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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A051150 Generalized Stirling number triangle of first kind.

Original entry on oeis.org

1, -5, 1, 50, -15, 1, -750, 275, -30, 1, 15000, -6250, 875, -50, 1, -375000, 171250, -28125, 2125, -75, 1, 11250000, -5512500, 1015000, -91875, 4375, -105, 1, -393750000, 204187500, -41037500, 4230625
Offset: 1

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Author

Keywords

Comments

a(n,m) = R_n^m(a=0, b=5) in the notation of the given 1961 and 1962 references.
a(n,m) is a Jabotinsky matrix, i.e., the monic row polynomials E(n,x) := Sum_{m=1..n} a(n,m)*x^m = Product_{j=0..n-1} (x - 5*j), n >= 1, and E(0,x) := 1 are exponential convolution polynomials (see A039692 for the definition and a Knuth reference).
This is the signed Stirling1 triangle A008275 with diagonal d >= 0 (main diagonal d = 0) scaled with 5^d.

Examples

			Triangle a(n,m) (with rows n >= 1 and columns m = 1..n) begins:
        1;
       -5,      1;
       50,    -15,      1;
     -750,    275,    -30,   1;
    15000,  -6250,    875, -50,    1;
  -375000, 171250, -28125, 2125, -75, 1;
  ...
3rd row o.g.f.: E(3,x) = 50*x - 15*x^2 + x^3.
		

Crossrefs

First (m=1) column sequence: A052562(n-1).
Row sums (signed triangle): A008546(n-1)*(-1)^(n-1).
Row sums (unsigned triangle): A008548(n).
Cf. A008275 (Stirling1 triangle, b=1), A039683 (b=2), A051141 (b=3), A051142 (b=4).

Formula

a(n, m) = a(n-1, m-1) - 5*(n-1)*a(n-1, m) for n >= m >= 1; a(n, m) := 0 for n < m; a(n, 0) := 0 for n >= 1; a(0, 0) = 1.
E.g.f. for the m-th column of the signed triangle: (log(1 + 5*x)/5)^m/m!.
a(n, m) = S1(n, m)*5^(n-m), with S1(n, m) := A008275(n, m) (signed Stirling1 triangle).

A051151 Generalized Stirling number triangle of first kind.

Original entry on oeis.org

1, -6, 1, 72, -18, 1, -1296, 396, -36, 1, 31104, -10800, 1260, -60, 1, -933120, 355104, -48600, 3060, -90, 1, 33592320, -13716864, 2104704, -158760, 6300, -126, 1, -1410877440, 609700608, -102114432, 8772624, -423360, 11592, -168
Offset: 1

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Author

Keywords

Comments

a(n,m) = R_n^m(a=0, b=6) in the notation of the given 1961 and 1962 references.
a(n,m) is a Jabotinsky matrix, i.e., the monic row polynomials E(n,x) := Sum_{m=1..n} a(n,m)*x^m = Product_{j=0..n-1} (x-6*j), n >= 1, and E(0,x) := 1 are exponential convolution polynomials (see A039692 for the definition and a Knuth reference).
This is the signed Stirling1 triangle A008275 with diagonal d >= 0 (main diagonal d = 0) scaled with 6^d.

Examples

			Triangle a(n,m) (with rows n >= 1 and columns m = 1..n) begins:
        1;
       -6,      1;
       72,    -18,      1;
    -1296,    396,    -36,    1;
    31104, -10800,   1260,  -60,   1;
  -933120, 355104, -48600, 3060, -90, 1;
   ...
3rd row o.g.f.: E(3,x) = 72*x - 18*x^2 + x^3.
		

Crossrefs

First (m=1) column sequence is: A047058(n-1).
Row sums (signed triangle): A008543(n-1)*(-1)^(n-1).
Row sums (unsigned triangle): A008542(n).
Cf. A008275 (Stirling1 triangle, b=1), A039683 (b=2), A051141 (b=3), A051142 (b=4), A051150 (b=5).

Formula

a(n, m) = a(n-1, m-1) - 6*(n-1)*a(n-1, m), n >= m >= 1; a(n, m) := 0, n < m; a(n, 0) := 0 for n >= 1; a(0, 0) = 1.
E.g.f. for the m-th column of the signed triangle: ((log(1 + 6*x)/6)^m)/m!.
a(n, m) = S1(n, m)*6^(n-m), with S1(n, m) := A008275(n, m) (signed Stirling1 triangle).

Extensions

Various sections edited by Petros Hadjicostas, Jun 08 2020

A051186 Generalized Stirling number triangle of first kind.

