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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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

A105868 Triangle read by rows, T(n,k) = C(n,k)*C(k,n-k).

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 0, 6, 1, 0, 0, 6, 12, 1, 0, 0, 0, 30, 20, 1, 0, 0, 0, 20, 90, 30, 1, 0, 0, 0, 0, 140, 210, 42, 1, 0, 0, 0, 0, 70, 560, 420, 56, 1, 0, 0, 0, 0, 0, 630, 1680, 756, 72, 1, 0, 0, 0, 0, 0, 252, 3150, 4200, 1260, 90, 1, 0, 0, 0, 0, 0, 0, 2772, 11550, 9240, 1980, 110, 1, 0, 0
Offset: 0

Views

Author

Paul Barry, Apr 23 2005

Keywords

Comments

Row sums are the central trinomial coefficients A002426.
Product of A007318 and this sequence is A008459.
Coefficient array for polynomials P(n,x) = x^n*F(1/2-n/2,-n/2;1;4/x). - Paul Barry, Oct 04 2008
Column sums give A001850. It appears that the sums along the antidiagonals of the triangle produce A182883. - Peter Bala, Mar 06 2013

Examples

			Triangle begins
  1;
  0,  1;
  0,  2,  1;
  0,  0,  6,  1;
  0,  0,  6, 12,  1;
  0,  0,  0, 30, 20, 1;
		

Crossrefs

Cf. A063007. A001850 (column sums), A182883.

Programs

  • Magma
    [[Binomial(n,k)*Binomial(k,n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jun 14 2015
  • Maple
    gf := 1/((1 - x*y)^2 - 4*y^2*x)^(1/2):
    yser := series(gf, y, 12): ycoeff := n -> coeff(yser, y, n):
    row := n -> seq(coeff(expand(ycoeff(n)), x, k), k=0..n):
    seq(row(n), n=0..7); # Peter Luschny, Oct 28 2020
  • Mathematica
    Flatten[Table[Binomial[n,k]Binomial[k,n-k],{n,0,20},{k,0,n}]] (* Harvey P. Dale, Nov 12 2014 *)

Formula

G.f.: 1/(sqrt((1-x*y)^2-4*x^2*y)). - Vladimir Kruchinin, Oct 28 2020

A145904 Square array read by antidiagonals: Hilbert transform of the Narayana numbers A001263.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 16, 7, 1, 1, 15, 40, 31, 9, 1, 1, 21, 85, 105, 51, 11, 1, 1, 28, 161, 295, 219, 76, 13, 1, 1, 36, 280, 721, 771, 396, 106, 15, 1, 1, 45, 456, 1582, 2331, 1681, 650, 141, 17, 1
Offset: 0

Views

Author

Peter Bala, Oct 31 2008

Keywords

Comments

Refer to A145905 for the definition of the Hilbert transform of a lower triangular array. For the Hilbert transform of A008459, the array of type B Narayana numbers, see A108625.
This seems to be a duplicate of A273350. - Alois P. Heinz, Jun 04 2016. This could probably be proved by showing that the g.f.s are the same. - N. J. A. Sloane, Jul 02 2016

Examples

			The array begins
n\k|..0.....1.....2.....3.....4.....5
=====================================
0..|..1.....1.....1.....1.....1.....1
1..|..1.....3.....5.....7.....9....11
2..|..1.....6....16....31....51....76
3..|..1....10....40...105...219...396
4..|..1....15....85...295...771..1681
5..|..1....21...161...721..2331..6083
...
Row 2: (1 + 3x + x^2)/(1 - x)^3 = 1 + 6x + 16x^2 + 31x^3 + ... .
Row 3: (1 + 6x + 6x^2 + x^3)/(1 - x)^4 = 1 + 10x + 40x^2 + 105x^3 + ... .
		

Crossrefs

Cf. A001263, A005891 (row 2), A063490 (row 3), A108625 (Hilbert transform of h-vectors of type B associahedra).
Cf. also A273350.

