A238385 -id:A238385 - cp's OEIS frontend

A104712 Pascal's triangle, with the first two columns removed.

1, 3, 1, 6, 4, 1, 10, 10, 5, 1, 15, 20, 15, 6, 1, 21, 35, 35, 21, 7, 1, 28, 56, 70, 56, 28, 8, 1, 36, 84, 126, 126, 84, 36, 9, 1, 45, 120, 210, 252, 210, 120, 45, 10, 1, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1, 78, 286, 715

Offset: 2

Gary W. Adamson, Mar 19 2005

A000295 (Eulerian numbers) gives the row sums.
Write A004736 and Pascal's triangle as infinite lower triangular matrices A and B; then A*B is this triangle.
From Peter Luschny, Apr 10 2011: (Start)
A slight variation has a combinatorial interpretation: remove the last column and the second one from Pascal's triangle. Let P(m, k) denote the set partitions of {1,2,..,n} with the following properties:
(a) Each partition has at least one singleton block;
(c) k is the size of the largest block of the partition;
(b) m = n - k + 1 is the number of parts of the partition.
Then A000295(n) = Sum_{k=1..n} card(P(n-k+1,k)).
For instance, A000295(4) = P(4,1) + P(3,2) + P(2,3) + P(1,4) = card({1|2|3|4}) + card({1|2|34, 1|3|24,1|4|23, 2|3|14, 2|4|13, 3|4|12}) + card({1|234, 2|134, 3|124, 4|123}) = 1 + 6 + 4 = 11.
This interpretation can be superimposed on the sequence by changing the offset to 1 and adding the value 1 in front. The triangle then starts
  1;
  1,  3;
  1,  6,  4;
  1, 10, 10,  5;
  1, 15, 20, 15,  6;
  ...
(End)
Diagonal sums are A001924(n+1). - Philippe Deléham, Jan 11 2014
Relation to K-theory: T acting on the column vector (d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168, A111492, A238363, and A135278. - Tom Copeland, Apr 11 2014

			The triangle a(n, k) begins:
  n\k  2   3   4    5    6    7    8   9  10 11 12 13
  2:   1
  3:   3   1
  4:   6   4   1
  5:  10  10   5    1
  6:  15  20  15    6    1
  7:  21  35  35   21    7    1
  8:  28  56  70   56   28    8    1
  9:  36  84 126  126   84   36    9   1
  10: 45 120 210  252  210  120   45  10   1
  11: 55 165 330  462  462  330  165  55  11  1
  12: 66 220 495  792  924  792  495 220  66 12  1
  13: 78 286 715 1287 1716 1716 1287 715 286 78 13  1
... reformatted. - _Wolfdieter Lang_, Mar 20 2015

G. C. Greubel, Rows n=2..100 of triangle, flattened
D. Dugger, A Geometric Introduction to K-Theory
Candice A. Marshall, Construction of Pseudo-Involutions in the Riordan Group, Dissertation, Morgan State University, 2017.
T. Saito, The discriminant and the determinant of a hypersurface of even dimension (p. 4), arXiv:1110.1717 [math.AG], 2011-2012.

Cf. A000295, A007318, A008292, A104713, A027641/A027642 (first Bernoulli numbers B-), A164555/A027642 (second Bernoulli numbers B+), A176327/A176289.

/* As triangle */ [[Binomial(n, k): k in [2..n]]: n in [2..10]]; // G. C. Greubel, May 15 2018

t[n_, k_] := Binomial[n, k]; Table[ t[n, k], {n, 2, 13}, {k, 2, n}] // Flatten (* Robert G. Wilson v, Apr 16 2011 *)

for(n=2, 10, for(k=2,n, print1(binomial(n,k), ", "))) \\ G. C. Greubel, May 15 2018

