cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A143405 Number of forests of labeled rooted trees of height at most 1, with n labels, where any root may contain >= 1 labels, also row sums of A143395, A143396 and A143397.

Original entry on oeis.org

1, 1, 4, 17, 89, 552, 3895, 30641, 265186, 2497551, 25373097, 276105106, 3199697517, 39297401197, 509370849148, 6943232742493, 99217486649933, 1482237515573624, 23093484367004715, 374416757914118941, 6304680593346141746, 110063311977033807187
Offset: 0

Views

Author

Alois P. Heinz, Aug 12 2008

Keywords

Comments

a(n) is the number of the partitions of an n-set where each block is endowed with a nonempty subset. - Emanuele Munarini, Sep 15 2016

Examples

			a(2) = 4, because there are 4 forests for 2 labels: {1,2}, {1}{2}, {1}<-2, {2}<-1.
		

Crossrefs

Programs

  • Maple
    a:= n-> add(add(binomial(n, t)*Stirling2(t, k)*k^(n-t), t=k..n), k=0..n):
    seq(a(n), n=0..30);
    # second Maple program:
    a:= proc(n) option remember; `if`(n=0, 1, add(
          a(n-j)*binomial(n-1, j-1)*(2^j-1), j=1..n))
        end:
    seq(a(n), n=0..23);  # Alois P. Heinz, Oct 05 2019
  • Mathematica
    CoefficientList[Series[Exp[Exp[t] (Exp[t] - 1)], {t, 0, 12}], t] Range[0, 12]! (* Emanuele Munarini, Sep 15 2016 *)
    Table[Sum[Binomial[n, k] 2^k BellB[k] BellB[n - k, -1], {k, 0, n}], {n, 0, 12}] (* Emanuele Munarini, Sep 15 2016 *)
    Table[Sum[BellY[n, k, 2^Range[n] - 1], {k, 0, n}], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 09 2016 *)
  • PARI
    a(n) = sum(k=0, n, k!*sum(j=0, k\2, 1/(j!*(k-2*j)!))*stirling(n, k, 2)); \\ Seiichi Manyama, May 14 2022

Formula

a(n) = Sum_{k=0..n} Sum_{t=k..n} C(n,t) * Stirling2(t,k)*k^(n-t).
a(n) = Sum_{k=0..n} Sum_{t=0..k} C(n,k) * Stirling2(k,t)*t^(n-k).
a(n) = Sum_{k=0..n} Sum_{t=0..k} C(n,k-t) * Stirling2(n-(k-t),t)*t^(k-t).
E.g.f.: exp(exp(x)*(exp(x)-1)). - Vladeta Jovovic, Dec 08 2008
a(n) = sum(binomial(n,k)*2^k*bell(k)*S(n-k,-1),k=0..n), where bell(n) are the Bell numbers (A000110) and S(n,x) = sum(Stirling2(n,k)*x^k,k=0..n) are the Stirling (or exponential) polynomials. - Emanuele Munarini, Sep 15 2016
Identity: sum(binomial(n,k)*a(k)*bell(n-k),k=0..n) = 2^n*bell(n). - Emanuele Munarini, Sep 15 2016
a(n) = Sum_{k=0..n} A047974(k) * Stirling2(n,k). - Seiichi Manyama, May 14 2022
a(n) ~ exp(exp(2*z) - exp(z) - n) * (n/z)^(n + 1/2) / sqrt(2*(1 + 2*z)*exp(2*z) - (1 + z)*exp(z)), where z = LambertW(n)/2 - 1/(1 + 2/LambertW(n) - 4 * n^(1/2) * (1 + LambertW(n)) / LambertW(n)^(3/2)). - Vaclav Kotesovec, Jul 03 2022
a(n) ~ 2^n * n^n / (sqrt(1 + LambertW(n)) * LambertW(n)^n * exp(n + 1/8 - n/LambertW(n) + sqrt(n/LambertW(n)))). - Vaclav Kotesovec, Jul 08 2022

A111577 Galton triangle T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 21, 12, 1, 1, 85, 105, 22, 1, 1, 341, 820, 325, 35, 1, 1, 1365, 6081, 4070, 780, 51, 1, 1, 5461, 43932, 46781, 14210, 1596, 70, 1, 1, 21845, 312985, 511742, 231511, 39746, 2926, 92, 1, 1, 87381, 2212740, 5430405, 3521385, 867447, 95340, 4950, 117, 1
Offset: 1

Views

Author

Gary W. Adamson, Aug 07 2005

Keywords

Comments

In triangles of analogs to Stirling numbers of the second kind, the multipliers of T(n-1,k) in the recurrence are terms in arithmetic sequences: in Pascal's triangle A007318, the multiplier = 1. In triangle A008277, the Stirling numbers of the second kind, the multipliers are in the set (1,2,3...). For this sequence here, the multipliers are from A016777.
Riordan array [exp(x), (exp(3x)-1)/3]. - Paul Barry, Nov 26 2008
From Peter Bala, Jan 27 2015: (Start)
Working with an offset of 0, this is the triangle of connection constants between the polynomial basis sequences {x^n}, n>=0 and {n!*3^n*binomial((x - 1)/3,n)}, n>=0. An example is given below.
Call this array M and let P denote Pascal's triangle A007318, then P * M = A225468, P^2 * M = A075498. Also P^(-1) * M is a shifted version of A075498.
This triangle is the particular case a = 3, b = 0, c = 1 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. (End)
Named after the English scientist Francis Galton (1822-1911). - Amiram Eldar, Jun 13 2021
This is the array of (r, β)-Stirling numbers for r = 1 and β = 3. See Corcino. - Peter Bala, Feb 26 2025

Examples

			T(5,3) = T(4,2) + 7*T(4,3) = 21 + 7*12 = 105.
The triangle starts in row n = 1 as:
  1;
  1,  1;
  1,  5,   1;
  1, 21,  12,  1;
  1, 85, 105, 22, 1;
Connection constants: Row 4: [1, 21, 12, 1] so
x^3 = 1 + 21*(x - 1) + 12*(x - 1)*(x - 4) + (x - 1)*(x - 4)*(x - 7). - _Peter Bala_, Jan 27 2015
From _Peter Bala_, Feb 26 2025: (Start)
The array factorizes as
/1                \     /1               \/1              \/1             \
|1   1            |     |1   1           ||0  1           ||0  1          |
|1   5    1       |  =  |1   4   1       ||0  1   1       ||0  0  1       | ...
|1  21   12   1   |     |1  13   7   1   ||0  1   4  1    ||0  0  1  1    |
|1  85  105  22  1|     |1  44  34  10  1||0  1  13  7  1 ||0  0  1  4  1 |
|...              |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - x), x/(1 - 3*x)). Cf. A193843. (End)
		