Programs

  • Mathematica
    Table[1/(# + 1)*Sum[Binomial[# + 1, i - 1] Binomial[# + 1, i] Binomial[# + k - i + 1, k + 1 - i], {i, 0, k + 1}] &[m - k], {m, 0, 9}, {k, 0, m}] // Flatten (* Michael De Vlieger, Jan 15 2018 *)
  • Maxima
    taylor(((y-1)*sqrt(((x^2+2*x+1)*y-x^2+2*x-1)/(y-1))+(-x-1)*y-x+1)/(2*x^2*y),x,0,10,y,0,10);
    T(n,m,k):=1/(n+1)*sum(binomial(n+1,i-1)*binomial(n+1,i)*binomial(n+m-i+1,m+1-i),i,0,m+1); /* Vladimir Kruchinin, Jan 15 2018 */

Formula

Row n generating function: 1/(n+1) * 1/(1-x) * Jacobi_P(n,1,1,(1+x)/(1-x)) = N_n(x)/(1-x)^n where N_n(x) denotes the shifted Narayana polynomial N_n(x) = sum{k = 1..n} A001263(k)*x^(k-1) of degree n-1.
Conjectural column n generating function: N_n(x^2)/(1-x)^(2n+1).
The entries in row n are given by the values of a polynomial function p_n(x) at x = 0,1,2,... . The first few are p_1(x) = 2x + 1, p_2(x) = (5x^2 + 5x + 2)/2, p_3(x) = (2x + 1)*(7x^2 + 7x + 6)/6 and p_4(x) = (7x^4 + 14x^3 + 21x^2 + 14x + 4)/4. These polynomials appear to have their zeros on the line Re x = -1/2; that is, the polynomials p_n(-x) appear to satisfy a Riemann hypothesis. The corresponding result for A108625 is true (see A142995 for details).
Contribution from Paul Barry, Jan 06 2009: (Start)
The g.f. for the corresponding number triangle is:
1/(1-x-xy-x^2y/(1-x-x^2y/(1-x-xy-x^2y/(1-x-x^2y/(1-x-xy-x^2y.... (a continued fraction). (End)
This g.f. satisfies x^2*y*g^2 - (1-x-x*y)*g + 1 = 0. - R. J. Mathar, Jun 16 2016
G.f.: ((y-1)*sqrt(((x^2+2*x+1)*y-x^2+2*x-1)/(y-1))+(-x-1)*y-x+1)/(2*x^2*y). - Vladimir Kruchinin, Jan 15 2018
T(n,m) = 1/(n+1)*Sum_{i=0..m+1} C(n+1,i-1)*C(n+1,i)*C(n+m-i+1,m+1-i). - Vladimir Kruchinin, Jan 15 2018

A375451 Expansion of g.f. A(x) satisfying 0 = Sum_{k=0..n} (-1)^k * binomial(n,k)^2 * ([x^k] A(x)^n) for n >= 1.

Original entry on oeis.org

1, 1, 3, 21, 264, 5100, 138595, 5021209, 233863116, 13628372628, 972514037307, 83479400425677, 8490972592164813, 1010263560882000981, 139051185192340895094, 21926159523172792097194, 3927328317712845680689864, 793059545751159815604109176, 179339266160209677707004583560
Offset: 0

Views

Author

Paul D. Hanna, Sep 10 2024

Keywords

Comments

Note that 0 = Sum_{k=0..n} (-1)^k * binomial(n,k) * ([x^k] G(x)^n) is satisfied by G(x) = 1/(1-x) for n >= 1.

