A000255 -id:A000255 - cp's OEIS frontend

A000261 a(n) = na(n-1) + (n-3)a(n-2), with a(1) = 0, a(2) = 1.

0, 1, 3, 13, 71, 465, 3539, 30637, 296967, 3184129, 37401155, 477471021, 6581134823, 97388068753, 1539794649171, 25902759280525, 461904032857319, 8702813980639617, 172743930157869827, 3602826440828270029, 78768746000235327495, 1801366114380914335441

Offset: 1

N. J. A. Sloane

nonn

With offset 1, permanent of (0,1)-matrix of size n X (n+d) with d=3 and n zeros not on a line. This is a special case of Theorem 2.3 of Seok-Zun Song et al. Extremes of permanents of (0,1)-matrices, pp. 201-202. - Jaap Spies, Dec 12 2003
a(n+2)=:b(n), n>=1, enumerates the ways to distribute n beads, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and three indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as a beadless cords contribute each a factor 1 in the counting, e.g., b(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads.
This produces for b(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A001710(n+2)}. See the necklaces and cords problem comment in A000153. Therefore also the recurrence b(n) = (n+2)*b(n-1) + (n-1)*b(n-2) with b(-1)=0 and b(0)=1 holds. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010). - Wolfdieter Lang, Jun 02 2010

			Necklaces and 3 cords problem. For n=4 one considers the following weak 2 part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively sf(4)*1,binomial(4,3)*sf(3)*c3(1), (binomial(4,2)*sf(2))*c3(2), and 1*c3(4) with the subfactorials sf(n):=A000166(n) (see the necklace comment there) and the c3(n):=A001710(n+2) = (n+2)!/2! numbers for the pure 3 cord problem (see the remark on the e.g.f. for the k cords problem in A000153; here for k=3: 1/(1-x)^3). This adds up as 9 + 4*2*3 + (6*1)*12 + 360 = 465 = b(4) = A000261(6). - _Wolfdieter Lang_, Jun 02 2010
G.f. = x^2 + 3*x^3 + 13*x^4 + 71*x^5 + 465*x^6 + 3539*x^7 + 30637*x^8 + ...

Brualdi, Richard A. and Ryser, Herbert J., Combinatorial Matrix Theory, Cambridge NY (1991), Chapter 7.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 188.
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=1..102
Roland Bacher, Counting Packings of Generic Subsets in Finite Groups, Electr. J. Combinatorics, 19 (2012), #P7. - From _N. J. A. Sloane_, Feb 06 2013
Seok-Zun Song et al., Extremes of permanents of (0,1)-matrices, Special issue on the Combinatorial Matrix Theory Conference (Pohang, 2002). Linear Algebra Appl. 373 (2003), pp. 197-210.

Cf. A000255, A000153, A001909, A001910, A090010, A055790, A090012-A090016.
Cf. A086764(n+1,3), n>=1.
Cf. A000153 (necklaces and two cords). - Wolfdieter Lang, Jun 02 2010

a:= proc(n) a(n):= `if`(n<3, n-1, n*a(n-1) +(n-3)*a(n-2)) end:
seq(a(n), n=1..30);  # Alois P. Heinz, Nov 03 2012
a := n -> `if`(n=1,0,hypergeom([4,-n+2],[],1))*(-1)^(n); seq(round(evalf(a(n), 100)), n=1..22); # Peter Luschny, Sep 20 2014

nn=20;Prepend[Range[0,nn]!CoefficientList[Series[Exp[-x]/ (1-x)^4, {x,0,nn}],x],0]  (* Geoffrey Critzer, Nov 03 2012 *)
a[ n_] := SeriesCoefficient[ x^2 HypergeometricPFQ[ {1, 4}, {}, x / (1 + x)] / (1 + x), {x, 0, n}]; (* Michael Somos, May 04 2014 *)
a[ n_] := If[ n < 2, 0, With[{m = n - 1}, Round[ Gamma[m] (m^3 + 6 m^2 + 8 m + 1) Exp[-1]/6]]]; (* Michael Somos, May 04 2014 *)

E.g.f.: exp(-x)/(1-x)^4 for offset -1.
For offset -1: (1/6)*Sum_{k=0..n} (-1)^k*(n-k+1)*(n-k+2)*(n-k+3)*n!/k! = (1/6)*(A000166(n)+3*A000166(n+1)+3*A000166(n+2)+A000166(n+3)). - Vladeta Jovovic, Jan 07 2003
a(n+1) = round( GAMMA(n)*(n^3+6*n^2+8*n+1)*exp(-1)/6 ) for n>0. - Mark van Hoeij, Nov 11 2009
G.f.: x^2*hypergeom([1,4],[],x/(x+1))/(x+1). - Mark van Hoeij, Nov 07 2011
E.g.f. for offset -1: 1/(exp(x)*(1-x)^4) = 1/E(0) where E(k) = 1 - 4*x/(1 + 3*x/(2 - 3*x + 4*x/(3 - 2*x + 3*x/(4 - x - 4/(1 + x^3*(k+1)/E(k+1)))))); (continued fraction, 3rd kind, 6-step). - Sergei N. Gladkovskii, Sep 21 2012
a(n) = hypergeometric([4,-n+2],[],1)*(-1)^n for n>=2. - Peter Luschny, Sep 20 2014

More terms from Vladeta Jovovic, Jan 07 2003

A001910 a(n) = na(n-1) + (n-5)a(n-2).

Original entry on oeis.org

0, 1, 5, 31, 227, 1909, 18089, 190435, 2203319, 27772873, 378673901, 5551390471, 87057596075, 1453986832381, 25762467303377, 482626240281739, 9530573107600319, 197850855756232465, 4307357140602486869, 98125321641110663023, 2334414826276390013171

Offset: 3

N. J. A. Sloane

nonn

With offset 1, permanent of (0,1)-matrix of size n X (n+d) with d=5 and n zeros not on a line. This is a special case of Theorem 2.3 of Seok-Zun Song et al. Extremes of permanents of (0,1)-matrices, pp. 201-202. - Jaap Spies, Dec 12 2003
a(n+4)=:b(n), n>=1, enumerates the ways to distribute n beads labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=5 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as a beadless cords contribute each a factor 1 in the counting, e.g., b(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads.
This produces for b(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A001720(n+4) = (n+4)!/4!}. See the necklaces and cords problem comment in A000153. Therefore also the recurrence b(n) = (n+4)*b(n-1) + (n-1)*b(n-2) with b(-1)=0 and b(0)=1 holds. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010). - Wolfdieter Lang, Jun 02 2010

			Necklaces and 5 cords problem. For n=4 one considers the following weak 2 part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively sf(4)*1, binomial(4,3)*sf(3)*c5(1), (binomial(4,2)*sf(2))*c5(2), and 1*c5(4) with the subfactorials sf(n):=A000166(n) (see the necklace comment there) and the c5(n):=A001720(n+4) numbers for the pure 5 cord problem (see the remark on the e.g.f. for the k cords problem in A000153; here for k=5: 1/(1-x)^5). This adds up as 9 + 4*2*5 + (6*1)*30 + 1680 = 1909 = b(4) = A001910(8). - _Wolfdieter Lang_, Jun 02 2010

Brualdi, Richard A. and Ryser, Herbert J., Combinatorial Matrix Theory, Cambridge NY (1991), Chapter 7.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 188.
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 = 3..100
Seok-Zun Song et al., Extremes of permanents of (0,1)-matrices, Special issue on the Combinatorial Matrix Theory Conference (Pohang, 2002). Linear Algebra Appl. 373 (2003), pp. 197-210.