Crossrefs

Programs

  • Maple
    A111577 := proc(n,k) option remember; if k = 1 or k = n then 1; else procname(n-1,k-1)+(3*k-2)*procname(n-1,k) ; fi; end:
    seq( seq(A111577(n,k),k=1..n), n=1..10) ; # R. J. Mathar, Aug 22 2009
  • Mathematica
    T[, 1] = 1; T[n, n_] = 1;
    T[n_, k_] := T[n, k] = T[n-1, k-1] + (3k-2) T[n-1, k];
    Table[T[n, k], {n, 1, 10}, {k, 1, n}] (* Jean-François Alcover, Jun 13 2019 *)

Formula

T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k).
E.g.f.: exp(x)*exp((y/3)*(exp(3x)-1)). - Paul Barry, Nov 26 2008
Let f(x) = exp(1/3*exp(3*x) + x). Then, with an offset of 0, the row polynomials R(n,x) are given by R(n,exp(3*x)) = 1/f(x)*(d/dx)^n(f(x)). Similar formulas hold for A008277, A039755, A105794, A143494 and A154537. - Peter Bala, Mar 01 2012
T(n, k) = 1/(3^k*k!)*Sum_{j=0..k}((-1)^(k-j)*binomial(k,j)*(3*j+1)^n). - Peter Luschny, May 20 2013
From Peter Bala, Jan 27 2015: (Start)
T(n,k) = Sum_{i = 0..n-1} 3^(i-k+1)*binomial(n-1,i)*Stirling2(i,k-1).
O.g.f. for n-th diagonal: exp(-x/3)*Sum_{k >= 0} (3*k + 1)^(k+n-1)*((x/3*exp(-x))^k)/k!.
O.g.f. column k (with offset 0): 1/( (1 - x)*(1 - 4*x)*...*(1 - (3*k + 1)*x) ). (End)

Extensions

Edited and extended by R. J. Mathar, Aug 22 2009

A105794 Inverse of a generalized Stirling number triangle of first kind.

Original entry on oeis.org

1, -1, 1, 1, -1, 1, -1, 1, 0, 1, 1, -1, 1, 2, 1, -1, 1, 0, 5, 5, 1, 1, -1, 1, 10, 20, 9, 1, -1, 1, 0, 21, 70, 56, 14, 1, 1, -1, 1, 42, 231, 294, 126, 20, 1, -1, 1, 0, 85, 735, 1407, 924, 246, 27, 1, 1, -1, 1, 170, 2290, 6363, 6027, 2400, 435, 35, 1
Offset: 0

Views

Author

Paul Barry, Apr 20 2005

Keywords

Comments

Inverse of number triangle A105793. Row sums are the generalized Bell numbers A000296.
T(n,k)*(-1)^(n-k) gives the inverse Sheffer matrix of A094645. In the umbral notation (cf. Roman, p. 17, quoted in A094645) this is Sheffer for (1-t,-log(1-t)). - Wolfdieter Lang, Jun 20 2011
From Peter Bala, Jul 10 2013: (Start)
For a graph G and a positive integer k, the graphical Stirling number S(G;k) is the number of partitions of the vertex set of G into k nonempty independent sets (an independent set in G is a subset of the vertices, no two elements of which are adjacent). Omitting the first two rows and columns of this triangle produces the triangle of graphical Stirling numbers of cycles on n vertices (interpreting a 2-cycle as a single edge). In comparison, the classical Stirling numbers of the second kind, A008277, are the graphical Stirling numbers of path graphs on n vertices. See Galvin and Than. See also A227341.
This is the triangle of connection constants for expressing the polynomial sequence (x-1)^n in terms of the falling factorial polynomials, that is, (x-1)^n = Sum_{k = 0..n} T(n,k)*x_(k), where x_(k) = x*(x-1)*...*(x-k+1) denotes the falling factorial.
The row polynomials are a particular case of the Actuarial polynomials - see Roman 4.3.4. (End)

Examples

			The triangle starts with
  n=0:  1;
  n=1: -1,  1;
  n=2:  1, -1, 1;
  n=3: -1,  1, 0, 1;
  n=4:  1, -1, 1, 2, 1;
  n=5: -1,  1, 0, 5, 5, 1;
  ... - _Wolfdieter Lang_, Jun 20 2011
		

References

  • S. Roman, The umbral calculus, Pure and Applied Mathematics 111, Academic Press Inc., New York, 1984. Reprinted by Dover in 2005.

Crossrefs

Cf. A000296 (row sums), A105793 (matrix inverse), A227341.

Programs

  • Maple
    B:= Matrix(12,12,shape=triangular[lower],(n,k) -> combinat:-stirling1(n-1,k-1)+(n-1)*combinat:-stirling1(n-2,k-1)):
    A:= B^(-1):
    seq(seq(A[i,j],j=1..i),i=1..12); # Robert Israel, Jan 19 2015
    T := (n, k) -> add((-1)^(n - i)*binomial(n, i)*Stirling2(i, k), i=0..n):
    seq(seq(T(n, k), k=0..n), n=0..9);  # Peter Luschny, Feb 15 2025
  • Mathematica
    Table[Sum[(-1)^(n - i)*Binomial[n, i] StirlingS2[i, k], {i, 0, n}], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Oct 14 2019 *)