Original entry on oeis.org

1, -7, 1, 98, -21, 1, -2058, 539, -42, 1, 57624, -17150, 1715, -70, 1, -2016840, 657874, -77175, 4165, -105, 1, 84707280, -29647548, 3899224, -252105, 8575, -147, 1, -4150656720, 1537437132, -220709524, 16252369, -672280, 15778, -196, 1
Offset: 1

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Author

Keywords

Comments

T(n,m) = R_n^m(a=0, b=7) in the notation of the given 1962 reference.
T(n,m) is a Jabotinsky matrix, i.e., the monic row polynomials E(n,x) := Sum_{m=1..n} T(n,m)*x^m = Product_{j=0..n-1} (x-7*j), n >= 1, and E(0,x) := 1 are exponential convolution polynomials (see A039692 for the definition and a Knuth reference).
From Petros Hadjicostas, Jun 07 2020: (Start)
For integers n, m >= 0 and complex numbers a, b (with b <> 0), the numbers R_n^m(a,b) were introduced by Mitrinovic (1961) and further examined by Mitrinovic and Mitrinovic (1962).
They are defined via Product_{r=0..n-1} (x - (a + b*r)) = Sum_{m=0..n} R_n^m(a,b)*x^m for n >= 0. As a result, R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b) for n >= m >= 1 with R_1^0(a,b) = a, R_1^1(a,b) = 1, R_n^m(a,b) = 0 for n < m, and R_0^0(a,b) = 1.
With a = 0 and b = 1, we get the Stirling numbers of the first kind S1(n,m) = R_n^m(a=0, b=1) = A048994(n,m).
We have R_n^m(a,b) = Sum_{k=0}^{n-m} (-1)^k * a^k * b^(n-m-k) * binomial(m+k, k) * S1(n, m+k) for n >= m >= 0.
For the current array, T(n,m) = R_n^m(a=0, b=7) but with no zero row or column. (End)

Examples

			Triangle T(n,m) (with rows n >= 1 and columns m = 1..n) begins:
         1;
        -7,      1;
        98,    -21,      1;
     -2058,    539,    -42,    1;
     57624, -17150,   1715,  -70,    1;
  -2016840, 657874, -77175, 4165, -105, 1;
  ...
3rd row o.g.f.: E(3,x) = Product_{j=0..2} (x - 7*j) = 98*x - 21*x^2 + x^3.
		

Crossrefs

Cf. A000142, A045754 (unsigned row sums), A049209 (row sums), A051188.
The b=1..6 triangles are: A008275 (Stirling1 triangle), A039683, A051141, A051142, A051150, A051151.

Programs

  • Magma
    [7^(n-k)*StirlingFirst(n,k): k in [1..n], n in [1..12]]; // G. C. Greubel, Feb 22 2022
    
  • Mathematica
    Table[7^(n-k)*StirlingS1[n, k], {n,12}, {k,n}]//Flatten (* G. C. Greubel, Feb 22 2022 *)
  • Sage
    flatten([[(-7)^(n-k)*stirling_number1(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Feb 22 2022

Formula

T(n, m) = T(n-1, m-1) - 7*(n-1)*T(n-1, m) for n >= m >= 1, T(n, m) = 0 for n < m, T(n, 0) = 0 for n >= 1, and T(0, 0) = 1.
T(n, 1) = A051188(n-1).
Sum_{k=0..n} T(n, k) = (-1)^(n-1)*A049209(n-1).
Sum_{k=0..n} (-1)^(n-k)*T(n, k) = A045754(n).
E.g.f. for m-th column of signed triangle: (log(1 + 7*x)/7)^m/m!.
T(n,m) = 7^(n-m)*S1(n,m) with the (signed) Stirling1 triangle S1(n,m) = A008275(n,m).
Bivariate e.g.f.-o.g.f.: Sum_{n,m >= 1} T(n,m)*x^n*y^m/n! = exp((y/7)*log(1 + 7*x)) - 1 = (1 + 7*x)^(y/7) - 1. - Petros Hadjicostas, Jun 07 2020
T(n, 0) = (-7)^(n-1)*A000142(n-1). - G. C. Greubel, Feb 22 2022

A051187 Generalized Stirling number triangle of the first kind.

Original entry on oeis.org

1, -8, 1, 128, -24, 1, -3072, 704, -48, 1, 98304, -25600, 2240, -80, 1, -3932160, 1122304, -115200, 5440, -120, 1, 188743680, -57802752, 6651904, -376320, 11200, -168, 1, -10569646080, 3425697792, -430309376, 27725824, -1003520, 20608, -224, 1
Offset: 1