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 21*x^3 + 264*x^4 + 5100*x^5 + 138595*x^6 + 5021209*x^7 + 233863116*x^8 + ...
The table of coefficients of x^k in A(x)^n begins:
  n=1: [1, 1,  3,  21,  264,  5100,  138595, ...];
  n=2: [1, 2,  7,  48,  579, 10854,  289415, ...];
  n=3: [1, 3, 12,  82,  954, 17352,  453657, ...];
  n=4: [1, 4, 18, 124, 1399, 24696,  632656, ...];
  n=5: [1, 5, 25, 175, 1925, 33001,  827900, ...];
  n=6: [1, 6, 33, 236, 2544, 42396, 1041046, ...];
  ...
from which we may illustrate the defining property given by
0 = Sum_{k=0..n} (-1)^k * binomial(n,k)^2 * ([x^k] A(x)^n).
Using the coefficients in the table above, we see that
  n=1: 0 = 1*1 - 1*1;
  n=2: 0 = 1*1 - 4*2 + 1*7;
  n=3: 0 = 1*1 - 9*3 + 9*12 - 1*82;
  n=4: 0 = 1*1 - 16*4 + 36*18 - 16*124 + 1*1399;
  n=5: 0 = 1*1 - 25*5 + 100*25 - 100*175 + 25*1925 - 1*33001;
  n=6: 0 = 1*1 - 36*6 + 225*33 - 400*236 + 225*2544 - 36*42396 + 1*1041046;
  ...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0);
    A[#A] = sum(k=0,#A-1, (-1)^(#A-k) * binomial(#A-1,k)^2 * polcoef(Ser(A)^(#A-1),k) )/(#A-1) ); A[n+1]}
    for(n=0,20,print1(a(n),", "))

Formula

a(n) ~ c * d^n * n!^2, where d = 0.691660276122579707675... = 4/BesselJZero(0,1)^2 = 4/A115368^2 and c = 3.8999463598998648630203... - Vaclav Kotesovec, Sep 10 2024

A132818 The matrix product A127773 * A001263 of infinite lower triangular matrices.

Original entry on oeis.org

1, 3, 3, 6, 18, 6, 10, 60, 60, 10, 15, 150, 300, 150, 15, 21, 315, 1050, 1050, 315, 21, 28, 588, 2940, 4900, 2940, 588, 28, 36, 1008, 7056, 17640, 17640, 7056, 1008, 36, 45, 1620, 15120, 52920, 79380, 52920, 15120, 1620, 45, 55, 2475, 29700, 138600, 291060
Offset: 1

Views

Author

Gary W. Adamson, Sep 02 2007

Keywords

Examples

			First few rows of the triangle are:
1;
3, 3;
6, 18, 6;
10, 60, 60, 10;
15, 150, 300, 150, 15;
21, 315, 1050, 1050, 315, 21;
...
		

Crossrefs

Programs

  • Maple
    A132818 := proc(n,k)
        (n+1-k)*binomial(n+1,k)*binomial(n,k-1)/2 ;
    end proc: # R. J. Mathar, Jul 29 2015

Formula

T(n,k) = A000217(n) * A001263(n,k).
Let a(n) = A006472(n), the 'triangular' factorial numbers. Then a(n)/(a(k)*a(n-k)) produces the present triangle (with a different offset). - Peter Bala, Dec 07 2011
T(n,k) = 1/2*(n+1-k)*C(n+1,k)*C(n,k-1), for n,k >= 1. O.g.f.: x*y/((1-x-x*y)^2 - 4*x^2*y)^(3/2) = x*y + x^2*(3*y + 3*y^2) + x^3*(6*y + 18*y^2 + 6*y^3) + .... Cf. A008459 with o.g.f.: x*y/((1-x-x*y)^2 - 4*x^2*y)^(1/2). Sum {k = 1..n-1} T(n,k)*2^(n-k) = A002695(n). - Peter Bala, Apr 10 2012

Extensions

Corrected by R. J. Mathar, Jul 29 2015

A136715 Triangle T(n,k), 1 <= k <= n, read by rows: T(n,k) is the number of permutations of the set {2,4,6,...,2n} with k excedances. Equivalently, T(n,k) is the number of permutations in the symmetric group S_n having k multiplicative 2-excedances.

