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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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)

A154537 Triangle T(n,m) read by rows: let p(n,x) = exp(-x) * Sum_{m >= 0} (2*m + 1)^n * x^m/m!; then T(n,m) = [x^m] p(n,x).

Original entry on oeis.org

1, 1, 2, 1, 8, 4, 1, 26, 36, 8, 1, 80, 232, 128, 16, 1, 242, 1320, 1360, 400, 32, 1, 728, 7084, 12160, 6320, 1152, 64, 1, 2186, 36876, 99288, 81200, 25312, 3136, 128, 1, 6560, 188752, 768768, 929376, 440832, 91392, 8192, 256, 1, 19682, 956880, 5758880, 9901920, 6707904, 2069760, 305664, 20736, 512
Offset: 0

Views

Author

Roger L. Bagula, Jan 11 2009

Keywords

Comments

Row sums are A126390.
These numbers are related to Stirling numbers of the second kind as MacMahon numbers A060187 are related to Eulerian numbers.
Let p and q denote operators acting on a function f(x) by pf(x) = x*f(x) and qf(x) = d/dx(f(x)). Let A be the anticommutator operator qp + pq. Then A^n = Sum_{k = 0..n} T(n,k) p^k q^k. For example, A^3(f) = f + 26*x*df/dx + 36*x^2*d^2(f)/dx^2 + 8*x^3*d^3(f)/dx^3. - Peter Bala, Jul 24 2014
From Peter Bala, May 21 2023: (Start)
Compare the definition of the polynomial p(n,x) with Dobiński's formula for the Bell polynomials (row polynomials of A008277 for n >= 1): Bell(n,x) = exp(-x) * Sum_{m >= 0} m^n * x^m/m!.
Boyadzhiev has shown that Bell(n,x) = d/dx( exp(-x) * Sum_{m >= 0} (1^n + 2^n + ... + (m-1)^n) * x^m/m! ). The corresponding result for this table is that the n-th row polynomial p(n,x) = d/dx( exp(-x) * Sum_{m >= 0} (1^n + 3^n + ... + (2*m-1)^n) * x^m/m! ). (End)

Examples

			Triangle begins:
  {1},
  {1, 2},
  {1, 8, 4},
  {1, 26, 36, 8},
  {1, 80, 232, 128, 16},
  {1, 242, 1320, 1360, 400, 32},
  {1, 728, 7084, 12160, 6320, 1152, 64},
  {1, 2186, 36876, 99288, 81200, 25312, 3136, 128},
  {1, 6560, 188752, 768768, 929376, 440832, 91392, 8192, 256},
  {1, 19682, 956880, 5758880, 9901920, 6707904, 2069760, 305664, 20736, 512},
  ...
Boas-Buck recurrence for column m = 2, and n = 4: T(4,2) = (1/2)*[4*3*T(3, 2) + 2*6*(-2)^2*Bernoulli(2)*T(2,2)] = (1/2)*(12*36 + 12*4*(1/6)*4) = 232. - _Wolfdieter Lang_, Aug 11 2017
		

Crossrefs

Programs

  • Mathematica
    p[x_, n_] = Sum[(2*m + 1)^n*x^m/m!, {m, 0, Infinity}]/(Exp[x]);
    Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}]
    Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}];
    Flatten[%]

Formula

From Peter Bala, Oct 28 2011: (Start)
T(n,k) = 1/k!*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n.
Recurrence relation: T(n,k) = 2*T(n-1,k-1) + (2*k+1)*T(n-1,k).
T(n,k) = (2^k)*A039755(n,k).
E.g.f.: exp(x + y*(exp(2*x) - 1)) = 1 + (1 + 2*y)*x + (1 + 8*y + 4*y^2)*x^2/2! + .... (End)
T(n, k) = Sum_{m=0..n} binomial(n, m)*2^m*Stirling2(m, k), 0 <= k <= n, where Stirling2 is A048993. - Wolfdieter Lang, Apr 13 2017
Boas-Buck recurrence for column sequence m: T(n,k) = (1/(n - k))*[n*(1 + m)*T(n-1,k) + k*Sum_{p=m..n-2} binomial(n,p)*(-2)^(n-p)*Bernoulli(n-p)*T(p,k)], for n > m >= 0, with input T(m,m) = 2^m. See a comment in A282629, also for references, and an example below. - Wolfdieter Lang, Aug 11 2017

Extensions

Edited by N. J. A. Sloane, Jan 12 2009

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

A103220 a(n) = n*(n+1)*(3*n^2+n-1)/6.