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Author

Keywords

Comments

T(n,m)= R_n^m(a=0, b=8) in the notation of the given 1962 reference.
T(n,m) is a Jabotinsky matrix, i.e. the monic row polynomials E(n,x) := Sum_{m=1..n} T(n,m)*x^m = Product_{j=0..n-1} (x - 8*j), n >= 1, and E(0,x) := 1 are exponential convolution polynomials (see A039692 for the definition and a Knuth reference).
From Petros Hadjicostas, Jun 07 2020: (Start)
For integers n, m >= 0 and complex numbers a, b (with b <> 0), the numbers R_n^m(a,b) were introduced by Mitrinovic (1961) and further examined by Mitrinovic and Mitrinovic (1962). Such numbers are related to the work of Nörlund (1924).
They are defined via Product_{r=0..n-1} (x - (a + b*r)) = Sum_{m=0..n} R_n^m(a,b)*x^m for n >= 0. As a result, R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b) for n >= m >= 1 with R_1^0(a,b) = a, R_1^1(a,b) = 1, R_n^m(a,b) = 0 for n < m, and R_0^0(a,b) = 1.
With a = 0 and b = 1, we get the Stirling numbers of the first kind S1(n,m) = R_n^m(a=0, b=1) = A048994(n,m).
We have R_n^m(a,b) = Sum_{k=0}^{n-m} (-1)^k * a^k * b^(n-m-k) * binomial(m+k, k) * S1(n, m+k) for n >= m >= 0.
For the current array, T(n,m) = R_n^m(a=0, b=8) but with no zero row or column. (End)

Examples

			Triangle T(n,m) (with rows n >= 1 and columns m = 1..n) begins:
          1;
         -8,         1;
        128,       -24,       1;
      -3072,       704,     -48,       1;
      98304,    -25600,    2240,     -80,     1;
   -3932160,   1122304, -115200,    5440,  -120,    1;
  188743680, -57802752, 6651904, -376320, 11200, -168, 1;
  ...
3rd row o.g.f.: E(3,x) = Product_{j=0..2} (x - 8*j) = 128*x - 24*x^2 + x^3.
		

Crossrefs

First (m=1) column sequence is: A051189(n-1).
Row sums (signed triangle): A049210(n-1)*(-1)^(n-1).
Row sums (unsigned triangle): A045755(n).
The b=1..7 triangles are: A008275 (Stirling1 triangle), A039683, A051141, A051142, A051150, A051151, A051186.

Formula

T(n, m) = T(n-1, m-1) - 8*(n-1)*T(n-1, m) for n >= m >= 1; T(n, m) := 0 for n < m; T(n, 0) := 0 for n >= 1; T(0, 0) = 1.
E.g.f. for the m-th column of the signed triangle: (log(1 + 8*x)/8)^m/m!.
From Petros Hadjicostas, Jun 07 2020: (Start)
T(n,m) = 8^(n-m)*Stirling1(n,m) = 8^(n-m)*A048994(n,m) = 8^(n-m)*A008275(n,m) for n >= m >= 1.
Bivariate e.g.f.-o.g.f.: Sum_{n,m >= 1} T(n,m)*x^n*y^m/n! = exp((y/8)*log(1 + 8*x)) - 1 = (1 + 8*x)^(y/8) - 1. (End)

A102625 Triangle read by rows: T(n,k) is the sum of the weights of all vertices labeled k at depth n in the Catalan tree (1 <= k <= n+1, n >= 0).

Original entry on oeis.org

1, 1, 2, 3, 6, 6, 15, 30, 36, 24, 105, 210, 270, 240, 120, 945, 1890, 2520, 2520, 1800, 720, 10395, 20790, 28350, 30240, 25200, 15120, 5040, 135135, 270270, 374220, 415800, 378000, 272160, 141120, 40320, 2027025, 4054050, 5675670, 6486480
Offset: 0

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Author

Emeric Deutsch, Jan 31 2005

Keywords

Comments

The Catalan tree is defined as follows: the root is labeled 1 and each vertex labeled i has i+1 children labeled 1,2,...,i+1. The weight of a vertex v is the product of all labels on the path from the root to v. Row n contains n+1 terms. Row sums and column 1 yield the double factorials (A001147). T(n,n+1)=(n+1)!, T(n,n)=n(n+1)!/2 (A001286; Lah numbers).
This table counts permutations of the multiset {1,1,2,2,...,n,n} satisfying the condition "the first appearance of i + 1 follows the first appearance of i" by the position of the first appearance of n. Specifically, T(n+1,k) is the number of such permutations for which n first occurs in position 2n+1-k. For example, with n=2 and k=1, T(3,1)=6 counts 121323, 121332, 122313, 122331, 112323, 112332. - David Callan, Nov 29 2007
T(n+1,k) is also the number of rooted complete binary forests with n labeled leaves and k labeled roots. This follows by comparing exponential generating functions; see Example 5.2.6 and Proposition 5.1.3 of Stanley's "Enumerative Combinatorics 2." - Timothy Y. Chow, Mar 28 2017