Cf. A000255, A000153, A000261, A001909, A001910, A055790, A090012-A090016, A086764.

A001909 (necklaces and four cords).

a := n -> `if`(n=3,0, hypergeom([6,-n+4],[],1))*(-1)^n;
seq(round(evalf(a(n),100)), n=3..20); # Peter Luschny, Sep 20 2014

t = {0, 1}; Do[AppendTo[t, n*t[[-1]] + (n - 5) t[[-2]]], {n, 5, 20}]; t (* T. D. Noe, Aug 17 2012 *)

a(n) = A086764(n+1,5), n>=3.
E.g.f. with offset -1: (exp(-x)/(1-x))*(1-x)^5 = exp(-x)/(1-x)^6. - Wolfdieter Lang, Jun 02 2010
G.f.: x*hypergeom([1,6],[],x/(x+1))/(x+1). - Mark van Hoeij, Nov 07 2011
a(n) = hypergeometric([6,-n+4],[],1)*(-1)^n for n >=4. - Peter Luschny, Sep 20 2014

A068106 Euler's difference table: triangle read by rows, formed by starting with factorial numbers (A000142) and repeatedly taking differences. T(n,n) = n!, T(n,k) = T(n,k+1) - T(n-1,k).

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 2, 3, 4, 6, 9, 11, 14, 18, 24, 44, 53, 64, 78, 96, 120, 265, 309, 362, 426, 504, 600, 720, 1854, 2119, 2428, 2790, 3216, 3720, 4320, 5040, 14833, 16687, 18806, 21234, 24024, 27240, 30960, 35280, 40320, 133496, 148329, 165016, 183822, 205056, 229080, 256320, 287280, 322560, 362880

Offset: 0

N. J. A. Sloane, Apr 12 2002

Triangle T(n,k) (n >= 1, 1 <= k <= n) giving number of ways of winning with (n-k+1)st card in the generalized "Game of Thirteen" with n cards.
From Emeric Deutsch, Apr 21 2009: (Start)
T(n-1,k-1) is the number of non-derangements of {1,2,...,n} having largest fixed point equal to k. Example: T(3,1)=3 because we have 1243, 4213, and 3241.
Mirror image of A047920.
(End)

			Triangle begins:
[0]    1;
[1]    0,    1;
[2]    1,    1,    2;
[3]    2,    3,    4,    6;
[4]    9,   11,   14,   18,   24;
[5]   44,   53,   64,   78,   96,  120;
[6]  265,  309,  362,  426,  504,  600,  720;
[7] 1854, 2119, 2428, 2790, 3216, 3720, 4320, 5040.

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
W. Y. C. Chen et al., Higher-order log-concavity in Euler's difference table, Discrete Math., 311 (2011), 2128-2134.
P. R. de Montmort, On the Game of Thirteen (1713), reprinted in Annotated Readings in the History of Statistics, ed. H. A. David and A. W. F. Edwards, Springer-Verlag, 2001, pp. 25-29.
Emeric Deutsch and S. Elizalde, The largest and the smallest fixed points of permutations, arXiv:0904.2792 [math.CO], 2009.
D. Dumont, Matrices d'Euler-Seidel, Sem. Loth. Comb. B05c (1981) 59-78.
Philip Feinsilver and John McSorley, Zeons, Permanents, the Johnson scheme, and Generalized Derangements, arXiv:1710.00788 [math.CO], (2017); see page 29.
P. Feinsilver and J. McSorley, Zeons, Permanents, the Johnson scheme, and Generalized Derangements, International Journal of Combinatorics, 2011 (2011).
Fanja Rakotondrajao, k-Fixed-Points-Permutations, Integers: Electronic journal of combinatorial number theory 7 (2007) A36.
Index entries for sequences related to factorial numbers

Row sums give A002467.
Diagonals give A000142, A001563, A001564, A001565, A001688, A001689, A023043, A023044, A023045, A023046, A023047 (factorials and k-th differences, k=1..10).
See A047920 and A086764 for other versions.
T(2*n, n) is A033815.
Columns k=0..10 give A000166, A000255, A055790, A277609, A277563, A280425, A280920, A284204, A284205, A284206, A284207.

a068106 n k = a068106_tabl !! n !! k
a068106_row n = a068106_tabl !! n
a068106_tabl = map reverse a047920_tabl
-- Reinhard Zumkeller, Mar 05 2012

d[0] := 1: for n to 15 do d[n] := n*d[n-1]+(-1)^n end do: T := proc (n, k) if k <= n then sum(binomial(k, j)*d[n-j], j = 0 .. k) else 0 end if end proc: for n from 0 to 9 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form; Emeric Deutsch, Jul 18 2009

t[n_, k_] := Sum[(-1)^j*Binomial[n-k, j]*(n-j)!, {j, 0, n}]; Flatten[ Table[ t[n, k], {n, 0, 9}, {k, 0, n}]] (* Jean-François Alcover, Feb 21 2012, after Philippe Deléham *)
T[n_, k_] := n! HypergeometricPFQ[{k-n}, {-n}, -1];
Table[T[n, k], {n,0,9}, {k,0,n}] // Flatten (* Peter Luschny, Oct 05 2017 *)

T(n, k) = Sum_{j>= 0} (-1)^j*binomial(n-k, j)*(n-j)!. - Philippe Deléham, May 29 2005
From Emeric Deutsch, Jul 18 2009: (Start)
T(n,k) = Sum_{j=0..k} d(n-j)*binomial(k, j), where d(i) = A000166(i) are the derangement numbers.
Sum_{k=0..n} (k+1)*T(n,k) = A000166(n+2) (the derangement numbers). (End)
T(n, k) = n!*hypergeom([k-n], [-n], -1). - Peter Luschny, Oct 05 2017
D-finite recurrence for columns: T(n,k) = n*T(n-1,k) + (n-k)*T(n-2,k). - Georg Fischer, Aug 13 2022

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 01 2003

Edited by N. J. A. Sloane, Sep 24 2011

A187235 Number of ways to place n nonattacking semi-bishops on an n X n board.

Original entry on oeis.org

1, 5, 51, 769, 15345, 381065, 11323991, 391861841, 15476988033, 687029386845, 33861652925595, 1834814222811361, 108411291759763681, 6936921762461326545, 477881176664541171375, 35264213540563039871265, 2775185864375851234241985, 232010235620834821000259765, 20534530616200868936398461635

Offset: 1

Vaclav Kotesovec, Mar 08 2011

Two semi-bishops do not attack each other if they are in the same NorthWest-SouthEast diagonal.