Formula

Term k in row n is given by {(-1)^(k+n) * (Sum_{j=0..k} (-1)^j * binomial(k,j) * (1-j)^n) / k! }; i.e., a finite difference. - Tom Copeland, Jun 05 2006
O.g.f. for row n = n-th finite difference of the Touchard (Bell) polynomials, T(x,j) and so the e.g.f. for these finite differences and therefore the sequence = exp{x*[exp(t)-1]-t} = exp{t*[T(x,.)-1]} umbrally. - Tom Copeland, Jun 05 2006
The e.g.f. A(x,t) = exp(x*(exp(t)-1)-t) satisfies the partial differential equation x*dA/dx - dA/dt = (1-x)*A.
Recurrence relation: T(n+1,k) = T(n,k-1) + (k-1)*T(n,k).
Let f(x) = ((x-1)/x)*exp(x). For n >= 1, the n-th row polynomial R(n,x) = x*exp(-x)*(x*d/dx)^(n-1)(f(x)) and satisfies the recurrence equation R(n+1,x) = (x-1)*R(n,x) + x*R'(n,x). - Peter Bala, Oct 28 2011
Let f(x) = exp(exp(x)-x). Then R(n,exp(x)) = 1/f(x)*(d/dx)^n(f(x)). Similar formulas hold for A008277, A039755, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012
From Peter Bala, Jul 10 2013: (Start)
T(n,k) = Sum_{i = 0..n-1} (-1)^i*Stirling2(n-1-i,k-1), for n >= 1, k >= 1.
The k-th column o.g.f. is (1/(1+x))*x^k/((1-x)*(1-2*x)*...*(1-(k-1)*x)) (the empty product occurring in the denominator when k = 0 and k = 1 is taken equal to 1).
Dobinski-type formula for the row polynomials: R(n,x) = exp(-x)*Sum_{k >= 0} (k-1)^n*x^k/k!.
Sum_{k = 0..n} binomial(n,k)*R(k,x) = n-th Bell polynomial (n-th row polynomial of A048993). (End)
From Peter Bala, Jan 13 2014: (Start)
T(n,k) = Sum_{i = 0..n} (-1)^(n-i)*binomial(n,i)*Stirling2(i,k).
T(n,k) = Sum_{i = 0..n} (-2)^(n-i)*binomial(n,i)*Stirling2(i+1,k+1).
Matrix product P^(-1) * S = P^(-2) * S1, where P = A007318, S = A048993 and S1 = A008277. (End)
From Werner Schulte, Feb 15 2025: (Start)
Conjecture 1: Sum_{k=0..n} (-1)^(n-k) * T(n, k) * (k+m)! / m! = (m+2)^n for m >= 0.
Conjecture 2: (-1)^n - B(n) = Sum_{k=1..n} (-1)^k * T(n, k) * (k-1)! / (k+1) where B(n) = B(n, 0) is n-th Bernoulli number. (End)

A286718 Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 3*x)^(-1/3), (-1/3)*log(1 - 3*x)). A generalized Stirling1 triangle.

Original entry on oeis.org

1, 1, 1, 4, 5, 1, 28, 39, 12, 1, 280, 418, 159, 22, 1, 3640, 5714, 2485, 445, 35, 1, 58240, 95064, 45474, 9605, 1005, 51, 1, 1106560, 1864456, 959070, 227969, 28700, 1974, 70, 1, 24344320, 42124592, 22963996, 5974388, 859369, 72128, 3514, 92, 1, 608608000, 1077459120, 616224492, 172323696, 27458613, 2662569, 159978, 5814, 117, 1, 17041024000, 30777463360, 18331744896, 5441287980, 941164860, 102010545, 7141953, 322770, 9090, 145, 1
Offset: 0

Views

Author

Wolfdieter Lang, May 18 2017

Keywords

Comments

This is a generalization of the unsigned Stirling1 triangle A132393.
In general the lower triangular Sheffer matrix ((1 - d*x)^(-a/d), (-1/d)*log(1 - d*x)) is called here |S1hat[d,a]|. The signed matrix S1hat[d,a] with elements (-1)^(n-k)*|S1hat[d,a]|(n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[d,a] with elements S2[d,a](n, k)/d^k, where S2[d,a] is Sheffer (exp(a*x), exp(d*x) - 1).
In the Bala link the signed S1hat[d,a] (with row scaled elements S1[d,a](n,k)/d^n where S1[d,a] is the inverse matrix of S2[d,a]) is denoted by s_{(d,0,a)}, and there the notion exponential Riordan array is used for Sheffer array.
In the Luschny link the elements of |S1hat[m,m-1]| are called Stirling-Frobenius cycle numbers SF-C with parameter m.
From Wolfdieter Lang, Aug 09 2017: (Start)
The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,k)*x^k of the Sheffer triangle |S1hat[d,a]| satisfy, as special polynomials of the Boas-Buck class (see the reference), the identity (we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function)
(E_x - n*1)*R(d,a;n,x) = -n!*Sum_{k=0..n-1} d^k*(a*1 + d*beta(k)*E_x)*R(d,a;n-1-k,x)/(n-1-k)!, for n >= 0, with E_x = x*d/dx (Euler operator), and beta(k) = A002208(k+1)/A002209(k+1).
This entails a recurrence for the sequence of column k, for n > k >= 0: T(d,a;n,k) = (n!/(n - k))*Sum_{p=k..n-1} d^(n-1-p)*(a + d*k*beta(n-1-p))*T(d,a;p,k)/p!, with input T(d,a;k,k) = 1. For the present [d,a] = [3,1] case see the formula and example sections below. (End)
The inverse of the Sheffer triangular matrix S2[3,1] = A282629 is the Sheffer matrix S1[3,1] = (1/(1 + x)^(1/3), log(1 + x)/3) with rational elements S1[3,1](n, k) = (-1)^(n-m)*T(n, k)/3^n. - Wolfdieter Lang, Nov 15 2018

Examples

			The triangle T(n, k) begins:
n\k        0        1        2       3      4     5    6  7 8 ...
O:         1
1:         1        1
2:         4        5        1
3:        28       39       12       1
4:       280      418      159      22      1
5:      3640     5714     2485     445     35     1
6:     58240    95064    45474    9605   1005    51    1
7:   1106560  1864456   959070  227969  28700  1974   70  1
8:  24344320 42124592 22963996 5974388 859369 72128 3514 92 1
...
From _Wolfdieter Lang_, Aug 09 2017: (Start)
Recurrence: T(3, 1) = T(2, 0) + (3*3-2)*T(2, 1) = 4 + 7*5 = 39.
Boas-Buck recurrence for column k = 2 and n = 5:
T(5, 2) = (5!/3)*(3^2*(1 + 6*(3/8))*T(2,2)/2! + 3*(1 + 6*(5/12)*T(3, 2)/3! + (1 + 6*(1/2))* T(4, 2)/4!)) = (5!/3)*(9*(1 + 9/4)/2 + 3*(1 + 15/6)*12/6 + (1 + 3)*159/24) = 2485.
The beta sequence begins: {1/2, 5/12, 3/8, 251/720, 95/288, 19087/60480, ...}.
(End)
		

References

  • Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
  • Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

Crossrefs

S2[d,a] for [d,a] = [1,0], [2,1], [3,1], [3,2], [4,1] and [4,3] is A048993, A154537, A282629, A225466, A285061 and A225467, respectively.
S2hat[d,a] for these [d,a] values is A048993, A039755, A111577 (offset 0), A225468, A111578 (offset 0) and A225469, respectively.
|S1hat[d,a]| for [d,a] = [1,0], [2,1], [3,2], [4,1] and [4,3] is A132393, A028338, A225470, A290317 and A225471, respectively.
Column sequences for k = 0..4: A007559, A024216(n-1), A286721(n-2), A382984, A382985.
Diagonal sequences: A000012, A000326(n+1), A024212(n+1), A024213(n+1).
Row sums: A008544. Alternating row sums: A000007.
Beta sequence: A002208(n+1)/A002209(n+1).