Original entry on oeis.org

0, 1, 13, 58, 170, 395, 791, 1428, 2388, 3765, 5665, 8206, 11518, 15743, 21035, 27560, 35496, 45033, 56373, 69730, 85330, 103411, 124223, 148028, 175100, 205725, 240201, 278838, 321958, 369895, 422995, 481616, 546128, 616913, 694365, 778890
Offset: 0

Views

Author

Lambert Klasen (lambert.klasen(AT)gmx.de) and Gary W. Adamson, Jan 25 2005

Keywords

Comments

Row sums of A103219.
From Bruno Berselli, Dec 10 2010: (Start)
a(n) = n*A002412(n) - Sum_{i=0..n-1} A002412(i). More generally: n^2*(n+1)*(2*d*n-2*d+3)/6 - (Sum_{i=0..n-1} i*(i+1)*(2*d*i-2*d+3))/6 = n * (n+1) * (3*d*n^2-d*n+4*n-2*d+2)/12; in this sequence is d=2.
The inverse binomial transform yields 0, 1, 11, 22, 12, 0, 0 (0 continued). (End)
a(n-1) is also number of ways to place 2 nonattacking semi-queens (see A099152) on an n X n board. - Vaclav Kotesovec, Dec 22 2011
Also, one-half the even-indexed terms of the partial sums of A045947. - J. M. Bergot, Apr 12 2018

Crossrefs

Programs

  • Maple
    for(n=0,100,print1((3*n^4+4*n^3-n)/6,","))
  • Mathematica
    CoefficientList[Series[- x (1 + 8 x + 3 x^2) / (x - 1)^5, {x, 0, 40}], x] (* Vincenzo Librandi, May 12 2013 *)
    LinearRecurrence[{5,-10,10,-5,1},{0,1,13,58,170},40] (* Harvey P. Dale, Jan 23 2016 *)
  • PARI
    a(n)=n*(n+1)*(3*n^2+n-1)/6 \\ Charles R Greathouse IV, Oct 07 2015

Formula

G.f.: x*(1+8*x+3*x^2)/(1-x)^5.
a(n) = Sum_{i=1..n} Sum_{j=1..n} max(i,j)^2. - Enrique Pérez Herrero, Jan 15 2013
a(n) = a(n-1) + (2*n-1)*n^2 with a(0)=0, see A015237. - J. M. Bergot, Jun 10 2017
From Wesley Ivan Hurt, Nov 20 2021: (Start)
a(n) = Sum_{k=1..n} k * C(2*k,2).
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). (End)
From Peter Bala, Sep 03 2023: (Start)
a(n) = Sum_{1 <= i <= j <= n} (2*i - 1)*(2*j - 1).
Second subdiagonal of A039755. (End)

A145905 Square array read by antidiagonals: Hilbert transform of triangle A060187.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 9, 5, 1, 1, 27, 25, 7, 1, 1, 81, 125, 49, 9, 1, 1, 243, 625, 343, 81, 11, 1, 1, 729, 3125, 2401, 729, 121, 13, 1, 1, 2187, 15625, 16807, 6561, 1331, 169, 15, 1, 1, 6561, 78125, 117649, 59049, 14641, 2197, 225, 17, 1, 1, 19683, 390625, 823543
Offset: 0

Views

Author

Peter Bala, Oct 27 2008

Keywords

Comments

Definition of the Hilbert transform of a triangular array:
For many square arrays in the database the entries in a row are polynomial in the column index, of degree d say and hence the row generating function has the form P(x)/(1-x)^(d+1), where P is some polynomial function. Often the array whose rows are formed from the coefficients of these P polynomials is of independent interest. This suggests the following definition.
Let [L(n,k)]n,k>=0 be a lower triangular array and let R(n,x) := sum {k = 0 .. n} L(n,k)*x^k, denote the n-th row generating polynomial of L. Then we define the Hilbert transform of L, denoted Hilb(L), to be the square array whose n-th row, n >= 0, has the generating function R(n,x)/(1-x)^(n+1).
In this particular case, L is the array A060187, the array of Eulerian numbers of type B, whose row polynomials are the h-polynomials for permutohedra of type B. The Hilbert transform is an infinite Vandermonde matrix V(1,3,5,...).
We illustrate the Hilbert transform with a few examples:
(1) The Delannoy number array A008288 is the Hilbert transform of Pascal's triangle A007318 (view as the array of coefficients of h-polynomials of n-dimensional cross polytopes).
(2) The transpose of the array of nexus numbers A047969 is the Hilbert transform of the triangle of Eulerian numbers A008292 (best viewed in this context as the coefficients of h-polynomials of n-dimensional permutohedra of type A).
(3) The sequence of Eulerian polynomials begins [1, x, x + x^2, x + 4*x^2 + x^3, ...]. The coefficients of these polynomials are recorded in triangle A123125, whose Hilbert transform is A004248 read as square array.
(4) A108625, the array of crystal ball sequences for the A_n lattices, is the Hilbert transform of A008459 (viewed as the triangle of coefficients of h-polynomials of n-dimensional associahedra of type B).
(5) A142992, the array of crystal ball sequences for the C_n lattices, is the Hilbert transform of A086645, the array of h-vectors for type C root polytopes.
(6) A108553, the array of crystal ball sequences for the D_n lattices, is the Hilbert transform of A108558, the array of h-vectors for type D root polytopes.
(7) A086764, read as a square array, is the Hilbert transform of the rencontres numbers A008290.
(8) A143409 is the Hilbert transform of triangle A073107.