Conjecture: Number of parity preserving permutations of the set {1, 2, ..., 2n+1} with exactly n+1 cycles (see A246117). - Peter Luschny, Feb 09 2015

Vaclav Kotesovec, Table of n, a(n) for n = 1..350
V. Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 260-265.

Cf. A238261, A238262, A002465, A099152, A000255, A129256.

Table[If[n==1,1,Coefficient[Expand[Product[x+i,{i,1,n}]*Product[x+i,{i,1,n-1}],x],x,n-1]],{n,1,50}]
Table[(-1)^n*Sum[StirlingS1[n+1,j]*StirlingS1[n,n-j+1],{j,1,n}],{n,1,50}] (* Explicit formula, Vaclav Kotesovec, Mar 24 2011 *)

a(n) = {(-1)^n*sum(i=0, n, stirling(n,i,1) * stirling(n+1,n-i+1,1))} \\ Andrew Howroyd, May 09 2020

a(n)/(n-1)! ~ 0.24252191 * 4.9108149^n where the second constant is 1/(z*(1-z)) = 4.910814964..., where z=0.715331862959... is a root of the equation z=2*(z-1)*log(1-z).
For constants see A238261 and A238262. - Vaclav Kotesovec, Feb 21 2014
a(n) = (-1)^n * Sum_{i=0..n} Stirling1(n,i) * Stirling1(n+1,n-i+1). - Ryan Brooks, May 09 2020

A087981 E.g.f.: exp(-2*x) / (1-x)^2.

Original entry on oeis.org

1, 0, 2, 4, 24, 128, 880, 6816, 60032, 589312, 6384384, 75630080, 972387328, 13483769856, 200571078656, 3185540657152, 53800242216960, 962741176500224, 18195808235880448, 362183230599856128, 7572922094360723456, 165945771111208714240, 3802923921298533384192, 90965940197460917878784, 2267151124921333646884864

Offset: 0

Gordon F. Royle, Oct 28 2003

Permanent of an (n+1) X (n+1) (+1, -1)-matrix with exactly n -1's on the diagonal and 1's everywhere else.
It is conjectured by Kräuter and Seifter that for n >= 5 a(n-1) is the maximal possible value for the permanent of a nonsingular n X n (+1, -1)-matrix. I do not know for which values of n this has been confirmed - compare A087982. - N. J. A. Sloane
The Kräuter conjecture on permanents is true (see Budrevich and Guterman). - Sergei Shteiner, Jan 17 2020
The maximal possible value for the permanent of a singular n X n (+1, -1)-matrix is obviously n!.
Degree of the "hyperdeterminant" of a multilinear polynomial on (\P^1(\C))^n, or equivalently of an element of (\C^2)^{⊗ n}: see Gelfand, Kapranov and Zelevinsky. - Eric Rains, Mar 15 2004
(-1)^n * a(n) = Polynomials in A010027 evaluated at -1. - Ralf Stephan, Dec 15 2004
a(n) is the number of n X n (-1, 0, 1)-matrices containing in every row and every column exactly one -1 and one 1 such that the main diagonal does not contain 0's. - Vladimir Shevelev, Apr 01 2010
a(n) is the number of colored permutations with no fixed points of n elements where each cycle is one of two colors. - Michael Somos, Jan 19 2011
Binomial transform is A000255. Hankel transform is A059332. - Paul Barry, Apr 11 2011
Exponential self-convolution of subfactorials (A000166). - Vladimir Reshetnikov, Oct 07 2016

			G.f. = 1 + 2*x^2 + 4*x^3 + 24*x^4 + 128*x^5 + 880*x^6 + 6816*x^7 + ...
Since a(1) = 0, then, for n = 2, we have a(2) = -(-2)^3/4 = 2; further, for n = 3, we find a(3) = (3*6/5)*2 - (-2)^4/5 = 36/5 - 16/5 = 4. - _Vladimir Shevelev_, Apr 01 2010
a(4) = 24 because there are 6 derangements with one 4-cycle with 2^1 ways to color each derangement and 3 derangements with two 2-cycles with 2^2 ways to color each derangement. - _Michael Somos_, Jan 19 2011

I. M. Gelfand, M. M. Kapranov, and A. V. Zelevinsky, Discriminants, Resultants and Multidimensional Determinants, Birkhauser, 1994; see Corollary 2.10 in Chapter 14 (p. 457).

T. D. Noe, Table of n, a(n) for n = 0..100
Mikhail V. Budrevich and Alexander E. Guterman, Kräuter conjecture on permanents is true, arXiv:1810.04439 [math.CO], 2018.
Steven Finch, Rounds, Color, Parity, Squares, arXiv:2111.14487 [math.CO], 2021.
I. M. Gelfand, M. M. Kapranov, and A. V. Zelevinsky, Hyperdeterminants, Advances in Mathematics 96(2) (1992), 226-263; see Corollary 3.10 (p. 246).
Arnold R. Kräuter, Permanenten - Ein kurzer Überblick, Séminaire Lotharingien de Combinatoire, B09b (1983), 34 pp.
Arnold R. Kräuter and Norbert Seifter, Some properties of the permanent of (1,-1)-matrices, Linear and Multilinear Algebra 15 (1984), 207-223.
Norbert Seifter, Upper bounds for permanents of (1,-1)-matrices, Israel J. Math. 48 (1984), 69-78.
Edward Tzu-Hsia Wang, On permanents of (1,-1)-matrices, Israel J. Math. 18 (1974), 353-361.
Wikipedia, Hyperdeterminant
Index entries for sequences related to binary matrices

Cf. A087982, A087983, A176097.

Cf. A000007, A000023, A000166, A024000, A137775. - Michael Somos, Jan 19 2011

seq(simplify(KummerU(-n, -n-1, -2)), n = 0..24); # Peter Luschny, May 10 2022

Range[0, 20]! CoefficientList[Series[Exp[-2 x]/(1 - x)^2, {x, 0, 20}], x]
Table[(-2)^n HypergeometricPFQ[{2, -n}, {}, 1/2], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 07 2016 *)

{a(n) = if( n<0, 0, n! * polcoeff( exp( -2 * x + x * O(x^n) ) / ( 1 - x )^2, n ) )} /* Michael Somos, Jan 19 2011 */