Programs

  • Mathematica
    T[n_ /; n >= 1, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + (3*n-2)* T[n-1, k]; T[, -1] = 0; T[0, 0] = 1; T[n, k_] /; nJean-François Alcover, Jun 20 2018 *)

Formula

Recurrence: T(n, k) = T(n-1, k-1) + (3*n-2)*T(n-1, k), for n >= 1, k = 0..n, and T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k.
E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k (i.e., e.g.f. of the triangle) is (1 - 3*z)^{-(x+1)/3}.
E.g.f. of column k is (1 - 3*x)^(-1/3)*((-1/3)*log(1 - 3*x))^k/k!.
Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, x+3), with R(0, x) = 1.
Row polynomial R(n, x) = risefac(3,1;x,n) with the rising factorial
risefac(d,a;x,n) := Product_{j=0..n-1} (x + (a + j*d)). (For the signed case see the Bala link, eq. (16)).
T(n, k) = sigma^{(n)}{n-k}(a_0,a_1,...,a{n-1}) with the elementary symmetric functions with indeterminates a_j = 1 + 3*j.
T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*3^j, with the unsigned Stirling1 triangle |S1| = A132393.
Boas-Buck column recurrence (see a comment above): T(n, k) =
(n!/(n - k))*Sum_{p=k..n-1} 3^(n-1-p)*(1 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, with beta(k) = A002208(k+1)/A002209(k+1). See an example below. - Wolfdieter Lang, Aug 09 2017

A225470 Triangle read by rows, s_3(n, k) where s_m(n, k) are the Stirling-Frobenius cycle numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 2, 1, 10, 7, 1, 80, 66, 15, 1, 880, 806, 231, 26, 1, 12320, 12164, 4040, 595, 40, 1, 209440, 219108, 80844, 14155, 1275, 57, 1, 4188800, 4591600, 1835988, 363944, 39655, 2415, 77, 1, 96342400, 109795600, 46819324, 10206700, 1276009, 95200, 4186, 100, 1
Offset: 0

Views

Author

Peter Luschny, May 16 2013

Keywords

Comments

The Stirling-Frobenius subset numbers S_{m}(n,k), for m >= 1 fixed, regarded as an infinite lower triangular matrix, can be inverted by Sum_{k} S_{m}(n,k)*s_{m}(k,j)*(-1)^(n-k) = [j = n]. The inverse numbers s_{m}(k,j), which are unsigned, are the Stirling-Frobenius cycle numbers. For m = 1 this gives the classical Stirling cycle numbers A132393. The Stirling-Frobenius subset numbers are defined in A225468.
Triangle T(n,k), read by rows, given by (2, 3, 5, 6, 8, 9, 11, 12, 14, 15, ... (A007494)) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 14 2015

Examples

			Triangle starts:
  [n\k][    0,      1,     2,     3,    4,  5,  6]
  [0]       1,
  [1]       2,      1,
  [2]      10,      7,     1,
  [3]      80,     66,    15,     1,
  [4]     880,    806,   231,    26,    1,
  [5]   12320,  12164,  4040,   595,   40,  1,
  [6]  209440, 219108, 80844, 14155, 1275, 57,  1.
  ...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence (see Maple program): T(4, 2) = T(3, 1) + (3*4 - 1)*T(3, 2) = 66 + 11*15 = 231.
Boas-Buck type recurrence for column k = 2 and n = 4: T(4, 2) = (4!/2)*(3*(2 + 6*(5/12))*T(2, 2)/2! + 1*(2 + 6*(1/2))*T(3,2)/3!) = (4!/2)*(3*9/4 + 5*15/3!) = 231. (End)
		

Crossrefs

Cf. A225468; A132393 (m=1), A028338 (m=2), A225471(m=4).
Column k=0..4 give A008544, A024395(n-1), A286722(n-2), A383221, A383222.
T(n, n) ~ A000012; T(n, n-1) ~ A005449; T(n, n-2) ~ A024391; T(n, n-3) ~ A024392.
row sums ~ A032031; alternating row sums ~ A007559.
Cf. A132393.

Programs

  • Maple
    SF_C := proc(n, k, m) option remember;
    if n = 0 and k = 0 then return(1) fi;
    if k > n or  k < 0 then return(0) fi;
    SF_C(n-1, k-1, m) + (m*n-1)*SF_C(n-1, k, m) end:
    seq(print(seq(SF_C(n, k, 3), k = 0..n)), n = 0..8);
  • Mathematica
    SFC[0, 0, ] = 1; SFC[n, k_, ] /; (k > n || k < 0) = 0; SFC[n, k_, m_] := SFC[n, k, m] = SFC[n-1, k-1, m] + (m*n-1)*SFC[n-1, k, m]; Table[SFC[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 26 2013, after Maple *)

Formula

For a recurrence see the Maple program.
From Wolfdieter Lang, May 18 2017: (Start)
This is the Sheffer triangle (1/(1 - 3*x)^{-2/3}, -(1/3)*log(1-3*x)). See the P. Bala link where this is called exponential Riordan array, and the signed version is denoted by s_{(3,0,2)}.
E.g.f. of row polynomials in the variable x (i.e., of the triangle): (1 - 3*z)^{-(2+x)/3}.
E.g.f. of column k: (1-3*x)^(-2/3)*(-(1/3)*log(1-3*x))^k/k!, k >= 0.
Recurrence for row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k: R(n, x) = (x+2)*R(n-1,x+3), with R(0, x) = 1.
R(n, x) = risefac(3,2;x,n) := Product_{j=0..(n-1)} (x + (2 + 3*j)). (See the P. Bala link, eq. (16) for the signed s_{3,0,2} row polynomials.)
T(n, k) = Sum_{j=0..(n-m)} binomial(n-j, k)* S1p(n, n-j)*2^(n-k-j)*3^j, with S1p(n, m) = A132393(n, m). (End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 3^(n-1-p)*(2 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/A002209(k+1), beginning {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017