T(n,k) = binomial(n,k), for 2 <= k <= n.
From Peter Bala, Jul 16 2013: (Start)
The following remarks assume an offset of 0.
Riordan array (1/(1 - x)^3, x/(1 - x)).
O.g.f.: 1/(1 - t)^2*1/(1 - (1 + x)*t) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2 + ....
E.g.f.: (1/x*d/dt)^2 (exp(t)*(exp(x*t) - 1 - x*t)) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2/2! + ....
The infinitesimal generator for this triangle has the sequence [3,4,5,...] on the main subdiagonal and 0's elsewhere. (End)
As triangle T(n,k), 0<=k<=n: T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 2*T(n-2,k-1) + T(n-3,k) + T(n-3,k-1), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 11 2014
From Tom Copeland, Apr 11 2014: (Start)
A) The infinitesimal generator for this matrix is given in A132681 with m=2. See that entry for numerous relations to differential operators and the Laguerre polynomials of order m=2, i.e., Lag(n,t,2) = Sum_{j=0..n} binomial(n+2,n-j)*(-t)^j/j!.
B) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^3 }
C) O.g.f. of row e.g.f.s: exp[t*x/(1-x)]/(1-x)^3 = [Sum_{n>=0} x^n * Lag(n,-t,2)] = 1 + (3 + t)*x + (6 + 4t + t^2/2!)*x^2 + (10 + 10t + 5t^2/2! + t^3/3!)*x^3 + ....
D) E.g.f. of row o.g.f.s: [(1+t)*exp((1+t)*x) - (1+t+t*x)exp(x)]/t^2. (End)
O.g.f. for m-th row (m=n-2): [(1+x)^(m+2)-(1+(m+2)*x)]/x^2. - Tom Copeland, Apr 16 2014
Reverse T = [St2]*dP*[St1]- dP = [St2]*(exp(x*M)-I)*[St1]-(exp(x*M)-I) with top two rows of zeros removed, [St1]=padded A008275 just as [St2]=A048993=padded A008277, dP= A132440, M=A238385-I, and I=identity matrix. Cf. A238363. - Tom Copeland, Apr 26 2014
O.g.f. of column k (with k leading zeros): (x^k)/(1-x)^(k+1), k >= 2. - Wolfdieter Lang, Mar 20 2015

Edited and extended by David Wasserman, Jul 03 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

A055137 Regard triangle of rencontres numbers (see A008290) as infinite matrix, compute inverse, read by rows.

Original entry on oeis.org

1, 0, 1, -1, 0, 1, -2, -3, 0, 1, -3, -8, -6, 0, 1, -4, -15, -20, -10, 0, 1, -5, -24, -45, -40, -15, 0, 1, -6, -35, -84, -105, -70, -21, 0, 1, -7, -48, -140, -224, -210, -112, -28, 0, 1, -8, -63, -216, -420, -504, -378, -168, -36, 0, 1, -9, -80, -315, -720

Offset: 0

Christian G. Bower, Apr 25 2000

The n-th row consists of coefficients of the characteristic polynomial of the adjacency matrix of the complete n-graph.
Triangle of coefficients of det(M(n)) where M(n) is the n X n matrix m(i,j)=x if i=j, m(i,j)=i/j otherwise. - Benoit Cloitre, Feb 01 2003
T is an example of the group of matrices outlined in the table in A132382--the associated matrix for rB(0,1). The e.g.f. for the row polynomials is exp(x*t) * exp(x) *(1-x). T(n,k) = Binomial(n,k)* s(n-k) where s = (1,0,-1,-2,-3,...) with an e.g.f. of exp(x)*(1-x) which is the reciprocal of the e.g.f. of A000166. - Tom Copeland, Sep 10 2008
Row sums are: {1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...}. - Roger L. Bagula, Feb 20 2009
T is related to an operational calculus connecting an infinitesimal generator for fractional integro-derivatives with the values of the Riemann zeta function at positive integers (see MathOverflow links). - Tom Copeland, Nov 02 2012
The submatrix below the null subdiagonal is signed and row reversed A127717. The submatrix below the diagonal is A074909(n,k)s(n-k) where s(n)= -n, i.e., multiply the n-th diagonal by -n. A074909 and its reverse A135278 have several combinatorial interpretations. - Tom Copeland, Nov 04 2012
T(n,k) is the difference between the number of even (A145224) and odd (A145225) permutations (of an n-set) with exactly k fixed points. - Julian Hatfield Iacoponi, Aug 08 2024