Krauter and Seifter prove that the permanent of an n X n {-1, 1} matrix is divisible by 2^{n - [log_2(n)] - 1}.
Let c(n) be the permanent of the {-1, +1}-matrix of order n X n with n diagonal -1's only. Let a(n) be the permanent of the {-1, +1}-matrix of order (n+1) X (n+1) with n diagonal -1's only. Then by expanding along the first row (like determinant, but with no sign) we get c(n+1) = -c(n) + n a(n-1), a(n) = c(n) + n a(n-1), with c(2) = 2, a(2) = 2. {c(n)} has e.g.f. exp(-2x)/(1-x), see A000023. Also a(n) = c(n+1) + 2*c(n).
The following 4 formulas hold: a(n) = Sum_{k = 0..n} C(n, k)*D_k*D_{n-k}, where D_n = A000166(n); a(n) = n!*Sum_{j = 0..n} (n+1-j)*(-2)^j/j!; a(0) = 1, a(1) = 0 and, for n > 0, a(n+1) = n*(a(n) + 2*a(n-1)); a(0) = 1 and, for n > 0, a(n) = (n*(n+3)/(n+2))*a(n-1) - (-2)^(n+1)/(n+2). - Vladimir Shevelev, Apr 01 2010 [edited by Michael Somos, Jan 19 2011]
G.f.: 1/(1-2x^2/(1-2x-6x^2/(1-4x-12x^2/(1-6x-20x^2/(1-.../(1-2n*x-(n+1)(n+2)x^2/(1-... (continued fraction). - Paul Barry, Apr 11 2011
E.g.f.: 1/U(0) where U(k)=  1 - 2*x/( 1 + x/(2 - x - 4/( 2 + x*(k+1)/U(k+1)))) ; (continued fraction, 3rd kind, 4-step). - Sergei N. Gladkovskii, Oct 28 2012
G.f.: 1/Q(0), where Q(k) = 1 + 2*x - x*(k+2)/(1 - x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 22 2013
G.f.: 1/Q(0) where Q(k) = 1 - 2*k*x - x^2*(k + 1)*(k+2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 10 2013
G.f.: S(x)/x - 1/x = G(0)/x - 1/x, where S(x) = sum(k >= 0, k!*(x/(1+2*x))^k  ), G(k) = 1 + (2*k + 1)*x/( 1+2*x - 2*x*(1+2*x)*(k+1)/(2*x*(k+1) + (1+2*x)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 26 2013
a(n) = (-2)^n*hypergeom([2, -n], [], 1/2) = 4*(-2)^n*(1 - 2*hypergeom([1, -n-3], [], 1/2))/(n^2+3*n+2) = (4*(-2)^n + Gamma(n+4, -2)*exp(-2))/(n^2+3*n+2). - Vladimir Reshetnikov, Oct 07 2016
a(n) ~ sqrt(2*Pi) * n^(n+3/2) / exp(n+2). - Vaclav Kotesovec, Oct 08 2016
a(n) = KummerU(-n, -n - 1, -2). - Peter Luschny, May 10 2022

More terms from Jaap Spies, Oct 28 2003
Further terms from Gordon F. Royle, Oct 29 2003
Definition via e.g.f. from Eric Rains, Mar 15 2004
Changed the offset and terms to correspond to e.g.f, Michael Somos, Jan 19 2011

A180191 Number of permutations of [n] having at least one succession. A succession of a permutation p is a position i such that p(i+1)-p(i) = 1.

Original entry on oeis.org

0, 1, 3, 13, 67, 411, 2921, 23633, 214551, 2160343, 23897269, 288102189, 3760013027, 52816397219, 794536751217, 12744659120521, 217140271564591, 3916221952414383, 74539067188152941, 1493136645424092773, 31400620285465593339, 691708660911435955579

Offset: 1

Emeric Deutsch, Sep 07 2010

nonn

a(n) = A180190(n,1).
a(n+2) = p(n+2) where p(x) is the unique degree-n polynomial such that p(k) = k! for k = 1, ..., n+1. - Michael Somos, Jan 05 2012
From Jon Perry, Jan 04 2013: (Start)
Number of permutations of {1,...,n-1,n+1} with at least one indexed point p(k)=k with 1<=k<=n. Note that this means p(k)=n+1 is never an indexed point as kFor n>1, a(n) is the number of permutations of [n+1] that have a fixed point and contain 12; for example the a(3)=3 such permutations of {1,2,3,4} are 1234, 1243, and 3124.
(End)
For n > 0: row sums of triangle A116853. - Reinhard Zumkeller, Aug 31 2014

			x^2 + 3*x^3 + 13*x^4 + 67*x^5 + 411*x^6 + 2921*x^7 + 23633*x^8 + ...
a(3) = 3 because we have 123, 312, and 231; the permutations 132, 213, and 321 have no successions.
a(4) = 13 since p(x) = (3*x^2 - 7*x + 6) / 2 interpolates p(1) = 1, p(2) = 2, p(3) = 6, and p(4) = 13. - _Michael Somos_, Jan 05 2012

Alois P. Heinz, Table of n, a(n) for n = 1..200
Uriel Feige, Tighter bounds for online bipartite matching, 2018.

Cf. A000142, A000166, A000255, A002467, A028417, A180190.
Cf. A116853, A276975.
Column k=1 of A306234, A306461, and of A324362(n-1).

a180191 n = if n == 1 then 0 else sum $ a116853_row (n - 1)
-- Reinhard Zumkeller, Aug 31 2014

d[0] := 1: for n to 50 do d[n] := n*d[n-1]+(-1)^n end do: seq(factorial(n)-d[n]-d[n-1], n = 1 .. 22);

f[n_] := Sum[ -(-1)^k (n - k)! Binomial[n - 1, k], {k, 1, n}]; Array[f, 20] (* Robert G. Wilson v, Oct 16 2010 *)

{a(n) = if( n<2, 0, n--; subst( polinterpolate( vector( n, k, k!)), x, n+1))} /* Michael Somos, Jan 05 2012 */

a(n) = n! - d(n) - d(n-1), where d(j) = A000166(j) are the derangement numbers.
a(n) = n! - A000255(n-1) = A002467(n) - A000166(n-1). - Jon Perry, Jan 05 2013
a(n) = (n-1)! [x^(n-1)] (1-exp(-x))/(1-x)^2. - Alois P. Heinz, Feb 23 2019

A086764 Triangle T(n, k), read by row, related to Euler's difference table A068106 (divide column k of A068106 by k!).

Original entry on oeis.org

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

Offset: 0

Philippe Deléham, Aug 02 2003

The k-th column sequence, k >= 0, without leading zeros, enumerates the ways to distribute n beads, n >= 1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k+1 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords each contribute a factor 1, hence for n=0 one has 1. See A000255 for the description of a fixed cord with beads. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010). - Wolfdieter Lang, Jun 02 2010

			Formatted as a square array:
      1      3     7    13   21   31  43 57 ... A002061;
      2     11    32    71  134  227 356    ... A094792;
      9     53   181   465 1001 1909        ... A094793;
     44    309  1214  3539 8544             ... A094794;
    265   2119  9403 30637                  ... A023043;
   1854  16687 82508                        ... A023044;
  14833 148329                              ... A023045;
Formatted as a triangular array (mirror of A076731):
       1;
       0      1;
       1      1     1;
       2      3     2     1;
       9     11     7     3    1;
      44     53    32    13    4    1;
     265    309   181    71   21    5    1;
    1854   2119  1214   465  134   31    6   1;
   14833  16687  9403  3539 1001  227   43   7   1;
  133496 148329 82508 30637 8544 1909  356  57   8   1;

Indranil Ghosh, Rows 0..50, flattened
W. Y. C. Chen et al., Higher-order log-concavity in Euler's difference table, Discrete Math., 311 (2011), 2128-2134. (These are the numbers d^k_n.)
Fanja Rakotondrajao, k-Fixed-Points-Permutations, Integers: Electronic journal of combinatorial number theory 7 (2007) A36.