Examples

			Triangle starts:
   1;
   1,  2;
   3,  6,  6;
  15, 30, 36, 24;
  ...
Production matrix begins:
1, 2
1, 2, 3
1, 2, 3, 4
1, 2, 3, 4, 5
1, 2, 3, 4, 5, 6
1, 2, 3, 4, 5, 6, 7
1, 2, 3, 4, 5, 6, 7, 8
1, 2, 3, 4, 5, 6, 7, 8, 9
... - _Philippe Deléham_, Sep 30 2014
From _Peter Bala_, Apr 16 2017: (Start)
The Catalan tree starts          o1
                                / \
                               /   \
                              /     \
                             /       \
                            /         \
                           o1          o2
                          / \         /|\
                         /   \       / | \
                        /     \     /  |  \
                       o1      o2  o1  o2  o3
Level 2:
2 vertices labeled 1: total weight 1x1x1 + 1x2x1 = 3
2 vertices labeled 2: total weight 2x1x1 + 2x2x1 = 6
1 vertex labeled 3:   total weight 3x2x1         = 6
(End)
		

Crossrefs

Programs

  • Maple
    A102625:=proc(n,k) if k<=n+1 then k*(2*n-k+1)!/2^(n-k+1)/(n-k+1)! else 0 fi end proc:
    for n from 0 to 8 do seq(A102625(n,k),k=1..n+1) od; # yields sequence in triangular form
  • Mathematica
    t[n_, k_] := k*(2n-k+1)!/(2^(n-k+1)*(n-k+1)!); Table[t[n, k], {n, 0, 8}, {k, 1, n+1}] // Flatten (* Jean-François Alcover, Jan 21 2013 *)
  • PARI
    {T(n, k) = my(m = n-k+1); if( k<1 || k>n+1, 0, k * (n+m)! / (2^m * m!))}; /* Michael Somos, Aug 16 2016 */

Formula

T(n,k) = k*(2*n-k+1)!/[2^(n-k+1)*(n-k+1)!] (1 <= k <= n+1).
From Tom Copeland, Nov 11 2007: (Start)
Bivariate G.F.: exp[P(.,t)*x] = D_x {1 - [g(x)/(1+t*g(x))]} = 1 / {(1+g(x))*[1+t*g(x)]^2}, where g(x) = sqrt(1-2*x) - 1 and P(n,t) = Sum_{k=0..n} T(n,k) * t^k.
Also D_x g(x) = -(1-2*x)^(-1/2) = -exp[x*A001147(.)] = -exp[x *(2*(.)-1)!! ], so the coefficients of x^n/n! in the expansion of g(x) are -(2*(n-1)-1)!! = -A001147(n-1) for n > 0.
See A132382 for an array which is essentially the revert from which this G.f. may be derived and for connections to other arrays. (End)
E.g.f.: 1/(1 - x + x*sqrt(1-2*z)) = 1 + x*z + (x+2*x^2)*z^2/2! + (3*x+6*x^2+6*x^3)*z^3/3! + .... T(n,k) gives the number of plane recursive trees on n+2 nodes where the root has degree k (Bergeron et al., Corollary 5). - Peter Bala, Jul 09 2012
From Peter Bala, Jul 09 2014: (Start)
T(n,k) = k!*A001497(n,k) modulo offset differences.
The n-th row polynomial R(n,x) = (-1)^n/(x - 1)*( Sum_{k = 1..infinity} k*(k - 2)*...*(k - 2*n)*(x/(x - 1))^k ). Cf. the Dobinski-type formula for the row polynomials of A001497. (End)
From Tom Copeland, Aug 06 2016: (Start)
From the 2007 formulas above, an alternate g.f. for this entry is GF(x,t) = -g(x) / [1 + t*g(x)] = x + (1 + 2*t)*x^2/2! + (3 + 6*t + 6*t^2)*x^3/3! + ... with compositional inverse GFinv(x,t) = {1 - [1 - x / (1+t*x)]^2} / 2 = -(1/2)[x / (1+t*x)]^2 + x / (1+t*x) = Sum_{n>0} (-1)^(n+1) [(n-1)/2*t^(n-2) + t^(n-1)]*x^n, a series containing the Lah numbers A001286 when expressed as an e.g.f.
From A145271, with K(x,t) = 1 / dGinv(x,t)/dx = 1 + (1+2*t) x + (1+t+t^2) x^2 + x^3 / [1-(1-t)*x], then [K(x,t) d/dx]^n x evaluated at x=0 gives the n-th row polynomial of this entry.
Since the reciprocal of Bala's e.g.f. above generates a shifted, signed A001147, for the polynomials P(n,t) generated by Bala's e.g.f., umbrally (P(.,t) + a.)^n = 0 for n > 0 with a_0 = 1 and a_n = -t * A001147(n-1) for n > 0. E.g., (P(.,t) + a.)^2 = a_0 * P(2,t) + 2 a_1 * P(1,t) + a_2 * P(0,t) = 1 * (t + 2*t^2) + 2 * -t * t + -t * 1 = 0. (End)
From Peter Bala, Apr 16 2017: (Start)
T(n,k) = k*T(n-1,k-1) + (2*n - k)*T(n-1,k).
E.g.f.: t*x*c(x/2)/(1 - t*x*c(x/2)) = t*x + (t + 2*t^2)*x^2/2! + (3*t + 6*t^2 + 6*t^2)*x^3/3! + ..., where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. for the Catalan numbers A000108. Note that the related g.f. t*x*c(x)/(1 - t*x*c(x)) is the o.g.f. for A033184 (essentially the same as the Riordan array A106566) and enumerates the number of vertices labeled k on the n_th level of the Catalan tree (k >= 1, n >= 0). (End)