			1; 0,1; -1,0,1; -2,-3,0,1; -3,-8,-6,0,1; ...
(Bagula's matrix has a different sign convention from the list.)
From _Roger L. Bagula_, Feb 20 2009: (Start)
  { 1},
  { 0,   1},
  {-1,   0,    1},
  { 2,  -3,    0,    1},
  {-3,   8,   -6,    0,     1},
  { 4, -15,   20,  -10,     0,    1},
  {-5,  24,  -45,   40,   -15,    0,    1},
  { 6, -35,   84, -105,    70,  -21,    0,   1},
  {-7,  48, -140,  224,  -210,  112,  -28,   0,   1},
  { 8, -63,  216, -420,   504, -378,  168, -36,   0, 1},
  {-9,  80, -315,  720, -1050, 1008, -630, 240, -45, 0, 1}
(End)
R(3,x) = (-1)^3*Sum_{permutations p in S_3} sign(p)*(-x)^(fix(p)).
    p   | fix(p) | sign(p) | (-1)^3*sign(p)*(-x)^fix(p)
========+========+=========+===========================
  (123) |    3   |    +1   |      x^3
  (132) |    1   |    -1   |       -x
  (213) |    1   |    -1   |       -x
  (231) |    0   |    +1   |       -1
  (312) |    0   |    +1   |       -1
  (321) |    1   |    -1   |       -x
========+========+=========+===========================
                           | R(3,x) = x^3 - 3*x - 2
- _Peter Bala_, Aug 08 2011

Norman Biggs, Algebraic Graph Theory, 2nd ed. Cambridge University Press, 1993. p. 17.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p.184 problem 3.

Problem B6, The 66th William Lowell Putnam Mathematical Competition Saturday, Dec 03 2005
M. Bhargava, K. Kedlaya, and L. Ng, Solutions to the 66th William Lowell Putnam Mathematical Competition Saturday, Dec 03 2005
T. Copeland, Cycling through the Zeta Garden: Zeta functions for graphs, cycle index polynomials, and determinants
T. Copeland, Riemann zeta function at positive integers and an Appell sequence of polynomials related to fractional calculus

Cf. A005563, A005564 (absolute values of columns 1, 2).
Cf. A008290, A133314, A238363, A238385.
Cf. A000312.
Cf. A145224, A145225, A003221, A000387.

M[n_] := Table[If[i == j, x, 1], {i, 1, n}, {j, 1, n}]; a = Join[{{1}}, Flatten[Table[CoefficientList[Det[M[n]], x], {n, 1, 10}]]] (* Roger L. Bagula, Feb 20 2009 *)
t[n_, k_] := (k-n+1)*Binomial[n, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 29 2013, after Pari *)

T(n,k)=(1-n+k)*if(k<0 || k>n,0,n!/k!/(n-k)!)

