0

Emeric Deutsch, Apr 21 2009

nonn

a(n) is also the number of permutations of {1,2,...,n,n+1} having at least 2 fixed points. Example: a(3)=7 because we have 1234, 1243, 1324, 1432, 2134, 4231, and 3214.

			a(3)=7 because the non-derangements of {1,2,3} are 123, 132, 213, 321 with smallest fixed points 1, 1, 3, 2.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Emeric Deutsch and S. Elizalde, The largest and the smallest fixed points of permutations, arXiv:0904.2792 [math.CO], 2009.

Cf. A000166, A047920.

a[0] := 0: for n to 25 do a[n] := (n+1)*a[n-1]+n*(-1)^(n+1) end do: seq(a[n], n = 0 .. 21);

CoefficientList[Series[(1-(1+x^2)*E^(-x))/(1-x)^2, {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Oct 20 2012 *)

a(n) = (n+1)*a(n-1) +n*(-1)^(n+1); a(0)=0.
E.g.f.: (1-(1+x^2)*exp(-x))/(1-x)^2.
a(n) = (n+1)!+(-1)^n-2(n+1)*d(n),
a(n) = (n+1)!-(n+1)*d(n)-d(n+1), where d(n)=A000166(n) are the derangement numbers.
a(n) ~ n!*n*(1-2/e). - Vaclav Kotesovec, Oct 20 2012
a(n) = Sum_{k=0..n-1} (k+1) * A047920(n-1,k). - Alois P. Heinz, Sep 01 2021
D-finite with recurrence a(n) +(-n+1)*a(n-1) +(-2*n+1)*a(n-2) +(-n+1)*a(n-3)=0. - R. J. Mathar, Jul 26 2022

A047922 Triangle of numbers a(n,k) = number of terms in n X n determinant with 2 adjacent diagonals of k and k-1 0's (0<=k<=n).

Original entry on oeis.org

1, 1, 0, 2, 1, 0, 6, 4, 1, 1, 24, 18, 8, 5, 3, 120, 96, 54, 34, 23, 16, 720, 600, 384, 258, 182, 131, 96, 5040, 4320, 3000, 2136, 1566, 1168, 883, 675, 40320, 35280, 25920, 19320, 14664, 11274, 8756, 6859, 5413, 362880, 322560, 246960, 190800, 149160, 117696, 93582, 74902, 60301, 48800

Offset: 0

N. J. A. Sloane

			Triangle starts:
  1;
  1, 0;
  2, 1, 0;
  6, 4, 1, 1;
  ...

Alois P. Heinz, Rows n = 0..140, flattened
J. D. H. Dickson, Discussion of two double series arising from the number of terms in determinants of certain forms, Proc. London Math. Soc., 10 (1879), 120-122.
J. D. H. Dickson, Discussion of two double series arising from the number of terms in determinants of certain forms, Proc. London Math. Soc., 10 (1879), 120-122. [Annotated scanned copy]

Columns give A000142, A001563, A002775, A002776. Cf. A047920.

a:= proc(n, k) option remember; `if`(k=0, n!, `if`(n=k,
      `if`(n<3, (n-1)*(n-2)/2, (n-1)*(a(n-1$2)+a(n-2$2))
      +a(n-3$2)), a(n, k+1) +2*a(n-1, k) +a(n-2, k-1)))
    end:
seq(seq(a(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Jun 24 2017

a[n_, n_] := (-1)^n*HypergeometricPFQ[{1, -n, n+1}, {1/2}, 1/4]; a[n_, k_] := a[n, k] = a[n, k+1] + 2*a[n-1, k] + a[n-2, k-1]; Table[a[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 24 2015 *)

Right diagonal is A000271, column k=0 is A000142; other entries given by a(n, k) = a(n, k+1) + 2a(n-1, k) + a(n-2, k-1).