Columns: A000166, A000155, A000153, A000261, A001909, A001910, A176732, A176733, A176734, A176735, A176736.
Mirror image of A076731.
Cf. A002061, A003470, A008288, A023043, A023044, A023045, A068106.
Cf. A076731, A094792, A094793, A094794, A095000, A108625, A143007.
Cf. A143409, A143410, A143411, A143413.

A086764:= func< n,k | (&+[(-1)^j*Binomial(n-k,j)*Factorial(n-j): j in [0..n]])/Factorial(k) >;
[A086764(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023

T[n_,k_]:=(1/k!)*Sum[(-1)^j*Binomial[n-k,j]*(n-j)!,{j,0,n}];Flatten[Table[T[n,k],{n,0,11},{k,0,n}]] (* Indranil Ghosh, Feb 20 2017 *)
T[n_, k_] := (n!/k!) HypergeometricPFQ[{k-n},{-n},-1];
Table[T[n,k], {n,0,9}, {k,0,n}] // Flatten (* Peter Luschny, Oct 05 2017 *)

def A086764(n,k): return sum((-1)^j*binomial(n-k,j)*factorial(n-j) for j in range(n+1))//factorial(k)
flatten([[A086764(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

T(n, n) = 1; T(n+1, n) = n.
T(n+2, n) = A002061(n+1) = n^2 + n + 1; T(n+3, n) = n^3 + 3*n^2 + 5*n + 2.
T(n, k) = (k + 1)*T(n, k + 1) - T(n-1, k); T(n, n) = 1; T(n, k) = 0, if k > n.
T(n, k) = (n-1)*T(n-1, k) + (n-k-1)*T(n-2, k).
k!*T(n, k) = A068106(n, k). [corrected by Georg Fischer, Aug 13 2022]
Sum_{k>=0} T(n, k) = A003470(n+1).
T(n, k) = (1/k!) * Sum_{j>=0} (-1)^j*binomial(n-k, j)*(n-j)!. - Philippe Deléham, Jun 13 2005
From Peter Bala, Aug 14 2008: (Start)
The following remarks all relate to the array read as a square array: e.g.f for column k: exp(-y)/(1-y)^(k+1); e.g.f. for array: exp(-y)/(1-x-y) = (1 + x + x^2 + x^3 + ...) + (x + 2*x^2 + 3*x^3 + 4*x^4 + ...)*y + (1 + 3*x + 7*x^2 + 13*x^3 + ...)*y^2/2! + ... .
This table is closely connected to the constant e. The row, column and diagonal entries of this table occur in series formulas for e.
Row n for n >= 2: e = n!*(1/T(n,0) + (-1)^n*[1/(1!*T(n,0)*T(n,1)) + 1/(2!*T(n,1)*T(n,2)) + 1/(3!*T(n,2)*T(n,3)) + ...]). For example, row 3 gives e = 6*(1/2 - 1/(1!*2*11) - 1/(2!*11*32) - 1/(3!*32*71) - ...). See A095000.
Column 0: e = 2 + Sum_{n>=2} (-1)^n*n!/(T(n,0)*T(n+1,0)) = 2 + 2!/(1*2) - 3 !/(2*9) + 4!/(9*44) - ... .
Column k, k >= 1: e = (1 + 1/1! + 1/2! + ... + 1/k!) + 1/k!*Sum_{n >= 0} (-1)^n*n!/(T(n,k)*T(n+1,k)). For example, column 3 gives e = 8/3 + 1/6*(1/(1*3) - 1/(3*13) + 2/(13*71) - 6/(71*465) + ...).
Main diagonal: e = 1 + 2*(1/(1*1) - 1/(1*7) + 1/(7*71) - 1/(71*1001) + ...).
First subdiagonal: e = 8/3 + 5/(3*32) - 7/(32*465) + 9/(465*8544) - ... .
Second subdiagonal: e = 2*(1 + 2^2/(1*11) - 3^2/(11*181) + 4^2/(181*3539) - ...). See A143413.
Third subdiagonal: e = 3 - (2*3*5)/(2*53) + (3*4*7)/(53*1214) - (4*5*9)/(1214*30637) + ... .
For the corresponding results for the constants 1/e, sqrt(e) and 1/sqrt(e) see A143409, A143410 and A143411 respectively. For other arrays similarly related to constants see A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)). (End)
G.f. for column k is hypergeom([1,k+1],[],x/(x+1))/(x+1). - Mark van Hoeij, Nov 07 2011
T(n, k) = (n!/k!)*hypergeom([k-n], [-n], -1). - Peter Luschny, Oct 05 2017

More terms from David Wasserman, Mar 28 2005
Additional comments from Zerinvary Lajos, Mar 30 2006
Edited by N. J. A. Sloane, Sep 24 2011

A132382 Lower triangular array T(n,k) generator for group of arrays related to A001147 and A102625.

Original entry on oeis.org

1, -1, 1, -1, -2, 1, -3, -3, -3, 1, -15, -12, -6, -4, 1, -105, -75, -30, -10, -5, 1, -945, -630, -225, -60, -15, -6, 1, -10395, -6615, -2205, -525, -105, -21, -7, 1, -135135, -83160, -26460, -5880, -1050, -168, -28, -8, 1, -2027025, -1216215, -374220, -79380, -13230, -1890, -252, -36, -9, 1