Original entry on oeis.org

1, 1, 1, 0, 4, 2, 0, 4, 16, 4, 0, 0, 36, 72, 12, 0, 0, 36, 324, 324, 36, 0, 0, 0, 576, 2592, 1728, 144, 0, 0, 0, 576, 9216, 20736, 9216, 576, 0, 0, 0, 0, 14400, 115200, 172800, 57600, 2880, 0, 0, 0, 0, 14400, 360000, 1440000, 1440000, 360000, 14400
Offset: 1

Views

Author

Peter Bala, Jan 18 2008

Keywords

Comments

Let E_n denote the set {2,4,6,...,2n} and let p denote a bijection p:E_n -> E_n. We say the permutation p has an excedance at position i, 1 <= i <= n, if p(2i) > i. For example, if we represent p in one line notation by the vector (p(2),p(4),...,p(2n)), then the permutation (6,2,8,4,10) of E_5 has three excedances in total (at positions 1, 3 and 5). This array gives the number of permutations of the set E_n with k excedances. This is the viewpoint taken in [Jansson]. See A136716 for the corresponding array when the set E_n is replaced by the set O_n := {1,3,5,...,2n-1}.
Alternatively, we can work with permutations (p(1),p(2),...,p(n)) in the symmetric group S_n and define p to have a multiplicative 2-excedance at position i, 1 <= i <= n, if 2*p(i) > i. Then the (n,k)-th entry of this array gives the number of permutations in S_n with k multiplicative 2-excedances. Compare with A008292, the triangle of Eulerian numbers, which enumerates permutations by the usual excedance number and A136717 which enumerates permutations by multiplicative 3-excedances.
Let e(p)= |{i | 1 <= i < = n, 2*p(i) > i}| denote the number of multiplicative 2-excedances in the permutation p of S_n. This 2-excedance statistic e(p) on the symmetric group S_n is related to a descent statistic as follows.
Define a permutation p in S_n to have a descent from even at position i, 1 <= i <= n-1, if p(i) is even and p(i) > p(i+1). For example, the permutation (2,1,3,5,6,4) in S_6 has two descents from even (at position 1 and position 5). Array A134434 records the number of permutations of S_n with k descents from even.
Let d(p) = |{i | 1 <= i <= n-1, p(i) is even & p(i) > p(i+1)}| count the descents from even in the permutation p. Comparison of the formulas for the entries of this table with the formulas for the entries of A134434 shows that e(p) and d(p) are related by sum {p in S_n} x^e(p) = x^ceiling(n/2)* sum {p in S_n} x^d(p). Thus the shifted multiplicative 2-excedance statistic e(p) - ceiling(n/2) and the descent statistic d(p) are equidistributed on the symmetric group S_n.

Examples

			T(4,2) = 4; the four permutations in S_4 with two multiplicative 2-excedances are (3,4,1,2), (4,3,1,2), (3,1,4,2) and (4,1,3,2). Alternatively, the four permutations (6,8,2,4), (8,6,2,4), (6,2,8,4) and (8,2,6,4) of the set E_4 each have 2 excedances.
Triangle starts
n\k|..1....2....3....4....5....6
--------------------------------
1..|..1
2..|..1....1
3..|..0....4....2
4..|..0....4...16....4
5..|..0....0...36...72...12
6..|..0....0...36..324..324...36
		