G.f.: (x-n+1)*(x+1)^(n-1) = Sum_(k=0..n) T(n,k) x^k.
T(n, k) = (1-n+k)*binomial(n, k).
k-th column has o.g.f. x^k(1-(k+2)x)/(1-x)^(k+2). k-th row gives coefficients of (x-k)(x+1)^k. - Paul Barry, Jan 25 2004
T(n,k) = Coefficientslist[Det[Table[If[i == j, x, 1], {i, 1, n}, {k, 1, n}],x]. - Roger L. Bagula, Feb 20 2009
From Peter Bala, Aug 08 2011: (Start)
Given a permutation p belonging to the symmetric group S_n, let fix(p) be the number of fixed points of p and sign(p) its parity. The row polynomials R(n,x) have a combinatorial interpretation as R(n,x) = (-1)^n*Sum_{permutations p in S_n} sign(p)*(-x)^(fix(p)). An example is given below.
Note: The polynomials P(n,x) = Sum_{permutations p in S_n} x^(fix(p)) are the row polynomials of the rencontres numbers A008290. The integral results Integral_{x = 0..n} R(n,x) dx = n/(n+1) = Integral_{x = 0..-1} R(n,x) dx lead to the identities Sum_{p in S_n} sign(p)*(-n)^(1 + fix(p))/(1 + fix(p)) = (-1)^(n+1)*n/(n+1) = Sum_{p in S_n} sign(p)/(1 + fix(p)). The latter equality was Problem B6 in the 66th William Lowell Putnam Mathematical Competition 2005. (End)
From Tom Copeland, Jul 26 2017: (Start)
The e.g.f. in Copeland's 2008 comment implies this entry is an Appell sequence of polynomials P(n,x) with lowering and raising operators L = d/dx and R = x + d/dL log[exp(L)(1-L)] = x+1 - 1/(1-L) = x - L - L^2 - ... such that L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
P(n,x) = (1-L) exp(L) x^n = (1-L) (x+1)^n = (x+1)^n - n (x+1)^(n-1) = (x+1-n)(x+1)^(n-1) = (x+s.)^n umbrally, where (s.)^n = s_n = P(n,0).
The formalism of A133314 applies to the pair of entries A008290 and A055137.
The polynomials of this pair P_n(x) and Q_n(x) are umbral compositional inverses; i.e., P_n(Q.(x)) = x^n = Q_n(P.(x)), where, e.g., (Q.(x))^n = Q_n(x).
The exponential infinitesimal generator (infinigen) of this entry is the negated infinigen of A008290, the matrix (M) noted by Bala, related to A238363. Then e^M = [the lower triangular A008290], and e^(-M) = [the lower triangular A055137]. For more on the infinigens, see A238385. (End)
From the row g.f.s corresponding to Bagula's matrix example below, the n-th row polynomial has a zero of multiplicity n-1 at x = 1 and a zero at x = -n+1. Since this is an Appell sequence dP_n(x)/dx = n P_{n-1}(x), the critical points of P_n(x) have the same abscissas as the zeros of P_{n-1}(x); therefore, x = 1 is an inflection point for the polynomials of degree > 2 with P_n(1) = 0, and the one local extremum of P_n has the abscissa x = -n + 2 with the value (-n+1)^{n-1}, signed values of A000312. - Tom Copeland, Nov 15 2019
From Julian Hatfield Iacoponi, Aug 08 2024: (Start)
T(n,k) = A145224(n,k) - A145225(n,k).
T(n,k) = binomial(n,k)*(A003221(n-k)-A000387(n-k)). (End)

Additional comments from Michael Somos, Jul 04 2002

A132013 T(n,j) for an iterated mixed order Laguerre transform. Coefficients of the normalized generalized Laguerre polynomials (-1)^nn!L(n,1-n,x).

Original entry on oeis.org

1, -1, 1, 0, -2, 1, 0, 0, -3, 1, 0, 0, 0, -4, 1, 0, 0, 0, 0, -5, 1, 0, 0, 0, 0, 0, -6, 1, 0, 0, 0, 0, 0, 0, -7, 1, 0, 0, 0, 0, 0, 0, 0, -8, 1, 0, 0, 0, 0, 0, 0, 0, 0, -9, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -10, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -11, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -12, 1