A225468 Triangle read by rows, S_3(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 2, 1, 4, 7, 1, 8, 39, 15, 1, 16, 203, 159, 26, 1, 32, 1031, 1475, 445, 40, 1, 64, 5187, 12831, 6370, 1005, 57, 1, 128, 25999, 107835, 82901, 20440, 1974, 77, 1, 256, 130123, 888679, 1019746, 369061, 53998, 3514, 100, 1
Offset: 0

Views

Author

Peter Luschny, May 16 2013

Keywords

Comments

The definition of the Stirling-Frobenius subset numbers: S_m(n, k) = (Sum_{j=0..n} binomial(j, n-k)*A_m(n, j)) / (m^k*k!) where A_m(n, j) are the generalized Eulerian numbers. For m = 1 this gives the classical Stirling set numbers A048993. (See the links for details.)
From Peter Bala, Jan 27 2015: (Start)
Exponential Riordan array [ exp(2*z), 1/3*(exp(3*z) - 1)].
Triangle equals P * A111577 = P^(-1) * A075498, where P is Pascal's triangle A007318.
Triangle of connection constants between the polynomial basis sequences {x^n}n>=0 and { n!*3^n*binomial((x - 2)/3,n) }n>=0. An example is given below.
This triangle is the particular case a = 3, b = 0, c = 2 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. (End)

Examples

			[n\k][ 0,    1,     2,    3,    4,  5,  6]
[0]    1,
[1]    2,    1,
[2]    4,    7,     1,
[3]    8,   39,    15,    1,
[4]   16,  203,   159,   26,    1,
[5]   32, 1031,  1475,  445,   40,  1,
[6]   64, 5187, 12831, 6370, 1005, 57,  1.
Connection constants: Row 3: [8, 39, 15, 1] so
x^3 = 8 + 39*(x - 2) + 15*(x - 2)*(x - 5) + (x - 2)*(x - 5)*(x - 8). - _Peter Bala_, Jan 27 2015
		

Crossrefs

Cf. A048993 (m=1), A039755 (m=2), A225469 (m=4).

Programs

  • Maple
    SF_S := proc(n, k, m) option remember;
    if n = 0 and k = 0 then return(1) fi;
    if k > n or k < 0 then return(0) fi;
    SF_S(n-1, k-1, m) + (m*(k+1)-1)*SF_S(n-1, k, m) end:
    seq(print(seq(SF_S(n, k, 3), k=0..n)), n = 0..5);
  • Mathematica
    EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/(k!*m^k); Table[ SFS[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
  • Sage
    @CachedFunction
    def EulerianNumber(n, k, m) :
        if n == 0: return 1 if k == 0 else 0
        return (m*(n-k)+m-1)*EulerianNumber(n-1,k-1,m) + (m*k+1)*EulerianNumber(n-1,k,m)
    def SF_S(n, k, m):
        return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))/ (factorial(k)*m^k)
    for n in (0..6): [SF_S(n, k, 3) for k in (0..n)]

Formula

T(n, k) = (Sum_{j=0..n} binomial(j, n-k)*A_3(n, j)) / (3^k*k!) with A_3(n,j) = A225117.
For a recurrence see the Maple program.
T(n, 0) ~ A000079; T(n, 1) ~ A016127; T(n, 2) ~ A016297; T(n, 3) ~ A025999;
T(n, n) ~ A000012; T(n, n-1) ~ A005449; T(n, n-2) ~ A024212.
From Peter Bala, Jan 27 2015: (Start)
T(n,k) = Sum_{i = 0..n} (-1)^(n+i)*3^(i-k)*binomial(n,i)*Stirling2(i+1,k+1).
E.g.f.: exp(2*z)*exp(x/3*(exp(3*z) - 1)) = 1 + (2 + x)*z + (4 + 7*x + x^2)*z^2/2! + ....
T(n,k) = 1/(3^k*k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(3*j + 2)^n.
O.g.f. for n-th diagonal: exp(-2*x/3)*Sum_{k >= 0} (3*k + 2)^(k + n - 1)*((x/3*exp(-x))^k)/k!.
O.g.f. column k: 1/( (1 - 2*x)*(1 - 5*x)...(1 - (3*k + 2)*x) ). (End)
E.g.f. column k: exp(2*x)*((exp(3*x) - 1)/3)^k, k >= 0. See the Bala link for the S(3,0,2) exponential Riordan aka Sheffer triangle. - Wolfdieter Lang, Apr 10 2017

A225469 Triangle read by rows, S_4(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 3, 1, 9, 10, 1, 27, 79, 21, 1, 81, 580, 310, 36, 1, 243, 4141, 3990, 850, 55, 1, 729, 29230, 48031, 16740, 1895, 78, 1, 2187, 205339, 557571, 299131, 52745, 3689, 105, 1, 6561, 1439560, 6338620, 5044536, 1301286, 137592, 6524, 136, 1
Offset: 0

Views

Author

Peter Luschny, May 16 2013

Keywords

Comments

The definition of the Stirling-Frobenius subset numbers: S_m(n, k) = (sum_{j=0..n} binomial(j, n-k)*A_m(n, j)) / (m^k*k!) where A_m(n, j) are the generalized Eulerian numbers (see the links for details).
This is the Sheffer triangle (exp(3*x),(1/4)*(exp(4*x -1))). See the P. Bala link where this is called exponential Riordan array S_{(4,0,3)}. - Wolfdieter Lang, Apr 13 2017

Examples

			[n\k][ 0,     1,     2,     3,    4,   5,  6]
[0]    1,
[1]    3,     1,
[2]    9,    10,     1,
[3]   27,    79,    21,     1,
[4]   81,   580,   310,    36,    1,
[5]  243,  4141,  3990,   850,   55,  1,
[6]  729, 29230, 48031, 16740, 1895, 78,  1.
		

Crossrefs

Cf. A048993 (m=1), A039755 (m=2), A225468 (m=3).
Cf. Columns: A000244, A016138, A018054.