A113278 Triangle T, read by rows, such that the matrix square, T^2, forms a simple 2-diagonal triangle where [T^2](n,n) = 1 and [T^2](n+1,n) = 2*(n+1) for n>=0.

Original entry on oeis.org

1, 1, 1, -1, 2, 1, 3, -3, 3, 1, -15, 12, -6, 4, 1, 105, -75, 30, -10, 5, 1, -945, 630, -225, 60, -15, 6, 1, 10395, -6615, 2205, -525, 105, -21, 7, 1, -135135, 83160, -26460, 5880, -1050, 168, -28, 8, 1, 2027025, -1216215, 374220, -79380, 13230, -1890, 252, -36, 9, 1
Offset: 0

Views

Author

Paul D. Hanna, Oct 22 2005

Keywords

Examples

			Triangle begins:
  1;
  1,1;
  -1,2,1;
  3,-3,3,1;
  -15,12,-6,4,1;
  105,-75,30,-10,5,1;
  -945,630,-225,60,-15,6,1;
  10395,-6615,2205,-525,105,-21,7,1;
  ...
where T(n,k) = (-1)^(n-1-k)*A001147(n-1)*C(n,k).
The matrix square equals:
  1;
  2,1;
  0,4,1;
  0,0,6,1;
  0,0,0,8,1;
  0,0,0,0,10,1;
  0,0,0,0,0,12,1;
  ...
The matrix log, L, begins:
  0;
  1,0;
  -2,2,0;
  8,-6,3,0;
  -48,32,-12,4,0;
  384,-240,80,-20,5,0;
  -3840,2304,-720,160,-30,6,0;
  ...
where L(n,k) = (-1)^(n-1-k)*A000165(n-1)*C(n,k).
		

Crossrefs

Cf. A001147 (odd double factorials), A000165 (even double factorials).

Programs

  • Mathematica
    (* The function RiordanArray is defined in A256893. *)
    RiordanArray[Sqrt[1 + 2 #]&, #&, 10, True] // Flatten (* Jean-François Alcover, Jul 19 2019 *)
  • PARI
    {T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r==c,1,if(r==c+1,2*c)))); (sum(i=0,n+1,(sum(j=1,n+1,-(M^0-M)^j/j)/2)^i/i!))[n+1,k+1]}

Formula

Exponential Riordan array [sqrt(1 + 2*x),x] with e.g.f. sqrt(1+2*x)*exp(t*x) = 1 + (1+t)*x + (-1+2*t+t^2)*x^2/2! + ... . The n-th row polynomial R(n,x) is given by the type B Dobinski formula R(n,x) = exp(-x/2)*sum {k = 0..inf} (2*k+1)*(2*k-1)*...*(2*k+1-2*(n-1))*(x/2)^k/k!. Cf. A122848. - Peter Bala, Jun 23 2014

A203412 Triangle read by rows, a(n,k), n>=k>=1, which represent the s=3, h=1 case of a two-parameter generalization of Stirling numbers arising in conjunction with normal ordering.

Original entry on oeis.org

1, 1, 1, 4, 3, 1, 28, 19, 6, 1, 280, 180, 55, 10, 1, 3640, 2260, 675, 125, 15, 1, 58240, 35280, 10360, 1925, 245, 21, 1, 1106560, 658000, 190680, 35385, 4620, 434, 28, 1, 24344320, 14266560, 4090240, 756840, 100065, 9828, 714, 36, 1
Offset: 1

Views

Author

Mark Shattuck, Jan 01 2012

Keywords

Comments

Also the Bell transform of the triple factorial numbers A007559 which adds a first column (1,0,0 ...) on the left side of the triangle. For the definition of the Bell transform see A264428. See A051141 for the triple factorial numbers A032031 and A004747 for the triple factorial numbers A008544 as well as A039683 and A132062 for the case of double factorial numbers. - Peter Luschny, Dec 23 2015

Examples

			Triangle starts:
[    1]
[    1,     1]
[    4,     3,     1]
[   28,    19,     6,    1]
[  280,   180,    55,   10,   1]
[ 3640,  2260,   675,  125,  15,  1]
[58240, 35280, 10360, 1925, 245, 21, 1]
		