Offset: 0

Tom Copeland, Oct 30 2007

The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.s A(x) and B(x), or e.g.f.s EA(x) and EB(x).
1) b(0) = a(0), b(n) = a(n) - n*a(n-1) for n > 0
2) b(n) = n! Lag{n,(.)!*Lag[.,a(.),0],-1}, umbrally
3) b(n) = n! Sum_{j=0..min(1,n)} (-1)^j * binomial(n,j)*a(n-j)/(n-j)!
4) b(n) = (-1)^n n! Lag(n,a(.),1-n)
5) B(x) = (1-xDx) A(x) = [1-x*Lag(1,-xD:,0)] A(x)
6) EB(x) = (1-x) EA(x),
where D is the derivative w.r.t. x and Lag(n,x,m) is the associated Laguerre polynomial of order m. These formulas are easily generalized for repeated applications of the operator.
c = (1,-1,0,0,0,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314. The reciprocal sequence is d = (0!,1!,2!,3!,4!,...).
Consequently, the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A094587, which has the property that the terms at and below TI(m,m) are the associated sequence under the list partition transform for the inverse for T^(m+1) for m=0,1,2,3,... .
Row sums of T = [formula 3 with all a(n) = 1] = [binomial transform of c] = [coefficients of B(x) with A(x) = 1/(1-x)] = A024000 = (1,0,-1,-2,-3,...), with e.g.f. = [EB(x) with EA(x) = exp(x)] = (1-x) * exp(x) = exp(x)*exp(c(.)*x) = exp[(1+c(.))*x].
Alternating row sums of T = [formula 3 with all a(n) = (-1)^n] = [finite differences of c] = [coefficients of B(x) with A(x) = 1/(1+x)] = (1,-2,3,-4,...), with e.g.f. = [EB(x) with EA(x) = exp(-x)] = (1-x) * exp(-x) = exp(-x)*exp(c(.)*x) = exp[-(1-c(.))*x].
An e.g.f. for the o.g.f.s for repeated applications of T on A(x) is given by
exp[t*(1-xDx)] A(x) = e^t * Sum_{n=0,1,...} (-t*x)^n * Lag(n,-:xD:,0) A(x)
= e^t * exp{[-t*u/(1+t*u)]*:xD:} / (1+t*u) A(x) (eval. at u=x)
= e^t * A[x/(1+t*x)]/(1+t*x) .
See A132014 for more notes on repeated applications.

			First few rows of the triangle are
   1;
  -1,  1;
   0, -2,  1;
   0,  0, -3,  1;
   0,  0,  0, -4,  1;
   0,  0,  0,  0, -5,  1;
   0,  0,  0,  0,  0, -6,  1;
   0,  0,  0,  0,  0,  0, -7,  1;

G. C. Greubel, Rows n=0..100 of triangle, flattened
T. Copeland, Compositional inverse operators and Sheffer sequences, 2016.
M. Janjic, Some classes of numbers and derivatives, JIS 12 (2009) #09.8.3.
Wikipedia, Appell sequence

Cf. A235706, A132382, A132159, A094587, A132014, A154955, A235706.

c := n -> `if`(n=0,1,`if`(n=1,-1,0)):
T := (n,k) -> binomial(n,k)*c(n-k); # Peter Luschny, Nov 14 2016

Table[PadLeft[{-n, 1}, n+1], {n, 0, 13}] // Flatten (* Jean-François Alcover, Apr 29 2014 *)

row(n) = Vecrev((-1)^n*n!*pollaguerre(n, 1-n)); \\ Michel Marcus, Jul 26 2021

T(n,k) = binomial(n,k)*c(n-k), with the sequence c defined in the comments.
E.g.f.: exp(x*y)(1-x), which implies the row polynomials form an Appell sequence. More relations can be found in A132382. - Tom Copeland, Dec 03 2013
From Tom Copeland, Apr 21 2014: (Start)
Change notation letting L(n,m,x) = Lag(n,x,m).
Row polynomials: (-1)^n*n!*L(n,1-n,x) = -x^(n-1)*L(1,n-1,x) =
(-1)^n*(1/(1-n)!)*K(-n,1-n+1,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
For the row polynomials, the lowering operator = d/dx and the raising operator = x - 1/(1-D).
T = I - A132440 = 2*I - exp[A238385-I] = signed exp[A238385-I], where I = identity matrix.
Operationally, (-1)^n*n!*L(n,1-n,-:xD:) = -x^(n-1)*:Dx:^n*x^(1-n) = (-1)^n*x^(-1)*:xD:^n*x = (-1)^n*n!*binomial(xD+1,n) = (-1)^n*n!*binomial(1,n)*K(-n,1-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706. (End)
The unsigned row polynomials have e.g.f. (1+t)e^(xt) = exp(t*p.(x)), umbrally, and p_n(x) = (1+D) x^n. With q_n(x) the row polynomials of A094587, p_n(x) = u_n(q.(v.(x))), umbrally, where u_n(x) = (-1)^n v_n(-x) = (-1)^n Lah_n(x), the Lah polynomials with e.g.f. exp[x*t/(t-1)]. This has the matrix form unsigned [T] = [p] = [u]*[q]*[v]. Conversely, q_n(x) = v_n (p.(u.(x))). - Tom Copeland, Nov 10 2016
n-th row polynomial: n!*Sum_{k = 0..n} (-1)^k*binomial(n,k)*Lag(k,1,x). - Peter Bala, Jul 25 2021