Programs

  • Maple
    SF_S := proc(n, k, m) option remember;
    if n = 0 and k = 0 then return(1) fi;
    if k > n or k < 0 then return(0) fi;
    SF_S(n-1, k-1, m) + (m*(k+1)-1)*SF_S(n-1, k, m) end:
    seq(print(seq(SF_S(n, k, 4), k=0..n)), n = 0..5);
  • Mathematica
    EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFS[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]/(k!*m^k); Table[ SFS[n, k, 4], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
  • Sage
    @CachedFunction
    def EulerianNumber(n, k, m) :
        if n == 0: return 1 if k == 0 else 0
        return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m) + (m*k+1)*EulerianNumber(n-1, k, m)
    def SF_S(n, k, m):
        return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))/(factorial(k)*m^k)
    for n in (0..6): [SF_S(n, k, 4) for k in (0..n)]

Formula

T(n, k) = (sum_{j=0..n} binomial(j, n-k)*A_4(n, j)) / (4^k*k!) where A_4(n,j) = A225118.
For a recurrence see the Maple program.
T(n, 0) ~ A000244; T(n, 1) ~ A016138; T(n, 2) ~ A018054.
T(n, n) ~ A000012; T(n, n-1) ~ A014105.
From Wolfdieter Lang, Apr 13 2017: (Start)
E.g.f.: exp(3*z)*exp((x/4)*(exp(4*z -1))). Sheffer triangle (see a comment above).
E.g.f. column k: exp(3*x)*(exp(4*x) -1)^k/(4^k*k!), k >= 0 (Sheffer property).
O.g.f. column k: x^m/Product_{j=0..k} (1 - (3+4*j)*x), k >= 0.
(End)

A225471 Triangle read by rows, s_4(n, k) where s_m(n, k) are the Stirling-Frobenius cycle numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 3, 1, 21, 10, 1, 231, 131, 21, 1, 3465, 2196, 446, 36, 1, 65835, 45189, 10670, 1130, 55, 1, 1514205, 1105182, 290599, 36660, 2395, 78, 1, 40883535, 31354119, 8951355, 1280419, 101325, 4501, 105, 1, 1267389585, 1012861224, 308846124, 48644344, 4421494, 240856, 7756, 136, 1
Offset: 0

Views

Author

Peter Luschny, May 17 2013

Keywords

Comments

The Stirling-Frobenius cycle numbers are defined in A225470.
Triangle T(n,k), read by rows, given by (3, 4, 7, 8, 11, 12, 15, 16, 19, 20, ... (A014601)) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 14 2015

Examples

			[n\k][    0,       1,      2,     3,    4,  5,  6 ]
[0]       1,
[1]       3,       1,
[2]      21,      10,      1,
[3]     231,     131,     21,     1,
[4]    3465,    2196,    446,    36,    1,
[5]   65835,   45189,  10670,  1130,   55,  1,
[6] 1514205, 1105182, 290599, 36660, 2395, 78,  1.
...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence: T(4, 2) = T(3, 1) + (4*4 - 1)*T(3, 2) = 131 +15*21 = 446.
Boas-Buck recurrence for column k=2 and n=4: T(4, 2) = (4!/2)*(4*(3+8*(5/12)) *T(2, 2)/2! + 1*(3 + 8*(1/2))*T(3,2)/3!) = (4!/2)*(4*(19/3)/2  + 7*21/3!) =  446.
(End)
		

Crossrefs

Columns k=0..3 give A008545, A286723(n-1), A383702, A383703.
Cf. A132393 (m=1), A028338 (m=2), A225470 (m=3).

Programs

  • Mathematica
    T[0, 0] = 1; T[n_, k_] := Sum[Binomial[n - j, k]*Abs[StirlingS1[n, n - j]]* 3^(n - k - j)*4^j, {j, 0, n - k}];
    Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Wolfdieter Lang *)
  • Sage
    @CachedFunction
    def SF_C(n, k, m):
        if k > n or k < 0 : return 0
        if n == 0 and k == 0: return 1
        return SF_C(n-1, k-1, m) + (m*n-1)*SF_C(n-1, k, m)
    for n in (0..8): [SF_C(n, k, 4) for k in (0..n)]

Formula

For a recurrence see the Sage program.
T(n, 0) ~ A008545; T(n, n) ~ A000012; T(n, n-1) = A014105.
Row sums ~ A047053; alternating row sums ~ A001813.
From Wolfdieter Lang, May 29 2017: (Start)
This is the Sheffer triangle (1/(1 - 4*x)^{-3/4}, -(1/4)*log(1-4*x)). See the P. Bala link where this is called exponential Riordan array, and the signed version is denoted by s_{(4,0,3)}.
E.g.f. of row polynomials in the variable x (i.e., of the triangle): (1 - 4*z)^{-(3+x)/4}.
E.g.f. of column k: (1-4*x)^(-3/4)*(-(1/4)*log(1-4*x))^k/k!, k >= 0.
Recurrence for row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k: R(n, x) = (x+3)*R(n-1,x+4), with R(0, x) = 1.
R(n, x) = risefac(4,3;x,n) := Product_{j=0..(n-1)} (x + (3 + 4*j)). (See the P. Bala link, eq. (16) for the signed s_{4,0,3} row polynomials.)
T(n, k) = Sum_{j=0..(n-m)} binomial(n-j, k)* S1p(n, n-j)*3^(n-k-j)*4^j, with S1p(n, m) = A132393(n, m).
T(n, k) = sigma[4,3]^{(n)}_{n-k}, with the elementary symmetric functions sigma[4,3]^{(n)}_m of degree m in the n numbers 3, 7, 11, ..., 3+4*(n-1), with sigma[4,3]^{(n)}_0 := 1. (End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 4^(n-1-p)*(3 + 8*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/A002209(k+1), beginning with {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017

A111578 Triangle T(n, m) = T(n-1, m-1) + (4m-3)*T(n-1, m) read by rows 1<=m<=n.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 31, 15, 1, 1, 156, 166, 28, 1, 1, 781, 1650, 530, 45, 1, 1, 3906, 15631, 8540, 1295, 66, 1, 1, 19531, 144585, 126651, 30555, 2681, 91, 1, 1, 97656, 1320796, 1791048, 646086, 86856, 4956, 120, 1, 1, 488281, 11984820, 24604420, 12774510
Offset: 1

Views

Author

Gary W. Adamson, Aug 07 2005

Keywords

Comments

From Peter Bala, Jan 27 2015: (Start)
Working with an offset of 0, this is the exponential Riordan array [exp(z), (exp(4*z) - 1)/4].
This is the triangle of connection constants between the polynomial basis sequences {x^n}n>=0 and { n!*4^n * binomial((x - 1)/4,n) }n>=0. An example is given below.
Call this array M and let P denote Pascal's triangle A007318 then P^2 * M = A225469; P^(-1) * M is a shifted version of A075499.
This triangle is the particular case a = 4, b = 0, c = 1 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. (End)

Examples

			The triangle starts in row n=1 as:
  1;
  1,1;
  1,6,1;
  1,31,15,1;
  1,156,166,28,1;
Connection constants: Row 4: [1, 31, 15, 1] so
x^3 = 1 + 31*(x - 1) + 15*(x - 1)*(x - 5) + (x - 1)*(x - 5)*(x - 9). - _Peter Bala_, Jan 27 2015
		

Crossrefs

Cf. A111577, A008277, A039755, A016234 (3rd column).