Crossrefs

Programs

  • Maple
    A203412 := (n,k) -> (n!*3^n)/(k!*2^k)*add((-1)^j*binomial(k,j)*binomial(n-2*j/3-1, n), j=0..k): seq(seq(A203412(n,k),k=1..n),n=1..9); # Peter Luschny, Dec 21 2015
  • Mathematica
    Table[(n! 3^n)/(k! 2^k) Sum[ (-1)^j Binomial[k, j] Binomial[n - 2 j/3 - 1, n], {j, 0, k}], {n, 9}, {k, n}] // Flatten (* Michael De Vlieger, Dec 23 2015 *)
  • Sage
    # uses[bell_transform from A264428]
    triplefactorial = lambda n: prod(3*k + 1 for k in (0..n-1))
    def A203412_row(n):
        trifact = [triplefactorial(k) for k in (0..n)]
        return bell_transform(n, trifact)
    [A203412_row(n) for n in (0..8)] # Peter Luschny, Dec 21 2015

Formula

(1) Is given by the recurrence relation
a(n+1,k) = a(n,k-1)+(3*n-2*k)*a(n,k) if n>=0 and k>=1, along with the initial values a(n,0) = delta_{n,0} and a(0,k) = delta_{0,k} for all n,k>=0.
(2) Is given explicitly by
a(n,k) = (n!*3^n)/(k!*2^k)*Sum{j=0..k} (-1)^j*C(k,j)*C(n-2*j/3-1,n) for all n>=k>=1.
a(n,1) = A007559(n-1). - Peter Luschny, Dec 21 2015

A231846 Polynomials for total Pontryagin classes. Refinement of double Pochhammer triangle.

Original entry on oeis.org

1, 1, 2, 1, 8, 6, 1, 48, 32, 12, 12, 1, 384, 240, 160, 80, 60, 20, 1, 3840, 2304, 1440, 640, 720, 960, 120, 160, 180, 30, 1, 46080, 26880, 16128, 13440, 8064, 10080, 4480, 3360, 1680, 3360, 840, 280, 420, 42, 1, 645120, 368640, 215040, 172032, 80640, 107520, 129024, 107520, 40320, 35840, 21504, 40320, 17920, 26880, 1680, 3360, 8960, 3360, 448, 840, 56, 1
Offset: 0

Views

Author

Tom Copeland, Nov 14 2013

Keywords

Comments

The W. Lang link in A036039 explicitly gives the first several cycle index polynomials for the symmetric group S_n, or the partition polynomials for the refined Stirling numbers of the first kind. In line with the discussion in the Fecko link, null the indeterminates with odd indices, divide the 2n-th partition polynomial by the double factorial of odd numbers given in A001147, and re-index. The sum of the resulting row coefficients are also equal to A001147.

Examples

			In terms of the trace of a curvature form Tr(F^n)={n} or indeterminates c_n=[n]:
P_0 = 1,
P_1 = Tr(F^2) = {2}
    = c_1 = [1],
P_2 = 2Tr(F^4)+Tr(F^2)^2 = 2{4}+{2}^2
    = 2c_2+ (c_1)^2 = 2[2]+[1]^2,
P_3 = 8Tr(F^6)+6Tr(F^2)Tr(F^4)+Tr(F^2)^3= 8{6}+6{2}{4}+{2}^3
    = 8c_3+6c_1 c_2+(c_1)^3 = 8[3]+6[1][2]+[1]^3,
P_4 = 48{8}+32{2}{6}+12{4}^2+12{2}^2{4}+{2}^4
    = 48[4]+32[1][3]+12[2]^2+12[1]^2[2]+[1]^4,
P_5 = 384{10}+240{2}{8}+160{4}{6}+80{2}^2{6}
      + 60{2}{4}^2+20{2}^3{4}+{2}^5
    = 384[5]+240[1][4]+160[2][3]+80[1]^2[3]
      + 60[1][2]^2+20[1]^3[2]+[1]^5
P_6 = 3840[6]+2304[1][5]+1440[2][4]+640[3]^2+720[1]^2[4]
  +960[1][2][3]+120[2]^3+160[1]^3[3]+180[1]^2[2]^2+30[1]^4[2]+[1]^6
P_7 = 46080[7]+26880[1][6]+16128[2][5]+13440[3][4]+8064[1]^2[5]
  +10080[1][2][4]+4480[1][3]^2+3360[2]^2[3]+1680[1]^3[4]
  +3360[1]^2[2][3]+840[1][2]^3+280[1]^4[3]+420[1]^3[2]^2+42[1]^5[2]+[1]^7
....
Summing over partitions with the same number of blocks gives the unsigned double Pochhammer triangle A039683. Row sums are A001147. Multiplying P_n by the row sum gives the 2n-th partition polynomial of A036039 with its odd-indexed indeterminates nulled.
For c_1 = c_2 = x and c_n = 0 otherwise, see A119275. Let Omega(t) = xi(1/2 + i*t)/xi(1/2) where xi is the Landau version of the Riemann xi function, t is real, and i^2 = -1. The Taylor series coefficients vanish for odd order derivatives and, for even, are c_(2n) = Omega^(2n)(0) = (-1)^n * xi^(2n)(1/2) / xi(1/2) = A001147(n) * P_n as in the Example section with F^(2n) = -2 * Sum(1/x_k^(2n)) = -2 * Tr_(2n) where x_k is the imaginary part of the k-th zero of the Riemann zeta function and k ranges over all the zeros above the real axis. E.g., (see the Mathematics Stack Exchange question) summing over the first several thousands of zeros, c_4 = A001147(2)*P_2 = 3*[2*(-2*Tr_4) + (-2*Tr_2)^2] = 12*[-(0.000372) + (0.02311)^2] = .005962 and c_4 = xi^(4)*(1/2)/xi(1/2) = 0.002963/0.497 = 0.005962 (rounding off). Conversely, the Tr_(2n) can be calculated from the c_n using the Faber polynomials (A263916), as indicated in A036039. See Coffey for Taylor coefficients of Omega(t) about t = 0 and the MSE question for Tr_(2n). The traces are convergent and any zeros in the critical strip off the critical line would have a slightly more complicated real contribution to the traces but negligible to any practical order. - _Tom Copeland_, May 27 2020
		