Title modified by Tom Copeland, Apr 21 2014

A132014 T(n,j) for double application of an iterated mixed order Laguerre transform: Coefficients of Laguerre polynomial (-1)^nn!L(n,2-n,x).

Original entry on oeis.org

1, -2, 1, 2, -4, 1, 0, 6, -6, 1, 0, 0, 12, -8, 1, 0, 0, 0, 20, -10, 1, 0, 0, 0, 0, 30, -12, 1, 0, 0, 0, 0, 0, 42, -14, 1, 0, 0, 0, 0, 0, 0, 56, -16, 1, 0, 0, 0, 0, 0, 0, 0, 72, -18, 1, 0, 0, 0, 0, 0, 0, 0, 0, 90, -20, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 110, -22, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 132, -24, 1

Offset: 0

Tom Copeland, Oct 30 2007, Nov 05 2007, Nov 11 2007

The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.s A(x) and B(x), or e.g.f.s EA(x) and EB(x).
1) b(0) = a(0), b(1) = a(n) - 2 a(0), b(n) = a(n) - 2n a(n-1) + n(n-1) a(n-2) for n > 0.
2) b(n) = n! Lag{n,(.)!*Lag[.,a1(.),0],-1}, umbrally, where a1(n) = n! Lag{n,(.)!*Lag[.,a(.),0],-1}.
3) b(n) = n! Sum_{j=0..min(2,n)} (-1)^j * binomial(n,j)*a(n-j)/(n-j)!
4) b(n) = (-1)^n n! Lag(n,a(.),2-n)
5) B(x) = (1-xDx)^2 A(x)
6) B(x) = Sum_{j=0..2} {(-1)^j * binomial(2,j)*j!*x^j*Lag(j,-:xD:,0)} A(x)
where D is the derivative w.r.t. x, (:xD:)^j = x^j*D^j and Lag(n,x,m) is the associated Laguerre polynomial of order m.
7) EB(x) = (1-x)^2 EA(x)
8) T = S^2 = A132013^2 = A094587^(-2) = A132159^(-1).
c = (1,-2,2,0,0,...) is the sequence associated to T under the list partition transform and associated operations described in A133314. c are also the coefficients in formula 6. Thus T(n,k) = binomial(n,k)*c(n-k).
The reciprocal sequence to c is d = (1!,2!,3!,4!,...), so the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A132159.
These formulas are easily generalized for m applications of the basic operator n! Lag[n,(.)!*Lag[.,a(.),0],-1] by replacing 2 with m in formulas 3, 4, 5, 6 and 7.
The generalized c are given by the generalized coefficients of 6, i.e.,
c(n) = (-1)^n * binomial(m,n)*n! = (-1)^n * m!/(m-n)!.
The generalized d are given by the array at and below the term SI(m-1,m-1) in SI(n,k) = binomial(n,k) * (n-k)!, the inverse of S; i.e.,
d(n) = SI(m-1+n,m-1) = binomial(m-1+n,m-1) * n! = (m-1+n)!/(m-1)!.
As an aside, this shows that the signed falling factorials and the rising factorials form reciprocal pairs under the list partition transform of A133314.
Row sums of T = [formula 3 with all a(n) = 1] = [binomial transform of c] = [coefficients of B(x) with A(x) = 1/(1-x)] = (1,-1,-1,1,5,11,19,...),
with e.g.f. = [EB(x) with EA(x) = exp(x)] = (1-x)^2 * exp(x) = exp(x)*exp(c(.)*x) = exp[(1+c(.))*x].
Alternating row sums of T = [formula 3 with all a(n) = (-1)^n] = [finite differences of c] = [coefficients of B(x) with A(x) = 1/(1+x)] = (1,-3,7,-13,21,-31,...) = (-1)^n A002061(n+1),
with e.g.f. = [EB(x) with EA(x) = exp(-x)] = (1-x)^2 * exp(-x) = exp(- x)*exp(c(.)*x) = exp[-(1-c(.))*x].
See A132159 for a relation to the Poisson-Charlier polynomials. - Tom Copeland, Jan 15 2016