Examples

			Triangle A060187 (with an offset of 0) begins
1;
1, 1;
1, 6, 1;
so the entries in the first three rows of the Hilbert transform of
A060187 come from the expansions:
Row 0: 1/(1-x) = 1 + x + x^2 + x^3 + ...;
Row 1: (1+x)/(1-x)^2 = 1 + 3*x + 5*x^2 + 7*x^3 + ...;
Row 2: (1+6*x+x^2)/(1-x)^3 = 1 + 9*x + 25*x^2 + 49*x^3 + ...;
The array begins
n\k|..0....1.....2.....3......4
================================
0..|..1....1.....1.....1......1
1..|..1....3.....5.....7......9
2..|..1....9....25....49.....81
3..|..1...27...125...343....729
4..|..1...81...625..2401...6561
5..|..1..243..3125.16807..59049
...
		

Crossrefs

Cf. A008292, A039755, A052750 (first superdiagonal), A060187, A114172, A145901.

Programs

  • Maple
    T:=(n,k) -> (2*k + 1)^n: seq(seq(T(n-k,k),k = 0..n),n = 0..10);

Formula

T(n,k) = (2*k + 1)^n, (see equation 4.10 in [Franssens]). This array is the infinite Vandermonde matrix V(1,3,5,7, ....) having a LDU factorization equal to A039755 * diag(2^n*n!) * transpose(A007318).

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)

A016209 Expansion of 1/((1-x)(1-3x)(1-5x)).

Original entry on oeis.org

1, 9, 58, 330, 1771, 9219, 47188, 239220, 1205941, 6059229, 30384718, 152189310, 761743711, 3811110039, 19062724648, 95335146600, 476740303081, 2383895225649, 11920057258978, 59602029687090
Offset: 0

Views

Author

Keywords

Comments

For a combinatorial interpretation following from a(n) = A039755(n+2,2) = h^{(3)}A039755.%20-%20_Wolfdieter%20Lang">n, the complete homogeneous symmetric function of degree n in the symbols {1, 3, 5} see A039755. - _Wolfdieter Lang, May 26 2017

Examples

			a(2) = h^{(3)}_2 = 1^2 + 3^2 + 5^2 + 1^1*(3^1 + 5^1) + 3^1*5^1 = 58. - _Wolfdieter Lang_, May 26 2017
		

Crossrefs

Programs

  • Magma
    [(5^(n+2)-2*3^(n+2)+1)/8: n in [0..20]]; // Vincenzo Librandi, Sep 17 2011
  • Maple
    A016209 := proc(n) (5^(n+2)-2*3^(n+2)+1)/8; end proc: # R. J. Mathar, Mar 22 2011
  • Mathematica
    Join[{a=1,b=9},Table[c=8*b-15*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2011 *)
    CoefficientList[Series[1/((1-x)(1-3x)(1-5x)),{x,0,30}],x] (* or *) LinearRecurrence[ {9,-23,15},{1,9,58},30] (* Harvey P. Dale, Feb 20 2020 *)
  • PARI
    a(n)=if(n<0,0,n+=2; (5^n-2*3^n+1)/8)
    

Formula

a(n) = A039755(n+2, 2).
a(n) = (5^(n+2) - 2*3^(n+2)+1)/8 = a(n-1) + A005059(n+1) = 8*a(n-1) - 15*a(n-2) + 1 = (A003463(n+2) - A003462(n+2))/2. - Henry Bottomley, Jun 06 2000
G.f.: 1/((1-x)(1-3*x)(1-5*x)). See the name.
E.g.f.: (25*exp(5*x) - 18*exp(3*x) + exp(x))/8, from the e.g.f. of the third column (k=2) of A039755. - Wolfdieter Lang, May 26 2017

A214406 Triangle of second-order Eulerian numbers of type B.

