A258820 -id:A258820 - cp's OEIS frontend

A055151 Triangular array of Motzkin polynomial coefficients.

1, 1, 1, 1, 1, 3, 1, 6, 2, 1, 10, 10, 1, 15, 30, 5, 1, 21, 70, 35, 1, 28, 140, 140, 14, 1, 36, 252, 420, 126, 1, 45, 420, 1050, 630, 42, 1, 55, 660, 2310, 2310, 462, 1, 66, 990, 4620, 6930, 2772, 132, 1, 78, 1430, 8580, 18018, 12012, 1716, 1, 91, 2002, 15015, 42042

Offset: 0

Michael Somos, Jun 14 2000

T(n,k) = number of Motzkin paths of length n with k up steps. T(n,k)=number of 0-1-2 trees with n edges and k+1 leaves, n>0. (A 0-1-2 tree is an ordered tree in which every vertex has at most two children.) E.g., T(4,1)=6 because we have UDHH, UHDH, UHHD, HHUD, HUHD, HUDH, where U=(1,1), D(1,-1), H(1,0). - Emeric Deutsch, Nov 30 2003
Coefficients in series reversion of x/(1+H*x+U*D*x^2) corresponding to Motzkin paths with H colors for H(1,0), U colors for U(1,1) and D colors for D(1,-1). - Paul Barry, May 16 2005
Eigenvector equals A119020, so that A119020(n) = Sum_{k=0..[n/2]} T(n,k)*A119020(k). - Paul D. Hanna, May 09 2006
Row reverse of A107131. - Peter Bala, May 07 2012
Also equals the number of 231-avoiding permutations of n+1 for which descents(w) = peaks(w) = k, where descents(w) is the number of positions i such that w[i]>w[i+1], and peaks(w) is the number of positions i such that w[i-1]w[i+1]. For example, T(4,1) = 6 because 13245, 12435, 14235, 12354, 12534, 15234 are the only 231-avoiding permutations of 5 elements with descents(w) = peaks(w) = 1. - Kyle Petersen, Aug 02 2013
Apparently, a refined irregular triangle related to this triangle (and A097610) is given in the Alexeev et al. link on p. 12. This entry's triangle is also related through Barry's comment to A125181 and A134264. The diagonals of this entry are the rows of A088617. - Tom Copeland, Jun 17 2015
The row length sequence of this irregular triangle is A008619(n) = 1 + floor(n/2). - Wolfdieter Lang, Aug 24 2015

			The irregular triangle T(n,k) begins:
n\k 0  1   2    3   4  5 ...
0:  1
1:  1
2:  1  1
3:  1  3
4:  1  6   2
5:  1 10  10
6:  1 15  30    5
7:  1 21  70   35
8:  1 28 140  140  14
9:  1 36 252  420 126
10: 1 45 420 1050 630 42
... reformatted. - _Wolfdieter Lang_, Aug 24 2015

Miklos Bona, Handbook of Enumerative Combinatorics, CRC Press (2015), page 617, Corollary 10.8.2
T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 4.3.

Alois P. Heinz, Rows n = 0..200, flattened
N. Alexeev, J. Andersen, R. Penner, and P. Zograf, Enumeration of chord diagrams on many intervals and their non-orientable analogs, arXiv:1307.0967 [math.CO], 2013-2014.
Marcello Artioli, Giuseppe Dattoli, Silvia Licciardi, and Simonetta Pagnutti, Motzkin Numbers: an Operational Point of View, arXiv:1703.07262 [math.CO], 2017.
Paul Barry, The Gamma-Vectors of Pascal-like Triangles Defined by Riordan Arrays, arXiv:1804.05027 [math.CO], 2018.
Paul Barry, On the f-Matrices of Pascal-like Triangles Defined by Riordan Arrays, arXiv:1805.02274 [math.CO], 2018.
Colin Defant, Postorder Preimages, arXiv preprint arXiv:1604.01723 [math.CO], 2016.
Colin Defant, Troupes, Cumulants, and Stack-Sorting, arXiv:2004.11367 [math.CO], 2020.
Samuele Giraudo, Tree series and pattern avoidance in syntax trees, arXiv:1903.00677 [math.CO], 2019.
Thomas Grubb and Frederick Rajasekaran, Set Partition Patterns and the Dimension Index, arXiv:2009.00650 [math.CO], 2020. Mentions this sequence.
Paul W. Lapey and Aaron Williams, A Shift Gray Code for Fixed-Content Łukasiewicz Words, Williams College, 2022.
Shi-Mei Ma, On gamma-vectors and the derivatives of the tangent and secant functions, arXiv:1304.6654 [math.CO], 2013.
E. Marberg, Actions and identities on set partitions, arXiv preprint arXiv:1107.4173 [math.CO], 2011-2012.
MathOverflow, Motzkin polynomials and enumeration of chord diagrams.
Jean-Christophe Novelli and Jean-Yves Thibon, Noncommutative Symmetric Functions and Lagrange Inversion II: Noncrossing partitions and the Farahat-Higman algebra, arXiv:2106.08257 [math.CO], 2021-2022. See p. 32.
A. Postnikov, V. Reiner, and L. Williams, Faces of generalized permutohedra, arXiv:math/0609184 [math.CO], 2006-2007.
Tad White, Quota Trees, arXiv:2401.01462 [math.CO], 2024. See p. 20.
Claude Zeller and Robert Cordery, Light scattering as a Poisson process and first passage probability, arXiv:1906.11131 [cond-mat.stat-mech], 2019.