Offset: 0

Tom Copeland, Nov 11 2007, Nov 12 2007, Nov 19 2007, Dec 04 2007, Dec 06 2007

Let b(n) = LPT[ A001147 ] = -A001147(n-1) for n > 0 and 1 for n=0, where LPT represents the action of the list partition transform described in A133314.
Then T(n,k) = binomial(n,k) * b(n-k) .
Form the matrix of polynomials TB(n,k,t) = T(n,k) * t^(n-k) = binomial(n,k) * b(n-k) * t^(n-k) = binomial(n,k) * Pb(n-k,t),
beginning as
     1;
    -1,       1;
    -1*t,    -2,       1;
    -3*t^2,  -3*t,    -3,       1;
   -15*t^3, -12*t^2,  -6*t,    -4,    1;
  -105*t^4, -75*t^3, -30*t^2, -10*t, -5, 1;
Let Pc(n,t) = LPT(Pb(.,t)).
Then [TB(t)]^(-1) = TC(t) = [ binomial(n,k) * Pc(n-k,t) ] = LPT(TB),
whose first column is
Pc(0,t) = 1
Pc(1,t) = 1
Pc(2,t) = 2 + t
Pc(3,t) = 6 + 6*t + 3*t^2
Pc(4,t) = 24 + 36*t + 30*t^2 + 15*t^3
Pc(5,t) = 120 + 240*t + 270*t^2 + 210*t^3 + 105*t^4.
The coefficients of these polynomials are given by the reverse of A102625 with the highest order coefficients given by A001147 with an additional leading 1.
Note this is not the complete matrix TC. The complete matrix is formed by multiplying along the diagonal of the lower triangular Pascal matrix by these polynomials, embedding trees of coefficients in the matrix.
exp[Pb(.,t)*x] = 1 + [(1-2t*x)^(1/2) - 1] / (t-0) = [1 + a finite diff. of [(1-2t*x)^(1/2)] with step t] = e.g.f. of the first column of TB.
exp[Pc(.,t)*x] = 1 / { 1 + [(1-2t*x)^(1/2) - 1] / t } = 1 / exp[Pb(.,t)*x) = e.g.f. of the first column of TC.
TB(t) and TC(t), being inverse to each other, are the generators of an Abelian group.
TB(0) and TC(0) are generators for a subgroup representing the iterated Laguerre operator described in A132013 and A132014.
Let sb(t,m) and sc(t,m) be the associated sequences under the LPT to TB(t)^m = B(t,m) and TC(t)^m = C(t,m).
Let Esb(t,m) and Esc(t,m) be e.g.f.'s for sb(t,m) and sc(t,m), rB(t,m) and rC(t,m) be the row sums of B(t,m) and C(t,m) and aB(t,m) and aC(t,m) be the alternating row sums.
Then B(t,m) is the inverse of C(t,m), Esb(t,m) is the reciprocal of Esc(t,m) and sb(t,m) and sc(t,m) form a reciprocal pair under the LPT. Similar relations hold among the row sums and the alternating sign row sums and associated quantities.
All the group members have the form B(t,m) * C(u,p) = TB(t)^m * TC(u)^p = [ binomial(n,k) * s(n-k) ]
with associated e.g.f. Es(x) = exp[m * Pb(.,t) * x] * exp[p * Pc(.,u) * x] for the first column of the matrix, with terms s(n), so group multiplication is isomorphic to matrix multiplication and to multiplication of the e.g.f.'s for the associated sequences (see examples).
These results can be extended to other groups of integer-valued arrays by replacing the 2 by any natural number in the expression for exp[Pb(.,t)*x].
More generally,
[ G.f. for M = Product_{i=0..j} B[s(i),m(i)] * C[t(i),n(i)] ]
= exp(u*x) * Product_{i=0..j} { exp[m(i) * Pb(.,s(i)) * x] * exp[n(i) * Pc(.,t(i)) * x] }
= exp(u*x) * Product_{i=0..j} { 1 + [ (1 - 2*s(i)*x)^(1/2) - 1 ] / s(i) }^m(i) / { 1 + [ (1 - 2*t(i)*x)^(1/2) - 1 ] / t(i) }^n(i)
= exp(u*x) * H(x)
[ E.g.f. for M ] = I_o[2*(u*x)^(1/2)] * H(x).
M is an integer-valued matrix for m(i) and n(i) positive integers and s(i) and t(i) integers. To invert M, change B to C in Product for M.
H(x) is the e.g.f. for the first column of M and diagonally multiplying the Pascal matrix by the terms of this column generates M. See examples.
The G.f. for M, i.e., the e.g.f. for the row polynomials of M, implies that the row polynomials form an Appell sequence (see Wikipedia and Mathworld). - Tom Copeland, Dec 03 2013

			Some group members and associated arrays are
(t,m) :: Array :: Asc. Matrix :: Asc. Sequence :: E.g.f. for sequence
..............................................................................
(0,1).::.B..::..A132013.::.(1,-1,0,0,0,0,...).....::.s(x).=.1-x
(0,1).::.C..::..A094587.::.(0!,1!,2!,3!,...)......::.1./.s(x)
(0,1).::.rB.::.~A055137.::.(1,0,-1,-2,-3,-4,...)..::.exp(x).*.s(x)
(0,1).::.rC.::....-.....::..A000522...............::.exp(x)./.s(x)
(0,1).::.aB.::....-.....::.(1,-2,3,-4,5,-6,...)...::.exp(-x).*.s(x)
(0,1).::.aC.::..A008290.::..A000166...............::.exp(-x)./.s(x)
..............................................................................
(0,2).::.B..::..A132014.::.(1,-2,2,0,0,0,0...)....::.s(x).=.(1-x)^2
(0,2).::.C..::..A132159.::.(1!,2!,3!,4!,...)......::..1./.s(x).
(0,2).::.rB.::...-......::.(1,-1,-1,1,5,11,19,29,)::.exp(x).*.s(x).
(0,2).::.rC.::...-......::..A001339...............::.exp(x)./.s(x).
(0,2).::.aB.::...-......::.(-1)^n.A002061(n+1)....::.exp(-x).*.s(x).
(0,2).::.aC.::...-......::..A000255...............::.exp(-x)./.s(x).
..............................................................................
(1,1).::.B..::..T.......::.(1,-A001147(n-1))......::.s(x).=.(1-2x)^(1/2)
(1,1).::.C..::.~A113278.::..A001147...............::.1./.s(x)...
(1,1).::.rB.::...-......::..A055142...............::.exp(x).*.s(x).
(1,1).::.rC.::...-......::..A084262...............::.exp(x)./.s(x).
(1,1).::.aB.::...-......::.(1,-2,2,-4,-4,-56,...).::.exp(-x).*.s(x).
(1,1).::.aC.::...-......::..A053871...............::.exp(-x)./.s(x).
..............................................................................
(2,1).::.B..::...-......::.(1,-A001813)...........::.s=[1+(1-4x)^(1/2)]/2....
(2,1).::.C..::...-......::..A001761...............::.1./.s(x)..
(2,1).::.rB.::...-......::.(1,0,-3,-20,-183,...)..::.exp(x).*.s(x)..
(2,1).::.rC.::...-......::.(1,2,7,46,485,...).....::.exp(x)./.s(x).
(2,1).::.aB.::...-......::.(1,-2,1,-10,-79,...)...::.exp(-x).*.s(x).
(2,1).::.aC.::...-......::.(1,0,3,20,237,...).....::.exp(-x)./.s(x)
..............................................................................
(1,2).::.B..::.~A134082.::.(1,-2,0,0,0,0,...).....::.s(x).=.1.-.2x
(1,2).::.C..::....-.....::..A000165...............::.1./.s(x)..
(1,2).::.rB.::....-.....::.(1,-1,-3,-5,-7,-9,...).::.exp(x).*.s(x).
(1,2).::.rC.::....-.....::..A010844...............::.exp(x)./.s(x)..
(1,2).::.aB.::....-.....::.(1,-3,5,-7,9,-11,...)..::.exp(-x).*.s(x).
(1,2).::.aC.::....-.....::..A000354...............::.exp(-x)./.s(x).
..............................................................................
(The tilde indicates the match is not exact--specifically, there are differences in signs from the true matrices.)
Note the row sums correspond to binomial transforms of s(x) and the alternating row sums, to inverse binomial transforms, or, finite differences.
Some additional examples:
C(1,2)*B(0,1) = B(1,-2)*C(0,-1) = [ binomial(n,k)*A002866(n-k) ] with asc. e.g.f. (1-x) / (1-2x).
B(1,2)*C(0,1) = C(1,-2)*B(0,-1) = 2I - A094587 with asc. e.g.f. (1-2x) / (1-x).