			First few rows of the triangle are
   1;
  -2,   1;
   2,  -4,   1;
   0,   6,  -6,   1;
   0,   0,  12,  -8,   1;
   0,   0,   0,  20, -10,   1;

G. C. Greubel, Rows n=0..100 of triangle, flattened
M. Janjic, Some classes of numbers and derivatives, JIS 12 (2009) 09.8.3.
Wikipedia, Appell sequence

Cf. A132382, A110330, A132159, A094587, A132013, A235706.

m = 12; s = Exp[x*y]*(1 - x)^2 + O[x]^(m + 2) + O[y]^(m + 2); T[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {y, 0, k}]*n!; T[0, 0] = 1; Table[T[n, k], {n, 0, m}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 09 2015 *)

row(n) = Vecrev((-1)^n*n!*pollaguerre(n, 2-n)); \\ Michel Marcus, Feb 06 2021

T(n,k) = binomial(n,k)*c(n-k).
E.g.f. for row polynomials: exp(x*y)(1-x)^2. Implies the row polynomials form an Appell sequence (see Wikipedia). - Tom Copeland, Dec 03 2013
From Tom Copeland, Apr 21 2014: (Start)
Change notation letting L(n,m,x) = Lag(n,x,m).
Row polynomials: (-1)^n*n!*L(n,2-n,x) = (-1)^n*(-x)^(n-2)*2!*L(2,n-2,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).
For the row polynomials, the lowering operator = d/dx and the raising operator = x - 2/(1-D).
T = (I - A132440)^2 = [2*I - exp(A238385-I)]^2 = signed exp[2*(A238385-I)], where I = identity matrix.
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. (End)
n-th row polynomial: n!*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*Lag(k,2,x). - Peter Bala, Jul 25 2021

Title modified by Tom Copeland, Apr 21 2014

Missing term -18 inserted in 10th row by Jean-François Alcover, Jul 09 2015

A235706 (I + A132440)^3: Coefficients for normalized generalized Laguerre polynomials n!*Lag(n, 3-n, -x).

Original entry on oeis.org

1, 3, 1, 6, 6, 1, 6, 18, 9, 1, 0, 24, 36, 12, 1, 0, 0, 60, 60, 15, 1, 0, 0, 0, 120, 90, 18, 1, 0, 0, 0, 0, 210, 126, 21, 1, 0, 0, 0, 0, 0, 336, 168, 24, 1, 0, 0, 0, 0, 0, 0, 504, 216, 27, 1, 0, 0, 0, 0, 0, 0, 0, 720, 270, 30, 1

Offset: 0

Tom Copeland, Apr 20 2014

The associated Laguerre polynomials n!*Lag(n,3-n,-x) are related to the rook polynomials of a rectangular 3 X n chessboard by R(3,n,x) = n!*x^n*Lag(n,3-n,-1/x), which are also the matching polynomials, or generating function of the number of k-edge matchings, of the complete bipartite graph K(m,n), or biclique (cf. Wikipedia for details).

The formulas here and below can be naturally extended with 3 replaced by any positive integer m. For m = 1 and 2, see unsigned A132013 and A132014. The formulas there can be extrapolated to apply to this matrix.

			Triangle begins:
  1;
  3,  1;
  6,  6,  1;
  6, 18,  9,  1;
  0, 24, 36, 12,  1;
  0,  0, 60, 60, 15, 1;
  ...