Crossrefs

Formula

Recurrence relations:
T(2n,k) = (k+1-n)*T(2n-1,k) + (3n-k)*T(2n-1,k-1) for n >= 1;
T(2n+1,k) = (k-n)*T(2n,k) + (3n+2-k)*T(2n,k-1) for n >= 0. Boundary conditions: T(0,k) = 0 all k; T(n,0) = 0 all n; T(1,1) = 1.
The recurrence relations have the explicit solution T(2n,n+k) = [n!* C(n,k)]^2 and T(2n+1,n+k+1) = 1/(k+1)*[(n+1)!*C(n,k)]^2 = n!*(n+1)!*C(n,k)*C(n+1,k+1); or as a single formula, T(n,k) = floor(n/2)! * floor((n+1)/2)! * C(floor(n/2),k-floor((n+1)/2)) * C(floor((n+1)/2),k-floor(n/2)). Also T(2n,n+k) = n!^2 * A008459 (n,k); T(2n+1,n+k+1) = n!*(n+1)!* A103371 (n,k).
For the even numbered rows, define the shifted row polynomials F(2n,x) := x^(1-n)* sum {k = n..2n} T(2n,k)*x^k = n!^2 * x * (1 + C(n,1)^2*x + C(n,2)^2*x^2 + ... + C(n,n)^2*x^n). For the odd numbered rows, define the shifted row polynomials F(2n+1,x) := x^(-n)* sum {k = n+1..2n+1} T(2n+1,k)*x^k = n!*(n+1)!* ((n+1)*N(n+1,1)*x + n*N(n+1,2)*x^2 +(n-1)* N(n+1,3)*x^3 + ... + N(n+1,n+1)*x^(n+1)), where N(n,k) denotes the Narayana numbers. The first few values are F(1,x) = x, F(2,x) =x+x^2, F(3,x) = 4x+2x^2 and F(4,x) = 4x+16x^2+4x^2.
The recurrence relations yield the identities x*d/dx(F(2n-1,x)/(1-x)^(2n)) = F(2n,x)/(1-x)^(2n+1) and x*d/dx(1/x*F(2n,x)/(1-x)^(2n+1)) = F(2n+1,x)/(1-x)^(2n+2), for n = 1,2,3,... . An easy induction argument now gives the Taylor series expansions: F(2n,x)/(1-x)^(2n+1) = sum {m = 1..inf} (m*(m+1)*...*(m+n-1))^2*x^m; F(2n+1,x)/(1-x)^(2n+2) = sum {m = 1..inf} m*((m+1)*(m+2)*...*(m+n))^2*x^m.
For example, when n = 3 we have for row 6 the expansion (36x + 324x^2 + 324x^3 + 36x^4)/(1-x)^7 = 36x + 576x^2 + 3600x^3 + ... = (1.2.3)^2*x + (2.3.4)^2*x^2 + (3.4.5)^2*x^3 + ... and for row 7 the expansion (576x + 2592x^2 + 1728x^3 + 144x^4)/(1-x)^8 = 576x + 7200x^2 + 43200x^3 + ... = 1*(2.3.4)^2*x + 2*(3.4.5)^2*x^2 + 3*(4.5.6)^2*x^3 + ... .
Relation with the Jacobi polynomials P_n(a,b,x): F(2n,x) = n!^2*x*(1-x)^n *P_n(0,0,(1+x)/(1-x)), F(2n+1,x) = n!*(n+1)!*x*(1-x)^n *P_n(1,0,(1+x)/(1-x)).
Worpitzky-type identities:
Sum {k = n..2n} T(2n,k)*C(x+k,2n) = ((x+1)*(x+2)*...*(x+n))^2;
sum {k = n+1..2n+1} T(2n+1,k)*C(x+k,2n+1) = ((x+1)*(x+2)*...*(x+n))^2*(x+n+1);
and for the odd numbered rows read in reverse order, sum {k = n+1..2n+1} T(2n+1,3n+2-k)*C(x+k,2n+1) = (x+1)*((x+2)*(x+3)*...*(x+n+1))^2.

A145903 Generalized Narayana numbers for root systems of type D_n. Triangle of h-vectors of type D associahedra.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 6, 6, 1, 1, 12, 24, 12, 1, 1, 20, 70, 70, 20, 1, 1, 30, 165, 280, 165, 30, 1, 1, 42, 336, 875, 875, 336, 42, 1, 1, 56, 616, 2296, 3500, 2296, 616, 56, 1, 1, 72, 1044, 5292, 11466, 11466, 5292, 1044, 72, 1
Offset: 0

Views

Author

Peter Bala, Oct 28 2008

Keywords

Comments

The generalized Narayana numbers of type D_n (row n of this triangle) are defined as the entries of the h-vector of the simplicial complex dual to the generalized associahedron of type D_n [Fomin & Reading, p.60]. For the corresponding triangle of f-vectors see A080721. For Narayana numbers of root systems of type A and type B see A001263 and A008459 respectively.