[G.f. for TB(n,k,t)] = GTB(u,x,t) = exp(u*x) * { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } = exp[(u+Pb(.,t))*x] where TB(n,k,t) = (D_x)^n (D_u)^k /k! GTB(u,x,t) eval. at u=x=0.
[G.f. for TC(n,k,t)] = GTC(u,x,t) = exp(u*x) / { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } = exp[(u+Pc(.,t))*x] where TC(n,k,t) = (D_x)^n (D_u)^k /k! GTC(u,x,t) eval. at u=x=0.
[E.g.f. for TB(n,k,t)] = I_o[2*(u*x)^(1/2)] * { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } and
[E.g.f. for TC(n,k,t)] = I_o[2*(u*x)^(1/2)] / { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t }
where I_o is the zeroth modified Bessel function of the first kind, i.e.,
I_o[2*(u*x)^(1/2)] = Sum_{j>=0} (u^j/j!) * (x^j/j!).
So [e.g.f. for TB(n,k)] = I_o[2*(u*x)^(1/2)] * (1 - 2x)^(1/2).

More terms from Tom Copeland, Dec 05 2007

A132159 Lower triangular matrix T(n,j) for double application of an iterated mixed order Laguerre transform inverse to A132014. Coefficients of Laguerre polynomials (-1)^n * n! * L(n,-2-n,x).

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 24, 18, 6, 1, 120, 96, 36, 8, 1, 720, 600, 240, 60, 10, 1, 5040, 4320, 1800, 480, 90, 12, 1, 40320, 35280, 15120, 4200, 840, 126, 14, 1, 362880, 322560, 141120, 40320, 8400, 1344, 168, 16, 1, 3628800, 3265920, 1451520, 423360, 90720, 15120

Offset: 0

Tom Copeland, Nov 01 2007

The matrix operation b = T*a can be characterized several ways in terms of the coefficients a(n) and b(n), their o.g.f.'s A(x) and B(x), or their e.g.f.'s EA(x) and EB(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a1(.),-1],0], umbrally,
where a1(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0]
2) b(n) = (-1)^n * n! * Lag(n,a(.),-2-n)
3) b(n) = Sum_{j=0..n} (-1)^j * binomial(n,j) * binomial(-2,j) * j! * a(n-j)
4) b(n) = Sum_{j=0..n} binomial(n,j) * (j+1)! * a(n-j)
5) B(x) = (1-xDx))^(-2) A(x), formally
6) B(x) = Sum_{j>=0} (-1)^j * binomial(-2,j) * (xDx)^j A(x)
= Sum_{j>=0} (j+1) * (xDx)^j A(x)
7) B(x) = Sum_{j>=0} (j+1) * x^j * D^j * x^j A(x)
8) B(x) = Sum_{j>=0} (j+1)! * x^j * Lag(j,-:xD:,0) A(x)
9) EB(x) = Sum_{j>=0} x^j * Lag[j,(.)! * Lag[.,a1(.),-1],0]
10) EB(x) = Sum_{j>=0} Lag[j,a1(.),-1] * (-x)^j / (1-x)^(j+1)
11) EB(x) = Sum_{j>=0} x^n * Sum_{j=0..n} (j+1)!/j! * a(n-j) / (n-j)!
12) EB(x) = Sum_{j>=0} (-x)^j * Lag[j,a(.),-2-j]
13) EB(x) = exp(a(.)*x) / (1-x)^2 = (1-x)^(-2) * EA(x)
14) T = A094587^2 = A132013^(-2) = A132014^(-1)
where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the D operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and associated operations described in A133314. Thus T(n,k) = binomial(n,k)*c(n-k) . c are also the coefficients in formulas 4 and 8.
The reciprocal sequence to c is d = (1,-2,2,0,0,0,...), so the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A132014. (A121757 is the reverse of T.)
These formulas are easily generalized for m applications of the basic operator n! Lag[n,(.)!*Lag[.,a(.),-1],0] by replacing 2 by m in formulas 2, 3, 5, 6, 12, 13 and 14, or (j+1)! by (m-1+j)!/(m-1)! in 4, 8 and 11. For further discussion of repeated applications of T, see A132014.
The row sums of T = [formula 4 with a(n) all 1] = [binomial transform of c] = [coefficients of B(x) with A(x) = 1/(1-x)] = A001339. Therefore the e.g.f. of A001339 = [formula 13 with a(n) all 1] = exp(x)*(1-x)^(-2) = exp(x)*exp[c(.)*x)] = exp[(1+c(.))*x].
Note the reciprocal is 1/{exp[(1+c(.))*x]} = exp(-x)*(1-x)^2 = e.g.f. of signed A002061 with leading 1 removed], which makes A001339 and the signed, shifted A002061 reciprocal arrays under the list partition transform of A133314.
The e.g.f. for the row polynomials (see A132382) implies they form an Appell sequence (see Wikipedia). - Tom Copeland, Dec 03 2013
As noted in item 12 above and reiterated in the Bala formula below, the e.g.f. is e^(x*t)/(1-x)^2, and the Poisson-Charlier polynomials P_n(t,y) have the e.g.f. (1+x)^y e^(-xt) (Feinsilver, p. 5), so the row polynomials R_n(t) of this entry are (-1)^n P_n(t,-2). The associated Appell sequence IR_n(t) that is the umbral compositional inverse of this entry's polynomials has the e.g.f. (1-x)^2 e^(xt), i.e., the e.g.f. of A132014 (noted above), and, therefore, the row polynomials (-1)^n PC(t,2). As umbral compositional inverses, R_n(IR.(t)) = t^n =  IR_n(R.(t)), where, by definition, P.(t)^n = P_n(t), is the umbral evaluation. - Tom Copeland, Jan 15 2016
T(n,k) is the number of ways to place (n-k) rooks in a 2 x (n-1) Ferrers board (or diagram) under the Goldman-Haglund i-row creation rook mode for i=2. Triangular recurrence relation is given by T(n,k) = T(n-1,k-1) + (n+1-k)*T(n-1,k). - Ken Joffaniel M. Gonzales, Jan 21 2016

			First few rows of the triangle are
    1;
    2,  1;
    6,  4,  1;
   24, 18,  6, 1;
  120, 96, 36, 8, 1;

Nathaniel Johnston, Rows 0..100, flattened
Paul Barry, A note on number triangles that are almost their own production matrix, arXiv:1804.06801 [math.CO], 2018.
P. Feinsilver, Lie algebras, representations, and analytic semigroups through dual vector fields
Jay Goldman and James Haglund, Generalized rook polynomials, J. Combin. Theory A 91 (2000), 509-530.
M. Janjic, Some classes of numbers and derivatives, JIS 12 (2009) 09.8.3.
Wikipedia, Appell sequence
Wikipedia, Sheffer sequence

Cf. A008277, A094587, A132013, A132382.