Crossrefs

Cf. A263916.
The terms are indexed by partitions in the Abramowitz and Stegun order, A036036.

Programs

  • Mathematica
    rows[n_] := {{1}}~Join~With[{s = Exp[Sum[b[k] t^k/(2 k), {k, n}] + O[t]^(n+1)]}, Table[Expand@Coefficient[(2 k)!! s, t^k Product[b[t], {t, p}]], {k, n}, {p, Sort[Sort /@ IntegerPartitions[k]]}]];
    rows[8] // Flatten (* Andrey Zabolotskiy, Feb 19 2024 *)

Formula

From Tom Copeland, Oct 11 2016: (Start)
A generating function for the polynomials PB_n[b_2,b_4,..,b_(2n)] of this array is
exp[b_2 y^2/2 + b_4 y^4/4 + b_6 y^6/6 + ...] = Sum_{n >= 0} PB_n y^(2n) / A000165(n) = Sum_{n >= 0} St1[2n,0,b_2,0,b_4,0,..,b_(2n)] y^(2n) / (2n)! = Sum_{n >= 0} PB_n *(y/sqrt(2))^(2n) / n! with b_n = Tr(F^n), as in the examples, and St1(n,b_1,b_2,..,b_n), the partition polynomials of A036039. Then St1[2n,0,b_2,0,b_4,..,0,b_(2n)] = A001147(n) * PB_n.
The polynomials PC_n(c_1,c_2,..,c_n) of this array with c_k = b_(2k) are an Appell sequence in the indeterminate c_1 with lowering operator L = d/d(c_1), i.e., L*PC_n(c_1,..,c_n) = d(PC_n)/d(c_1) = n * PC_(n-1)[c_1,..,c_(n-1)].
With [PC.(c_1,c_2,..)]^n = PC_n(c_1,..,c_n), the e.g.f. is G(t,c_1,c_2,..) = exp[t*PC.(0,c_2,c_3,..)] * exp(t*c_1) = exp{t*[c_1 + PC.(0,c_2,c_3,..)]} = exp[t*PC.(c_1,c_2,..)] = exp[(1/2) * sum_{n > 0} c_n (2t)^n/n ] = exp[-log(1-2c.t) / 2], where, umbrally, (c.)^n = c_n.
The raising operator is R = d[log(G(L,c_1,c_2,..))]/dL = sum_{n >= 0} 2^n * c_(n+1) * (d/dc_1)^n = c./(1-2c.L), umbrally. R PC_n(c_1,..,c_n) = P_(n+1)[c_1,..,c_(n+1)].
Another generator: G(L,0,c_2,c_3,..) (c_1)^n = PC_n(c_1,c_2,..,c_n).
The Appell umbral compositional inverse sequence UPC_n to the PC_n sequence has e.g.f. UG(t,c_1,c_2,..) = [1 / G(t,0,c_2,c_3,..)] * exp(t*c_1) with lowering operator L, as above, and raising operator RU = c_1 - sum_{n > 0} 2^n * c_(n+1) * (d/dc_1)^n. It follows that UPC_n(c_1,c_2,..,c_n) = PC_n(c_1,-c_2,..,-c_n) and PC_n(PC.(c_1,c_2,..),-c_2,-c_3,..) = PC_n(PC.(c_1,-c_2,-c_3,..),c_2,c_3,..) = (c_1)^n, e.g., PC_2(PC.(c_1,-c_2,..),c_2) = 2 c_2 + (PC.(c_1,-c_2,..))^2 = 2 c_2 + PC_2(c_1,-c_2) = 2 c_2 + 2 (-c_2) + (c_1)^2 = (c_1)^2.
Letting c_1 = x and all other c_n = 1 gives the row polynomials of A055140.
(End)