Examples

			Root systems of type D_n are defined only for n >= 2. It seems convenient to complete the array to form a lower unit triangular matrix.
Triangle starts
n\k|..0....1....2....3....4....5....6
=====================================
0..|..1
1..|..1....1
2..|..1....2....1
3..|..1....6....6....1
4..|..1...12...24...12....1
5..|..1...20...70...70...20....1
6..|..1...30..165..280..165...30....1
...
		

References

  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Chapter 12.

Crossrefs

Formula

For n >= 2, T(n,k) = binomial(n,k)^2 - n/(n-1)*binomial(n-1,k-1)*binomial(n-1,k).

A183065 Triangle defined by g.f.: Sum_{n>=0} (4*n)!/n!^4 * x^(2*n)*y^n/(1-x-x*y)^(4*n+1), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 26, 1, 1, 123, 123, 1, 1, 364, 3246, 364, 1, 1, 845, 25210, 25210, 845, 1, 1, 1686, 120135, 606500, 120135, 1686, 1, 1, 3031, 430941, 6082475, 6082475, 430941, 3031, 1, 1, 5048, 1277668, 38698856, 137915470, 38698856, 1277668, 5048, 1, 1, 7929
Offset: 0

Views

Author

Paul D. Hanna, Dec 22 2010

Keywords

Comments

Compare the g.f. of this triangle with the g.f.s of triangles:
* A008459: Sum_{n>=0} (2n)!/n!^2 * x^(2n)*y^n/(1-x-xy)^(2n+1),
* A181543: Sum_{n>=0} (3n)!/n!^3 * x^(2n)*y^n/(1-x-xy)^(3n+1),
which have terms A008459(n,k) = C(n,k)^2 and A181543(n,k) = C(n,k)^3.

Examples

			G.f.: A(x,y) = 1/(1-x-xy) + 4!*x^2*y/(1-x-xy)^5 + (8!/2!^4)*x^4*y^2/(1-x-xy)^9 + (12!/3!^4)*x^6*y^3/(1-x-xy)^13 +...
Triangle begins:
1;
1, 1;
1, 26, 1;
1, 123, 123, 1;
1, 364, 3246, 364, 1;
1, 845, 25210, 25210, 845, 1;
1, 1686, 120135, 606500, 120135, 1686, 1;
1, 3031, 430941, 6082475, 6082475, 430941, 3031, 1;
1, 5048, 1277668, 38698856, 137915470, 38698856, 1277668, 5048, 1; ...
		

Crossrefs

Cf. A183066 (column 1), A183067 (row sums), A183068 (central terms).

Programs

  • PARI
    {T(n,k)=polcoeff(polcoeff(sum(m=0,n,(4*m)!/m!^4*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(4*m+1)),n,x),k,y)}

A371400 Triangle read by rows: T(n, k) = binomial(k + n, k)*binomial(2*n - k, n).

Original entry on oeis.org

1, 2, 2, 6, 9, 6, 20, 40, 40, 20, 70, 175, 225, 175, 70, 252, 756, 1176, 1176, 756, 252, 924, 3234, 5880, 7056, 5880, 3234, 924, 3432, 13728, 28512, 39600, 39600, 28512, 13728, 3432, 12870, 57915, 135135, 212355, 245025, 212355, 135135, 57915, 12870
Offset: 0

Views

Author

Peter Luschny, Mar 21 2024

Keywords

Comments

The main diagonal and column 0 of the triangle are the central binomial coefficients, which are the sums of the squares of Pascal's triangle entries. This sum representation can be generalized, and all terms can be seen as sums of coefficients of some polynomials. (See the Example section.)
To see this, consider T(n, k) as the value of the polynomials P(n, k)(x) at x = 1, where P(n, k)(x) = H([-n, -k], [1], x)*H([-n, -n + k], [1], x) and H denotes the hypergeometric sum 2F1. For instance column 0 is given by the row sums of A008459, and column 1 by the row sums of A371401.