Columns: A000142 (k=0), A001563 (k=1), A001286 (k=2), A005990 (k=3), A061206 (k=4), A062199 (k=5), A062148 (k=6).

a132159 n k = a132159_tabl !! n !! k
a132159_row n = a132159_tabl !! n
a132159_tabl = map reverse a121757_tabl
-- Reinhard Zumkeller, Mar 06 2014

/* As triangle */ [[Binomial(n,k)*Factorial(n-k+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Feb 10 2016

T := proc(n,k) return binomial(n,k)*factorial(n-k+1): end: seq(seq(T(n,k),k=0..n),n=0..10); # Nathaniel Johnston, Sep 28 2011

nn=10;f[list_]:=Select[list,#>0&];Map[f,Range[0,nn]!CoefficientList[Series[Exp[y x]/(1-x)^2,{x,0,nn}],{x,y}]]//Grid  (* Geoffrey Critzer, Feb 15 2013 *)

flatten([[binomial(n,k)*factorial(n-k+1) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, May 19 2021

T(n,k) = binomial(n,k)*c(n-k).
From Peter Bala, Jul 10 2008: (Start)
T(n,k) = binomial(n,k)*(n-k+1)!.
T(n,k) = (n-k+1)*T(n-1,k) + T(n-1,k-1).
E.g.f.: exp(x*y)/(1-y)^2 = 1 + (2+x)*y + (6+4*x+x^2)*y^2/2! + ... .
This array is the particular case P(2,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
  n\k|0....................1...............2.........3.....4
  ----------------------------------------------------------
  0..|1.....................................................
  1..|a....................1................................
  2..|a(a+b)...............2a..............1................
  3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
  4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
  ...
See A094587 for some general properties of these arrays.
Other cases recorded in the database include: P(1,0) = Pascal's triangle A007318, P(1,1) = A094587, P(2,0) = A038207, P(3,0) = A027465, P(1,3) = A136215 and P(2,3) = A136216. (End)
Let f(x) = (1/x^2)*exp(-x). The n-th row polynomial is R(n,x) = (-x)^n/f(x)*(d/dx)^n(f(x)), and satisfies the recurrence equation R(n+1,x) = (x+n+2)*R(n,x)-x*R'(n,x). Cf. A094587. - Peter Bala, Oct 28 2011
Exponential Riordan array [1/(1 - y)^2, y]. The row polynomials R(n,x) thus form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x). Define a polynomial sequence P(n,x) of binomial type by setting P(n,x) = Product_{k = 0..n-1} (2*x + k) with the convention that P(0,x) = 1. Then the present triangle is the triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (24, 18, 6, 1) so P(3,x + 1) = (2*x + 2)*(2*x + 3)*(2*x + 4) = 24 + 18*(2*x) + 6*(2*x)*(2*x + 1) + (2*x)*(2*x + 1)*(2*x + 2). Matrix square of triangle A094587. - Peter Bala, Aug 29 2013
From Tom Copeland, Apr 21 2014: (Start)
T = (I-A132440)^(-2) = {2*I - exp[(A238385-I)]}^(-2) = unsigned exp[2*(I-A238385)] = exp[A005649(.)*(A238385-I)], umbrally, where I = identity matrix.
The e.g.f. is exp(x*y)*(1-y)^(-2), so the row polynomials form an Appell sequence with lowering operator D=d/dx and raising operator x+2/(1-D).
With L(n,m,x) = Laguerre polynomials of order m, the row polynomials are (-1)^n * n! * L(n,-2-n,x) = (-1)^n*(-2!/(-2-n)!)*K(-n,-2-n+1,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
Operationally, (-1)^n*n!*L(n,-2-n,-:xD:) = (-1)^n*x^(n+2)*:Dx:^n*x^(-2-n) = (-1)^n*x^2*:xD:^n*x^(-2) = (-1)^n*n!*binomial(xD-2,n) = (-1)^n*n!*binomial(-2,n)*K(-n,-2-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706.
The generalized Pascal triangle Bala mentions is a special case of the fundamental generalized factorial matrices in A133314. (End)
From Peter Bala, Jul 26 2021: (Start)
O.g.f: 1/y * Sum_{k >= 0} k!*( y/(1 - x*y) )^k =  1 + (2 + x)*y + (6 + 4*x + x^2)*y^2 + ....
First-order recurrence for the row polynomials: (n - x)*R(n,x) = n*(n - x + 1)*R(n-1,x) - x^(n+1) with R(0,x) = 1.
R(n,x) = (x + n + 1)*R(n-1,x) - (n - 1)*x*R(n-2,x) with R(0,x) = 1 and R(1,x) = 2 + x.
R(n,x) = A087981 (x = -2), A000255 (x = -1), A000142 (x = 0), A001339 (x = 1), A081923 (x = 2) and A081924 (x = 3). (End)

Formula 3) in comments corrected by Tom Copeland, Apr 20 2014

Title modified by Tom Copeland, Apr 23 2014

A090010 Permanent of (0,1)-matrix of size n X (n+d) with d=6 and n zeros not on a line.

Original entry on oeis.org

6, 43, 356, 3333, 34754, 398959, 4996032, 67741129, 988344062, 15434831091, 256840738076, 4536075689293, 84731451264186, 1668866557980343, 34563571477305464, 750867999393119889, 17072113130285524982, 405423357986250112699, 10037458628015142154452

Offset: 1

Jaap Spies, Dec 13 2003

Brualdi, Richard A. and Ryser, Herbert J., Combinatorial Matrix Theory, Cambridge NY (1991), Chapter 7.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Seok-Zun Song et al., Extremes of permanents of (0,1)-matrices, Lin. Algebra and its Applic. 373 (2003), pp. 197-210.

Cf. A000255, A000153, A000261, A001909, A001910, A055790, A090012-A090016.

A090010 := proc(n,d) local r; if (n=1) then r := d elif (n=2) then r := d^2+d+1 else r := (n+d-1)*A090010(n-1,d)+(n-1)*A090010(n-2,d) fi; RETURN(r); end: seq(A090010(n,6),n=1..18);

Rest[CoefficientList[Series[E^(-x)/(1-x)^7,{x,0,20}],x]*Range[0,20]!] (* Vaclav Kotesovec, Oct 21 2012 *)

x='x+O('x^66); Vec(serlaplace(-1+exp(-x)/(1-x)^7)) \\ Joerg Arndt, May 11 2013

a(n) = (n+5)*a(n-1) + (n-1)*a(n-2), a(1)=6, a(2)=43
G.f.: -1+hypergeom([1,7],[],x/(x+1))/(x+1) - Mark van Hoeij, Nov 07 2011
E.g.f.: -1 + exp(-x)/(1-x)^7. - Vaclav Kotesovec, Oct 21 2012
a(n) ~ n!*n^6/(720*e). - Vaclav Kotesovec, Oct 21 2012

Corrected by Jaap Spies, Jan 26 2004

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A000261 a(n) = n*a(n-1) + (n-3)*a(n-2), with a(1) = 0, a(2) = 1.

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A001910 a(n) = n*a(n-1) + (n-5)*a(n-2).

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A068106 Euler's difference table: triangle read by rows, formed by starting with factorial numbers (A000142) and repeatedly taking differences. T(n,n) = n!, T(n,k) = T(n,k+1) - T(n-1,k).

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A187235 Number of ways to place n nonattacking semi-bishops on an n X n board.

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A087981 E.g.f.: exp(-2*x) / (1-x)^2.

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A180191 Number of permutations of [n] having at least one succession. A succession of a permutation p is a position i such that p(i+1)-p(i) = 1.

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A000261 a(n) = na(n-1) + (n-3)a(n-2), with a(1) = 0, a(2) = 1.

A001910 a(n) = na(n-1) + (n-5)a(n-2).