More terms from Larry Reeves (larryr(AT)acm.org), Sep 29 2000

A060475 Triangular array formed from successive differences of factorial numbers, then with factorials removed.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 2, 3, 2, 1, 3, 7, 11, 9, 1, 4, 13, 32, 53, 44, 1, 5, 21, 71, 181, 309, 265, 1, 6, 31, 134, 465, 1214, 2119, 1854, 1, 7, 43, 227, 1001, 3539, 9403, 16687, 14833, 1, 8, 57, 356, 1909, 8544, 30637, 82508, 148329, 133496, 1, 9, 73, 527, 3333, 18089, 81901, 296967, 808393, 1468457, 1334961

Offset: 0

Henry Bottomley, Mar 16 2001

T(n,k) is also the number of partial bijections (of an n-element set) with a fixed domain of size k and without fixed points. Equivalently, T(n,k) is the number of partial derangements with a fixed domain of size k in the symmetric inverse semigroup (monoid), I sub n. - Abdullahi Umar, Sep 14 2008

			Triangle begins
  1,
  1,  0,
  1,  1,  1,
  1,  2,  3,  2,
  1,  3,  7, 11,  9,
  1,  4, 13, 32, 53, 44,
  ...

G. C. Greubel, Rows n=0..100 of triangle, flattened
A. Laradji, and A. Umar, Combinatorial results for the symmetric inverse semigroup, Semigroup Forum 75 (2007), 221-236. - _Abdullahi Umar_, Sep 14 2008
L. Takacs, The Problem of Coincidences, Archive for History of Exact Sciences, Volume 21, No. 3, Sept. 1980. pp 229-244, paragraph 10 (Catalan).

Columns include A000012, A001477, A002061.
Diagonals include A000166, A000255, A000153, A000261, A001909, A001910.
Main diagonal is abs of A002119.
Similar to A076731.
Row sums equal A003470. - Johannes W. Meijer, Jul 27 2011
Cf. A047920, A008290.

[[Factorial(k)*(&+[(-1)^j*Binomial(n-j, k-j)/Factorial(j): j in [0..k]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Mar 04 2019

A060475 := proc(n,k): k! * add(binomial(n-j,k-j)*(-1)^j/j!, j=0..k) end:
seq(seq(A060475(n,k), k=0..n), n=0..7); # Johannes W. Meijer, Jul 27 2011
T := (n,k) -> KummerU(-k, -n, -1):
seq(seq(simplify(T(n, k)), k = 0..n), n = 0..10); # Peter Luschny, Jul 07 2022

t[n_, k_] := k!*Sum[Binomial[n - j, k - j]*(-1)^j/j!, {j, 0, k}]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Robert G. Wilson v, Aug 08 2011 *)

{T(n,k) = k!*sum(j=0,k, (-1)^j*binomial(n-j, k-j)/j!)};
for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Mar 04 2019

[[factorial(k)*sum((-1)^j*binomial(n-j, k-j)/factorial(j) for j in (0..k)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Mar 04 2019

T(n,k) = A047920(n,k)/(n-k)! = (n-1)*T(n-1,k-1) + (k-1)*T(n-2,k-2) = (n-k+1)*T(n, k-1) - T(n-1,k-1).
From Abdullahi Umar, Sep 14 2008: (Start)
T(n,k) = k! * Sum_{j=0..k} C(n-j,k-j)*(-1)^j/j!.
C(n,k)*T(n,k) = A144089(n, k). (End)
T(n,k) = A076732(n+1,k+1)/(k+1). - Johannes W. Meijer, Jul 27 2011
E.g.f. as a square array: A(x,y) = exp(-x)/(1 - x - y) = (1 + y + y^2 + y^3 + ...) + (y + 2*y^2 + 3*y^3 + 4*y^4 + ...)*x + (1 + 3*y + 7*y^2 + 13*y^3 + ...)*x^2/2! + (2 + 11*y + 32*y^2 + 71*y^3 + ...)*x^3/3! + .... Observe that (1 - y)*A(x*(1 - y),y) = exp(x*(y - 1))/(1 - x) is the e.g.f. for A008290. - Peter Bala, Sep 25 2013
T(n, k) = KummerU(-k, -n, -1). - Peter Luschny, Jul 07 2022

A129867 Row sums of triangular array T: T(j,k) = k*(j-k)! for k < j, T(j,k) = 1 for k = j; 1 <= k <= j.