Michel Marcus, Rows n = 0..50 of triangle, flattened

Cf. A007318, A008306 for a generalization, A132013, A132014, A132440, A238363, A238385.
....................................
With 0th row: 1
n-th row: n!*Lag(n,3-n,-x)
....................................
1st: 1!*Lag(1,2,-x)  = A062139(1,k,-x)
2nd: 2!*Lag(2,1,-x)  = A105278(2,k,x)
3rd: 3!*Lag(3,0,-x)  = A021009(3,k,-x)
4th: 4!*Lag(4,-1,-x) = A111596(4,k,-x)
5th: 5!*Lag(5,-2,-x) = cf. x^2*A062139(3,k,x)
6th: 6!*Lag(6,-3,-x) = cf. x^3*A062137(3,k,-x)
....................................
n-th row: x^(n-3)*3!*Lag(3,n-3,-x)
....................................
1st: x^(-2)*3!Lag(3,-2,-x) = cf. x^(-2)*[x^2*A062139(1,k,x)]
2nd: x^(-1)*3!Lag(3,-1,-x) = x^(-1)*A111596(3,k,-x)
3rd: x^0*3!Lag(3,0,-x)     = x^0*A021009(3,k,-x)
4th: x^1*3!Lag(3,1,-x)     = x^1*A105278(3,k,x)
5th: x^2*3!Lag(3,2,-x)     = x^2*A062139(3,k,-x)
6th: x^3*3!Lag(3,3,-x)     = x^3*A062137(3,k,-x)

/* As triangle */ [[Binomial(3, n-k)*Factorial(n)/Factorial(k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 28 2017

Table[Binomial[3, n - k] n! / k!, {n, 0, 9}, {k, 0, n}]//Flatten (* Vincenzo Librandi, Jul 28 2017 *)

T(n,k) = binomial(3,n-k)*n!/k!
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 28 2017

row(n) = Vecrev(n!*pollaguerre(n, 3-n, -x)); \\ Michel Marcus, Feb 06 2021

T(n,k) = binomial(3,n-k)*n!/k! = binomial(n,k)*3!/(3-n+k)!.
E.g.f.: exp(y*x)(1+y)^3, so this is an Appell sequence of polynomials with lowering operator L= D= d/dx and raising operator R = x + 3/(1+D).
E.g.f. of inverse matrix is exp(x*y)/(1+y)^3.
Multiply the n-th diagonal of the Pascal matrix A007318 by d(0)=1, d(1)=3, d(2)=6, d(3)=6, and d(n)=0 for n>3 to obtain T.
Row polynomials: n!*Lag(n,3-n,-x) = x^(n-3)*3!*Lag(3,n-3,-x) =
(3!/(3-n)!)*K(-n,3-n+1,-x) where K is Kummer's confluent hypergeometric function (as a limit of n+c as c tends to zero).
T = (I + A132440)^3 = exp[3*(A238385-I)]. I = identity matrix.
Operationally, n!Lag(n,3-n,-:xD:) = x^(n-3)*:Dx:^n*x^(3-n) = x^(-3)*:xD:^n*x^3 = n!*binomial(xD+3,n) = n!*binomial(3,n)*K(-n,3-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators.
n-th row polynomial: n!*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*Lag(k,3,-x). - Peter Bala, Jul 25 2021

cp's OEIS Frontend

A104712 Pascal's triangle, with the first two columns removed.

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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).

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A055137 Regard triangle of rencontres numbers (see A008290) as infinite matrix, compute inverse, read by rows.

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A132013 T(n,j) for an iterated mixed order Laguerre transform. Coefficients of the normalized generalized Laguerre polynomials (-1)^n*n!*L(n,1-n,x).

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A132014 T(n,j) for double application of an iterated mixed order Laguerre transform: Coefficients of Laguerre polynomial (-1)^n*n!*L(n,2-n,x).

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A235706 (I + A132440)^3: Coefficients for normalized generalized Laguerre polynomials n!*Lag(n, 3-n, -x).

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A132013 T(n,j) for an iterated mixed order Laguerre transform. Coefficients of the normalized generalized Laguerre polynomials (-1)^nn!L(n,1-n,x).

A132014 T(n,j) for double application of an iterated mixed order Laguerre transform: Coefficients of Laguerre polynomial (-1)^nn!L(n,2-n,x).