Examples

			Triangle starts:
[0]    1;
[1]    2,     2;
[2]    6,     9,     6;
[3]   20,    40,    40,    20;
[4]   70,   175,   225,   175,    70;
[5]  252,   756,  1176,  1176,   756,   252;
[6]  924,  3234,  5880,  7056,  5880,  3234,   924;
[7] 3432, 13728, 28512, 39600, 39600, 28512, 13728, 3432;
.
Because of the symmetry, only the sum representation of terms with k <= n/2 are shown.
0:                 [1]
1:               [1+1]
2:             [1+4+1],               [1+4+4]
3:           [1+9+9+1],            [1+9+21+9]
4:      [1+16+36+16+1],       [1+16+66+76+16],        [1+16+76+96+36]
5: [1+25+100+100+25+1], [1+25+160+340+205+25], [1+25+190+460+400+100]
		

Crossrefs

Column 0 and main diagonal are A000984.
Column 1 and subdiagonal are A097070.
Row sums are A045721.
The even bisection of the alternating row sums is A005809.
The central terms are A188662.

Programs

  • Maple
    T := (n, k) -> binomial(k + n, k) * binomial(2*n - k, n):
    seq(print(seq(T(n, k), k = 0..n)), n = 0..8);
  • Mathematica
    T[n_, k_] := Hypergeometric2F1[-n, -k, 1, 1] Hypergeometric2F1[-n, -n +k, 1, 1];
    Table[T[n, k], {n, 0, 7}, {k, 0, n}]

Formula

T(n, k) = A046899(n, k) * A092392(n, k).
T(n, k) = A046899(n, k) * A046899(n, n - k).
T(n, k) = A092392(n, k) * A092392(n, n - k).
T(n, k) = A371395(n, k) * (n + 1).
T(n, k) = hypergeom([-n, -k], [1], 1) * hypergeom([-n, -n + k], [1], 1).
2^n*Sum_{k=0..n} T(n, k)*(1/2)^k = A244038(n).
2^n*Sum_{k=0..n} T(n, k)*(-1/2)^k = A371399(n).

A101516 Antidiagonal sums of symmetric square array A101515 and also equals the binomial transform of a sequence formed from terms of A101514 repeated twice.

Original entry on oeis.org

1, 2, 4, 8, 17, 38, 91, 232, 632, 1824, 5571, 17892, 60355, 212898, 784416, 3008480, 11997341, 49612426, 212536067, 941213428, 4305049140, 20302469824, 98641434683, 493038167880, 2533414749409, 13366134856170, 72361098996208
Offset: 0

Views

Author

Paul D. Hanna, Dec 06 2004

Keywords

Comments

A101514 equals the main diagonal of A101515 shift one place right and also A101514 shifts one place left under the square binomial transform (A008459): A101514(n+1) = Sum_{k=0..n-1} C(n-1,k)^2*A101514(k).

Examples

			Given A101514 = [1,1,2,7,35,236,2037,21695,277966,4198635,...],
the binomial transform of A101514 terms repeated twice returns this sequence:
BINOMIAL[1,1,1,1,2,2,7,7,35,35,...] = [1,2,4,8,17,38,91,232,632,1824,...].
		

Crossrefs

Programs

  • PARI
    {a(n)=sum(k=0,n,binomial(n,k)* if(k\2==0,1,sum(j=0,k\2-1,binomial(k\2-1,j)^2* sum(i=0,2*j,(-1)^(2*j-i)*binomial(2*j,i)*a(i)))))}

Formula

G.f.: A(x) = G101514(x^2/(1-x)^2)/(1-x)^2, where G101514(x)= g.f. of A101514. a(n) = Sum_{k=0..n} C(n, k)*A101514([k/2]).
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