Programs

  • Mathematica
    T[n_, k_] := 1/(4^(k-1)*(k-1)!) * Sum[ (-1)^(k-j-1) * (4*j+1)^(n-1) * Binomial[k-1, j], {j, 0, k-1}]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 28 2015, after Peter Bala *)
  • Python
    def A096038(n,m):
        if n < 1 or m < 1 or m > n:
            return 0
        elif n <=2:
            return 1
        else:
            return A096038(n-1,m-1)+(4*m-3)*A096038(n-1,m)
    print( [A096038(n,m) for n in range(20) for m in range(1,n+1)] )
    # R. J. Mathar, Oct 11 2009

Formula

From Peter Bala, Jan 27 2015: (Start)
The following formulas assume an offset of 0.
T(n,k) = 1/(4^k*k!)*sum {j = 0..k} (-1)^(k-j)*binomial(k,j)*(4*j + 1)^n.
T(n,k) = sum {i = 0..n-1} 4^(i-k+1)*binomial(n-1,i)*Stirling2(i,k-1).
E.g.f.: exp(z)*exp(x/4*(exp(4*z) - 1)) = 1 + (1 + x)*z + (1 + 6*x + x^2)*z^2/2! + ....
O.g.f. for n-th diagonal: exp(-x/4)*sum {k >= 0} (4*k + 1)^(k + n - 1)*((x/4*exp(-x))^k)/k!.
O.g.f. column k: 1/( (1 - x)*(1 - 5*x)*...*(1 - (4*k + 1)*x) ). (End)

Extensions

Edited and extended by R. J. Mathar, Oct 11 2009

A286724 Triangle read by rows. A generalization of unsigned Lah numbers, called L[2,1].

Original entry on oeis.org

1, 2, 1, 8, 8, 1, 48, 72, 18, 1, 384, 768, 288, 32, 1, 3840, 9600, 4800, 800, 50, 1, 46080, 138240, 86400, 19200, 1800, 72, 1, 645120, 2257920, 1693440, 470400, 58800, 3528, 98, 1, 10321920, 41287680, 36126720, 12042240, 1881600, 150528, 6272, 128, 1, 185794560, 836075520, 836075520, 325140480, 60963840, 6096384, 338688, 10368, 162, 1, 3715891200, 18579456000, 20901888000, 9289728000, 2032128000, 243855360, 16934400, 691200, 16200, 200, 1
Offset: 0

Views

Author

Wolfdieter Lang, Jun 16 2017

Keywords

Comments

These generalized unsigned Lah numbers are the instance L[2,1] of the Sheffer triangles called L[d,a], with integers d >= 1 and integers 0 <= a < d with gcd(d,a) = 1. The standard unsigned Lah numbers are L[1,0] = A271703.
The Sheffer structure of L[d,a] is ((1 - d*t)^(-2*a/d), t/(1 - d*t)). This follows from the defining property
risefac[d,a](x, n) = Sum_{m=0..n} L[d,a](n, m)*fallfac[d,a](x, m), where risefac[d,a](x, n):= Product_{0..n-1} (x + (a+d*j)) for n >= 1 and risefac[d,a](x, 0) := 1, and fallfac[d,a](x, n):= Product_{0..n-1} (x - (a+d*j)) = for n >= 1 and fallfac[d,a](x, 0) := 1. Such rising and falling factorials arise in the generalization of Stirling numbers of both kinds S2[d,a] and S1[d,a]. See the Peter Bala link under A143395 for these falling factorials called there [t;a,b,c]_n with t=x, a=d, b=0, c=a.
In matrix notation: L[d,a] = S1phat[d,a].S2hat[d,a] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations with Sheffer structures S1phat[d,a] = ((1 - d*t)^(-a/d), -(1/d)*(log(1 - d*t))) and S2hat[d,a] = (exp(a*t), (1/d)*(exp(d*t) - 1). See, e.g., S1phat[2,1] = A028338 and S2hat[2,1] = A039755.
The a- and z-sequences for these Sheffer matrices have e.g.f.s 1 + d*t and ((1 + d*t)/t)*(1 - (1 + d*t)^(-2*a/d)), respectively. See a W. Lang link under A006232 for these types of sequences.
E.g.f. of row polynomials R[d,a](n, x) := Sum_{m=0..n} L[d,a](n, m)*x^m
(1 - d*x)^(-2*a/d)*exp(t*x/(1 - d*x)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - d*t)^(-2*a/d)*(t/(1 - d*t))^m/m, m >= 0.
Meixner type identity for (monic) row polynomials: (D_x/(1 + d*D_x)) * R[d,a](n, x) = n*R[d,a](n-1, x), n >= 1, with R[d,a](0, x) = 1. The series in the differentiations D_x = d/dx terminates.
General Sheffer recurrence for row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
R[d,a](n, x) = [(2*a+x)*1 + 2*d*(a + x)*D_x + d^2*x*(D_x)^2]*R[d,a](n-1, x), n >= 1, with R[d,a](0, x) = 1.
The inverse matrix L^(-1)[d,a] is Sheffer (g[d,a](-t), -f[d,a](-t)) with L[d,a] Sheffer (g[d,a](t), f[d,a](t)) from above. This means (see the column e.g.f. of Sheffer matrices) that L^(-1)[d,a](n, m) = (-1)^(n-m)*L[d,a](n, m). Therefore, the recurrence relations can easily be rewritten for L^(-1)[d,a] by replacing a -> -a and d -> -d.
fallfac[d,a](x, n) = Sum_{m=0..n} L^(-1)[d,a](n, m)*risefac[d,a](x, m), n >= 0.
From Wolfdieter Lang, Aug 12 2017: (Start)
The Sheffer row polynomials R[d,a](n, x) belong to the Boas-Buck class and satisfy therefore the Boas-Buck identity (see the reference, and we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function) (E_x - n*1)*R[d,a](n, x) = - n*(2*a*1 + d*E_x) * Sum_{k=0..n-1} d^k*R(d,a;n-1-k,x)/(n-1-k)!, with E_x = x*d/dx (Euler operator).
This implies a recurrence for the sequence of column m: L[d,a](n, m) = (n!*(2*a + d*m)/(n-m))*Sum_{p=0..n-1-m} d^p*L[d,a](n-1-p, m)/(n-1-p)!, for n > m>=0, and input L[d,a](m, m) = 1. For the present [d,a] = [2,1] instance see the formula and example sections. (End)
From Wolfdieter Lang, Sep 14 2017: (Start)
The diagonal sequences are 2^D*D!*(binomial(m+D, m))^2, m >= 0, for D >= 0 (main diagonal D = 0). From the o.g.f.s obtained via Lagrange's theorem. See the second W. Lang link below for the general Sheffer case.
The o.g.f. of the diagonal D sequence is 2^D*D!*Sum_{m=0..D} A008459(D, m)*x^m /(1- x)^(2*D + 1), D >= 0. (End)
It appears that this is also the matrix square of unsigned triangle of coefficients of Laguerre polynomials n!*L_n(x), abs(A021009(n, k)). - Ali Pourzand, Mar 10 2025 [This observation is correct. - Peter Luschny, Mar 10 2025]