Original entry on oeis.org

1, 2, 5, 14, 47, 200, 1073, 6986, 53219, 462332, 4500245, 48454958, 571411271, 7321388384, 101249656697, 1502852293010, 23827244817323, 401839065437636, 7182224591785949, 135607710526966262, 2696935204638786575

Offset: 1

Paul Curtz, May 24 2007

nonn

T read by rows is in A130469.
First differences are 1, 3, 9, 33, 153, 873, 5913, ... (see A007489), second differences are 2, 6, 24, 120, 720, 5040, ... (see A000142 ).
First terms of the sequences of m-th differences are 1, 2, 4, 14, 64, ... (see A055790, A047920, A068106).
Antidiagonal sums are 1, 1, 3, 8, 29, 135, ... (see A130470) with first differences 0, 2, 5, 21, 106, ... (see A130471).
Equals the row sums of irregular triangle A182961. - Paul D. Hanna, Mar 05 2012

			First seven rows of T are
[   1 ]
[   1,   1 ]
[   2,   2,   1 ]
[   6,   4,   3,   1 ]
[  24,  12,   6,   4,   1 ]
[ 120,  48,  18,   8,   5,   1 ]
[ 720, 240,  72,  24,  10,   6,   1 ]

Vincenzo Librandi, Table of n, a(n) for n = 1..100

Cf. A130469, A007489, A000142, A055790, A047920, A068106, A130470, A130471.

Cf. A182961.

m:=21; [ &+([ k*Factorial(j-k): k in [1..j-1] ] cat [ 1 ]): j in [1..m] ]; // Klaus Brockhaus, May 28 2007

Edited and extended by Klaus Brockhaus, May 28 2007

A076732 Table T(n,k) giving number of ways of obtaining exactly one correct answer on an (n,k)-matching problem (1 <= k <= n).

Original entry on oeis.org

1, 1, 0, 1, 2, 3, 1, 4, 9, 8, 1, 6, 21, 44, 45, 1, 8, 39, 128, 265, 264, 1, 10, 63, 284, 905, 1854, 1855, 1, 12, 93, 536, 2325, 7284, 14833, 14832, 1, 14, 129, 908, 5005, 21234, 65821, 133496, 133497, 1, 16, 171, 1424, 9545, 51264, 214459, 660064, 1334961, 1334960

Offset: 1

Mohammad K. Azarian, Oct 28 2002

Hanson et al. define the (n,k)-matching problem in the following realistic way. A matching question on an exam has k questions with n possible answers to choose from, each question having a unique answer. If a student guesses the answers at random, using each answer at most once, what is the probability of obtaining r of the k correct answers?

The T(n,k) represent the number of ways of obtaining exactly one correct answer, i.e., r=1, given k questions and n possible answers, 1 <= k <= n.

			Triangle begins
  1;
  1,0;
  1,2,3;
  1,4,9,8;
  ...

D. Hanson, K. Seyffarth and J. H. Weston, Matchings, Derangements, Rencontres, Mathematics Magazine, Vol. 56, No. 4, September 1983.