Original entry on oeis.org

1, 1, 1, 1, 8, 3, 1, 33, 71, 15, 1, 112, 718, 744, 105, 1, 353, 5270, 14542, 9129, 945, 1, 1080, 33057, 191384, 300291, 129072, 10395, 1, 3265, 190125, 2033885, 6338915, 6524739, 2071215, 135135, 1, 9824, 1038780, 18990320, 103829590, 204889344, 150895836, 37237680, 2027025
Offset: 0

Views

Author

Peter Bala, Jul 17 2012

Keywords

Comments

The second-order Eulerian numbers A008517 count Stirling permutations by ascents. A Stirling permutation of order n is a permutation of the multiset {1,1,2,2,...,n,n} such that for each i, 1 <= i <= n, the elements lying between the two occurrences of i are larger than i.
We define a signed Stirling permutation of order n to be a vector (x_0, x_1, ..., x_(2*n)) such that x_0 = 0 and (|x_1|, ... ,|x_(2*n)|) is a Stirling permutation of order n. We say that a signed Stirling permutation (x_0, x_1, ... , x_(2*n)) has an ascent at position j, 0 <= j <= 2*n-1, if |x_j| < |x_(j+1)|. We define T(n,k), the second-order Eulerian numbers of type B, as the number of signed Stirling permutations of order n having k ascents. An example is given below.

Examples

			Row 2: [1,8,3]:
Signed Stirling permutations of order 2
= = = = = = = = = = = = = = = = = = = =
..............ascents...................ascents
(0 2 2 1 1)......1.......(0 -2 -2 1 1).....1
(0 1 2 2 1)......2.......(0 1 -2 -2 1).....2
(0 1 1 2 2)......2.......(0 1 1 -2 -2).....1
(0 2 2 -1 -1)....1.......(0 -2 -2 -1 -1)...1
(0 -1 2 2 -1)....1.......(0 -1 -2 -2 -1)...1
(0 -1 -1 2 2)....1.......(0 -1 -1 -2 -2)...0
............................................
Triangle begins
.n\k.|..0.....1......2.......3......4........5......6
= = = = = = = = = = = = = = = = = = = = = = = = = = =
..0..|..1
..1..|..1.....1
..2..|..1.....8......3
..3..|..1....33.....71......15
..4..|..1...112....718.....744....105
..5..|..1...353...5270...14542...9129......945
..6..|..1..1080..33057..191384..300291..129072..10395
...
Recurrence example: T(4,2) = 11*T(3,1) + 5*T(3,2) = 11*33 + 5*71 = 718.
		

Crossrefs

Cf. A001813 (row sums), A008517, A039755, A185896, A034940.

Programs

  • Mathematica
    T[n_, k_] /; 0 < k <= n := T[n, k] = (4n-2k-1) T[n-1, k-1] + (2k+1) T[n-1, k]; T[, 0] = 1; T[, _] = 0;
    Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 11 2019 *)

Formula

T(n,k) = Sum_{i = 0..k} (-1)^(i-k)*binomial(2*n+1,k-i)*S(n+i,i), where S(n,k) = 1/(2^k*k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n = A039755(n,k).
It appears that Sum_{k = 0..n} (-1)^(k+1)*T(n,k)/((2*n-k)*binomial(2*n,k)) = (-1)^n *(2^n-2)*Bernoulli(n)/n.
Recurrence equation: T(n,k) = (4*n-2*k-1)*T(n-1,k-1) + (2*k+1)*T(n-1,k), for n,k >= 0.
The row polynomials R(n,x) may be calculated by means of the recurrence equation R(0,x) = 1 and for n >= 0, R(n,x^2) = (1 - x^2)^(2*n)*d/dx( x/(1-x^2)^(2*n-1)*R(n-1,x^2) ). Equivalently, x*R(n,x^2)/(1 - x^2)^(2*n+1)) = D^n(x), where D is the differential operator x/(1 - x^2)*d/dx.
Another recurrence is R(n+1,x) = 2*x*(1 - x)*d/dx(R(n,x)) + (1 + (4*n+1)*x)*R(n,x). It follows that the row polynomials R(n,x) have only real zeros (apply Liu and Wang, Corollary 1.2 with f(x) = R(n,x) and g(x) = R'(n,x)).
For n >= 0, the rational functions Q(n,x) := R(n,x)/(1 - x)^(2*n+1) are the o.g.f.'s for the diagonals of the type B Stirling numbers of the second kind A039755. They appear to satisfy the semi-orthogonality property Integral_{x = 0..oo} (1 - x)*Q(n,x)*Q(m,x) dx = (-1)^n*(2^(n+m) - 2)*Bernoulli(n+m)/(n+m), for n, m >= 0 but excluding the case (n,m) = (0,0). A similar result holds for the row polynomials of A185896.
Row sums are A001813.
Define functions F(n,z) := Sum_{k >= 0} (2*k+1)^(k+n)*z^k/k!, n = 0,1,2,.... Then exp(-x/2)*F(n,x/2*exp(-x)) = R(n,x)/(1 - x)^(2*n+1). - Peter Bala, Jul 26 2012

Extensions

Missing 1 in data inserted by Jean-François Alcover, Nov 11 2019
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