Examples

			The triangle T(n, m) begins:
  n\m        0         1         2         3        4       5      6     7   8 9
  0:         1
  1:         2         1
  2:         8         8         1
  3:        48        72        18         1
  4:       384       768       288        32        1
  5:      3840      9600      4800       800       50       1
  6:     46080    138240     86400     19200     1800      72      1
  7:    645120   2257920   1693440    470400    58800    3528     98     1
  8:  10321920  41287680  36126720  12042240  1881600  150528   6272   128   1
  9: 185794560 836075520 836075520 325140480 60963840 6096384 338688 10368 162 1
  ...
From _Wolfdieter Lang_, Aug 12 2017: (Start)
Recurrence for column elements with m >= 1, and input column m = 0: T(3, 2) = (3/2)*T(2, 1) + 2*3*T(2, 2) = (3/2)*8 + 6 = 18.
Four term recurrence: T(3, 2) = T(2, 1) + 2*5*T(2, 2) - 4*2^2*T(1, 2) = 8 + 10 + 0 = 18.
Meixner type identity, n=2: 2*R(1, x) = (D_x - 2*(D_x)^2)*R(2, x), 2*(2 + x) = (8 + 2*x) - 2*2.
Sheffer recurrence: R(2, x) = (2 + x)*(2 + x) + 4*(1 + x)*1 + 0 = 8 + 8*x + x^2.
Boas-Buck recurrence for column m = 2 and n = 4: T(4, 2) = (2*4!*3/2)*(1*T(3, 2)/3! + 2*T(2, 2)/2!) = 4!*3*(18/3! + 1) = 288. (End)
Diagonal sequence D = 1: o.g.f. 2*1!*(1 + 1*x)/(1- x)^3 generating
{2*(binomial(m+1, m))^2}_{m >= 0} = {2, 8, 18, 32, ...}. - _Wolfdieter Lang_, Sep 14 2017
		

References

  • Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
  • Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.
  • Steven Roman, The Umbral Calculus, Academic press, Orlando, London, 1984, p. 50.

Crossrefs

Column sequences (no leading zeros): A000165, A014479, A286725.
Diagonal sequences: A000012, 2*A000290(m+1), 8*A000537(n+1), 48*A001249, 384*A288876. - Wolfdieter Lang, Sep 14 2017
Row sums are A025167. - Michael Somos, Sep 27 2017

Programs

  • Maple
    T := (n, k) -> ifelse(n < k, 0, ifelse(k = 0, n!*2^n, (n/k)*T(n-1, k-1) + 2*n*T(n-1, k))): seq(seq(T(n, k), k = 0..n), n = 0..10);  # Peter Luschny, Mar 10 2025
  • Mathematica
    T[ n_, k_] := Coefficient[ Integrate[ Exp[-x^2 - y x] HermiteH[n, x]^2, {x, -Infinity, Infinity}] / (Sqrt[Pi] Exp[y^2 / 4]), y, 2 k]; (* Michael Somos, Sep 27 2017 *)
  • SageMath
    # Using the function A021009_triangle, displays as a matrix. Following the observation of Ali Pourzand.
    print(A021009_triangle(9)^2)  # Peter Luschny, Mar 10 2025

Formula

T(n, m) = L[2,1](n, m) = Sum_{k=m..n} A028338(n, k)*A039755(k, m).
Three term recurrence for column elements with m >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 2*n*T(n-1, m) with T(n, m) = 0 for n < m and the column m = 0 is T(n, 0) = (2*n)!! = n*2^n = A000165(n). (From the a- and z-sequences {1, 2, repeat(0)} and {2, repeat(0)}, respectively.)
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(2*n-1)*T(n-1, m) - 4*(n-1)^2*T(n-2, m), n >= m >= 0, with T(0, 0) = 1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
E.g.f. of row polynomials R(n, x) = R[2,1](n, x) (i.e., e.g.f. of the triangle): (1/(1-2*t))*exp(x*t/(1-2*t)).
E.g.f. of column m sequences: (t^m/(1-2*t)^(m+1))/m!, m >= 0.
Meixner type identity: Sum_{k=0..n-1} (-1)^k*2^k*(D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx.
Sheffer recurrence: R(n, x) = [(2 + x)*1 + 4*(1 + x)*D_x + 4*x*(D_x)^2]*R(n-1, x), n >= 1, and R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment above): T(n, m) = (2*n!*(1 + m)/(n-1))*Sum_{p=0..n-1-m} 2^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, and input T(m, m) = 1. - Wolfdieter Lang, Aug 12 2017
Explicit form (from the diagonal sequences with the o.g.f.s given as a comment above): T(n, m) = 2^(n-m)*(n-m)!*(binomial(n, n-m))^2 for n >= m >= 0. - Wolfdieter Lang, Sep 23 2017
Let R(n,x) denote the n-th row polynomial. Then x^n*R(n,x) = x^n o x^n, where o denotes the deformed Hadamard product of power series defined in Bala, Section 3.1. - Peter Bala, Jan 18 2018
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