Columns: A000012(n), 2*A001477(n-2), 3*A002061(n-2), 4*A094792(n-4), 5*A094793(n-5), 6*A094794(n-6), 7*A094795(n-7); A000240(n), A000166(n). - Johannes W. Meijer, Jul 27 2011

A076732:=proc(n,k): (k/(n-k)!)*A047920(n,k) end: A047920:=proc(n,k): add(((-1)^j)*binomial(k-1,j)*(n-1-j)!, j=0..k-1) end: seq(seq(A076732(n,k), k=1..n), n=1..10); # Johannes W. Meijer, Jul 27 2011

A000240[n_] := Subfactorial[n] - (-1)^n;
T[n_, k_] := T[n, k] = Switch[k, 1, 1, n, A000240[n], _, k*T[n-1, k-1] + T[n-1, k]];
Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 14 2023 *)

T(n,k) = F(n,k)*Sum{((-1)^j)*C(k-1, j)*(n-1-j)! (j=0 to k-1)}, where F(n,k) = k/(n-k)!, for 1 <= k <= n.
From Johannes W. Meijer, Jul 27 2011: (Start)
T(n,k) = k*T(n-1,k-1) + T(n-1,k) with T(n,1) = 1 and T(n,n) = A000240(n). [Hanson et al.]
T(n,k) = (n-1)*T(n-1,k-1) + (k-1)*T(n-2,k-2) + (1-k)*A076731(n-2,k-2) + A076731(n-1,k-1) with T(0,0) = T(n,0) = 0 and T(n,1) = 1. [Hanson et al.]
T(n,k) = k*A060475(n-1,k-1).
T(n,k) = (k/(n-k)!)*A047920(n-1,k-1).
Sum_{k=1..n} T(n,k) = A193463(n); row sums.
Sum_{k=1..n} T(n,k)/k = A003470(n-1). (End)

Edited and information added by Johannes W. Meijer, Jul 27 2011

A116853 Difference triangle of factorial numbers read by upward diagonals.

Original entry on oeis.org

1, 1, 2, 3, 4, 6, 11, 14, 18, 24, 53, 64, 78, 96, 120, 309, 362, 426, 504, 600, 720, 2119, 2428, 2790, 3216, 3720, 4320, 5040, 16687, 18806, 21234, 24024, 27240, 30960, 35280, 40320

Offset: 1

Gary W. Adamson, Feb 24 2006

This is a subsequence of Euler's difference table A068106 and of A047920 (in a different ordering), since 0! = 1 was left out here. - Georg Fischer, Mar 23 2019

			Starting with 1, 2, 6, 24, 120 ... we take the first difference row (A001563), second, third, etc. Reorient into a flush left format, getting:
[1]    1;
[2]    1,   2;
[3]    3,   4,   6;
[4]   11,  14,  18,  24;
[5]   53,  64,  78,  96, 120;
[6]  309, 362, 426, 504, 600, 720;
...

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened

Cf. A000142 (factorial numbers).
Cf. A000255 (first column and inverse binomial transform of A000142).
N-th forward differences of A000142: A001563 (1st), A001564 (2nd), A001565 (3rd), A001688 (4th), A001689 (5th).
Cf. A047920 (with 0!, different order), A068106 (with 0!), A180191 (row sums), A246606 (central terms).

a116853 n k = a116853_tabl !! (n-1) !! (k-1)
a116853_row n = a116853_tabl !! (n-1)
a116853_tabl = map reverse $ f (tail a000142_list) [] where
   f (u:us) vs = ws : f us ws where ws = scanl (-) u vs
-- Reinhard Zumkeller, Aug 31 2014

rows = 8;
rr = Range[rows]!;
dd = Table[Differences[rr, n], {n, 0, rows-1}];
T = Array[t, {rows, rows}];
Do[Thread[Evaluate[Diagonal[T, -k+1]] = dd[[k, ;;rows-k+1]]], {k, rows}];
Table[t[n, k], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 21 2019 *)

Take successive difference rows of factorial numbers n! starting with n=1. Reorient into a triangle format.

A136301 Frequency of occurrence for each possible "probability of derangement" for a Secret Santa drawing in which each person draws a name in sequence and the only person who does not draw someone else's name is the one who draws the final name.