Extensions

Polynomials P_6 and P_7 added by Tom Copeland, Oct 11 2016
Correction to P_3 in Example by Tom Copeland, May 27 2020
Terms in rows 6-7 reordered, row 8 added by Andrey Zabolotskiy, Feb 19 2024

A265606 Triangle read by rows: The Bell transform of the quartic factorial numbers (A007696).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 5, 3, 1, 0, 45, 23, 6, 1, 0, 585, 275, 65, 10, 1, 0, 9945, 4435, 990, 145, 15, 1, 0, 208845, 89775, 19285, 2730, 280, 21, 1, 0, 5221125, 2183895, 456190, 62965, 6370, 490, 28, 1, 0, 151412625, 62002395, 12676265, 1715490, 171255, 13230, 798, 36, 1
Offset: 0

Views

Author

Peter Luschny, Dec 30 2015

Keywords

Examples

			[1],
[0, 1],
[0, 1, 1],
[0, 5, 3, 1],
[0, 45, 23, 6, 1],
[0, 585, 275, 65, 10, 1],
[0, 9945, 4435, 990, 145, 15, 1],
[0, 208845, 89775, 19285, 2730, 280, 21, 1],
		

Crossrefs

Bell transforms of other multifactorials are: A000369, A004747, A039683, A051141, A051142, A119274, A132062, A132393, A203412.

Programs

  • Mathematica
    (* The function BellMatrix is defined in A264428. *)
    rows = 10;
    M = BellMatrix[Pochhammer[1/4, #] 4^# &, rows];
    Table[M[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 23 2019 *)
  • Sage
    # uses[bell_transform from A264428]
    def A265606_row(n):
        multifact_4_1 = lambda n: prod(4*k + 1 for k in (0..n-1))
        mfact = [multifact_4_1(k) for k in (0..n)]
        return bell_transform(n, mfact)
    [A265606_row(n) for n in (0..7)]

A155719 Triangle t(n,m)=A039757(n,m)+A039757(n,n-m) read by rows.

Original entry on oeis.org

2, 0, 0, 4, -8, 4, -14, 14, 14, -14, 106, -192, 172, -192, 106, -944, 1664, -720, -720, 1664, -944, 10396, -19560, 12644, -6960, 12644, -19560, 10396, -135134, 264158, -176358, 47334, 47334, -176358, 264158, -135134, 2027026, -4098304, 2925880
Offset: 0

Views

Author

Roger L. Bagula, Jan 25 2009

Keywords

Comments

Row sums are zero if n>0.
Building the symmetric form A(n,m)+A(n,n-m) as here is equivalent to tabulation of the coefficients of a polynomial p_n(x) of order n associated with A(.,.) plus its reverse: t(n,m) = [x^m] ( p_n(x)+x^n*p_n(1/x)), here with p_n(x)=product(x-(2i-1)). Note that the product of the polynomial p_n(x) = sum_{m>=0} A(n,m)*x^m and the polynomial p'n(x)= sum{m>=0} A(n,n-m)*x^m is given in terms of (terminating) generating function by a convolution, related to the reversal of the sense of the second index in the A(n,m). So the fact that one can obtain A(n,n-m) by using the reverse polynomial p'_n(x) = x^n/p_n(x) is by no means special to this sequence here. The consequence that A(n,m)+A(n,n-m) defines a left-right symmetric row is then obvious.

Examples

			2;
0, 0;
4, -8, 4;
-14, 14, 14, -14;
106, -192, 172, -192, 106;
-944, 1664, -720, -720, 1664, -944;
10396, -19560, 12644, -6960, 12644, -19560, 10396;
-135134, 264158, -176358, 47334, 47334, -176358, 264158, -135134;
2027026, -4098304, 2925880, -1062656, 416108, -1062656, 2925880, -4098304, 2027026;
-34459424, 71697024, -53806368, 20516768, -3948000, -3948000, 20516768, -53806368, 71697024, -34459424;
		

Crossrefs

Programs

  • Mathematica
    Clear[p, x, n, b, a, b0];
    p[x_, n_] := Product[x - (2*i + 1), {i, 0, Floor[n/2]}];
    Table[Expand[ CoefficientList[ExpandAll[p[x, n]], x] + Reverse[CoefficientList[ExpandAll[p[x, n]], x]]], {n, 0, 20, 2}];
    Flatten[%]
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