Original entry on oeis.org

1, 1, 1, 1, 5, 2, 1, 1, 13, 6, 13, 2, 6, 2, 1, 1, 29, 14, 73, 6, 42, 18, 29, 2, 18, 8, 14, 2, 6, 2, 1, 1, 61, 30, 301, 14, 186, 86, 301, 6, 102, 48, 186, 18, 102, 42, 61, 2, 42, 20, 86, 8, 48, 20, 30, 2, 18, 8, 14, 2, 6, 2, 1, 1, 125, 62, 1081, 30, 690, 330, 2069, 14, 414, 200, 1394

Offset: 3

Brian Parsonnet, Mar 22 2008

The sequence is best represented as a series of columns 1..n, where each column j has 2^(j-1) rows (see Example). For more details, see A136300.

The first column represents the case for 3 people (offset 3).

			Represented as a series of columns, where column j has 2^(j-1) rows, the sequence begins:
  row |j = 1   2   3   4   5 ...
  ----+-------------------------
    1 |    1   1   1   1   1 ...
    2 |        1   5  13  29 ...
    3 |        2   6  14  30 ...
    4 |        1  13  73 301 ...
    5 |            2   6  14 ...
    6 |            6  42 186 ...
    7 |            2  18  86 ...
    8 |            1  29 301 ...
    9 |                2   6 ...
   10 |               18 102 ...
   11 |                8  48 ...
   12 |               14 186 ...
   13 |                2  18 ...
   14 |                6 102 ...
   15 |                2  42 ...
   16 |                1  61 ...
   17 |                    2 ...
  ... |                  ... ...
.
If there are 5 people, numbered 1-5 according to the order in which they draw a name, and person #5 draws name #5, the first four people must draw 1-4 as a proper derangement, and there are 9 ways of doing so: 21435 / 23415 / 24135 / 31425 / 34125 / 34215 / 41235 / 43125 / 43215.
But the probability of each derangement depends on how many choices exist at each successive draw. The first person can draw from 4 possibilities (2,3,4,5). The second person nominally has 3 to choose from, unless the first person drew number 2, in which case person 2 may draw 4 possibilities (1,3,4,5), and so on. The probabilities of 21435 and 24135 are both then
        1/4 * 1/4 * 1/2 * 1/2 = 1/64.
More generally, if there are n people, at the i-th turn (i = 1..n), person i has either (n-i) or (n-i+1) choices, depending on whether the name of the person who is drawing has been chosen yet. A way to represent the two cases above is 01010, where a 0 indicates that the person's number is not yet drawn, and a 1 indicates it is.
For the n-th person to be forced to choose his or her own name, the last digit of this pattern must be 0, by definition. Similarly, the 1st digit must be a 0, and the second to last digit must be a 1. So all the problem patterns start with 0 and end with 10. For 5 people, that leaves 4 target patterns which cover all 9 derangements. By enumeration, that distribution can be shown to be (for the 3rd column = 5 person case):
        0-00-10 1 occurrences
        0-01-10 5 occurrences
        0-10-10 2 occurrences
        0-11-11 1 occurrences
1;
1, 1;
1, 5, 2, 1;
1, 13, 6, 13, 2, 6, 2, 1;
1, 29, 14, 73, 6, 42, 18, 29, 2, 18, 8, 14, 2, 6, 2, 1;

Brian Parsonnet, Table of n, a(n) for n = 3..257
Brian Parsonnet, Probability of Derangements

The application of this table towards final determination of the probabilities of derangements leads to sequence A136300, which is the sequence of numerators. The denominators are in A001044.
A048144 represents the peak value of all odd-numbers columns.
A000255 equals the sum of the bottom half of each column.
A000166 equals the sum of each column.
A047920 represents the frequency of replacements by person drawing at position n.
A008277, Triangle of Stirling numbers of 2nd kind, can be derived from A136301 through a series of transformations (see "Probability of Derangements.pdf").
Cf. A371761.

maxP = 15;
rows = Range[1, 2^(nP = maxP - 3)];
pasc = Table[
   Binomial[p + 1, i] - If[i >= p, 1, 0], {p, nP}, {i, 0, p}];
sFreq = Table[0, {maxP - 1}, {2^nP}]; sFreq[[2 ;; maxP - 1, 1]] = 1;
For[p = 1, p <= nP, p++,
  For[s = 1, s <= p, s++, rS = Range[2^(s - 1) + 1, 2^s];
        sFreq[[p + 2, rS]] = pasc[[p + 1 - s, 1 ;; p + 2 - s]] .
            sFreq[[s ;; p + 1, 1 ;; 2^(s - 1)]]]];
TableForm[ Transpose[ sFreq ] ]
(* Code snippet to illustrate the conjectured connection with A371761: *)
R[n_] := Table[Transpose[sFreq][[2^n]][[r]], {r, n + 1, maxP - 1}]
For[n = 0, n <= 6, n++, Print[n + 1, ": ", R[n]]] (* Peter Luschny, Apr 10 2024 *)

H(r,c) = Sum_{j=0..c-L(r)-1} H(T(r), L(r)+j) * M(c-T(r)-1, j) where M(y,z) = binomial distribution (y,z) when y - 1 > z and (y,z)-1 when y-1 <= z and T(r) = A053645 and L(r) = A000523.

Conjecture: Assume the table represented as in the Example section. Then row 2^n is row n + 1 of A371761. - Peter Luschny, Apr 10 2024

Edited by Brian Parsonnet, Mar 01 2011

A306535 Number of permutations p of [2n] having no index i with |p(i)-i| = n.

Original entry on oeis.org

1, 1, 9, 265, 14833, 1334961, 176214841, 32071101049, 7697064251745, 2355301661033953, 895014631192902121, 413496759611120779881, 228250211305338670494289, 148362637348470135821287825, 112162153835443422680893595673, 97581073836835777732377428235481

Offset: 0

Alois P. Heinz, Feb 22 2019

nonn

Also 0th term of the 2n-th forward differences of n!.

Alois P. Heinz, Table of n, a(n) for n = 0..225
Wikipedia, Permutation

Cf. A306512, A306675.

Cf. A000142, A047920, A068106.

b:= proc(n, k) b(n, k):= `if`(k=0, n!, b(n+1, k-1) -b(n, k-1)) end:
a:= n-> b(0, 2*n):
seq(a(n), n=0..23);
seq(simplify(KummerU(-2*n, -2*n, -1)), n=0..15); # Peter Luschny, May 10 2022

b[n_, k_] := b[n, k] = If[k == 0, n!, b[n + 1, k - 1] - b[n, k - 1]];
a[n_] := b[0, 2n];
a /@ Range[0, 23] (* Jean-François Alcover, Apr 02 2021, after Alois P. Heinz *)

a(n) = A306512(2n,n).
a(n) = (2n)! - A306675(n).
a(n) = KummerU(-2*n, -2*n, -1). - Peter Luschny, May 10 2022

cp's OEIS Frontend

A001565 3rd differences of factorial numbers.

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A033815 Number of standard permutations of [ a_1..a_n b_1..b_n ] (b_i is not immediately followed by a_i, for all i).

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A155521 Smallest fixed point summed over all non-derangement permutations of {1,2,...,n}.

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A047922 Triangle of numbers a(n,k) = number of terms in n X n determinant with 2 adjacent diagonals of k and k-1 0's (0<=k<=n).

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A060475 Triangular array formed from successive differences of factorial numbers, then with factorials removed.

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A129867 Row sums of triangular array T: T(j,k) = k*(j-k)! for k < j, T(j,k) = 1 for k = j; 1 <= k <= j.

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A076732 Table T(n,k) giving number of ways of obtaining exactly one correct answer on an (n,k)-matching problem (1 <= k <= n).