A107131 (row reversed), A080159 (with trailing zeros), A001006 = row sums, A000108(n) = T(2n, n), A001700(n) = T(2n+1, n), A119020 (eigenvector), A001263 (Narayana numbers), A089627 (gamma vectors of type B associahedra), A101280 (gamma vectors of type A permutohedra).
Cf. A125181, A134264, A088617, A161642, A258820, A003989, A008315, A091156, A011973, A097610, A126120.
Cf. A014531.

b:= proc(x, y) option remember;
      `if`(y>x or y<0, 0, `if`(x=0, 1, expand(
       b(x-1, y) +b(x-1, y+1) +b(x-1, y-1)*t)))
    end:
T:= n-> (p-> seq(coeff(p, t, i), i=0..degree(p)))(b(n, 0)):
seq(T(n), n=0..20);  # Alois P. Heinz, Feb 05 2014

m=(1-x-(1-2x+x^2-4x^2y)^(1/2))/(2x^2 y); Map[Select[#,#>0&]&, CoefficientList[ Series[m,{x,0,15}],{x,y}]]//Grid (* Geoffrey Critzer, Feb 05 2014 *)
p[n_] := Hypergeometric2F1[(1-n)/2, -n/2, 2, 4 x]; Table[CoefficientList[p[n], x], {n, 0, 13}] // Flatten (* Peter Luschny, Jan 23 2018 *)

{T(n, k) = if( k<0 || 2*k>n, 0, n! / ((n-2*k)! * k! * (k+1)!))}

{T(n, k) = if( k<0 || 2*k>n, 0, polcoeff( polcoeff( 2 / (1 - x + sqrt((1 - x)^2 - 4*y*x^2 + x * O(x^n))), n), k))} /* Michael Somos, Feb 14 2006 */

{T(n, k) = n++; if( k<0 || 2*k>n, 0, polcoeff( polcoeff( serreverse( x / (1 + x + y*x^2) + x * O(x^n)), n), k))} /* Michael Somos, Feb 14 2006 */

T(n,k) = n!/((n-2k)! k! (k+1)!) = A007318(n, 2k)*A000108(k). - Henry Bottomley, Jan 31 2003
E.g.f. row polynomials R(n,y): exp(x)*BesselI(1, 2*x*sqrt(y))/(x*sqrt(y)). - Vladeta Jovovic, Aug 20 2003
G.f. row polynomials R(n,y): 2 / (1 - x + sqrt((1 - x)^2 - 4 *y * x^2)).
From Peter Bala, Oct 28 2008: (Start)
The rows of this triangle are the gamma vectors of the n-dimensional (type A) associahedra (Postnikov et al., p. 38). Cf. A089627 and A101280.
The row polynomials R(n,x) = Sum_{k = 0..n} T(n,k)*x^k begin R(0,x) = 1, R(1,x) = 1, R(2,x) = 1 + x, R(3,x) = 1 + 3*x. They are related to the Narayana polynomials N(n,x) := Sum_{k = 1..n} (1/n)*C(n,k)*C(n,k-1)*x^k through x*(1 + x)^n*R(n, x/(1 + x)^2) = N(n+1,x). For example, for n = 3, x*(1 + x)^3*(1 + 3*x/(1 + x)^2) = x + 6*x^2 + 6*x^3 + x^4, the 4th Narayana polynomial.
Recursion relation: (n + 2)*R(n,x) = (2*n + 1)*R(n-1,x) - (n - 1)*(1 - 4*x)*R(n-2,x), R(0,x) = 1, R(1,x) = 1. (End)
G.f.: M(x,y) satisfies: M(x,y)= 1 + x M(x,y) + y*x^2*M(x,y)^2. - Geoffrey Critzer, Feb 05 2014
T(n,k) = A161642(n,k)*A258820(n,k) = (binomial(n,k)/A003989(n+1, k+1))* A258820(n,k). - Tom Copeland, Jun 18 2015
Let T(n,k;q) = n!*(1+k)/((n-2*k)!*(1+k)!^2)*hypergeom([k,2*k-n],[k+2],q) then T(n,k;0) = A055151(n,k), T(n,k;1) = A008315(n,k) and T(n,k;-1) = A091156(n,k). - Peter Luschny, Oct 16 2015
From Tom Copeland, Jan 21 2016: (Start)
Reversed rows of A107131 are rows of this entry, and the diagonals of A107131 are the columns of this entry. The diagonals of this entry are the rows of A088617. The antidiagonals (bottom to top) of A088617 are the rows of this entry.
O.g.f.: [1-x-sqrt[(1-x)^2-4tx^2]]/(2tx^2), from the relation to A107131.
Re-indexed and signed, this triangle gives the row polynomials of the compositional inverse of the shifted o.g.f. for the Fibonacci polynomials of A011973, x / [1-x-tx^2] = x + x^2 + (1+t) x^3 + (1+2t) x^4 + ... . (End)
Row polynomials are P(n,x) = (1 + b.y)^n = Sum{k=0 to n} binomial(n,k) b(k) y^k = y^n M(n,1/y), where b(k) = A126120(k), y = sqrt(x), and M(n,y) are the Motzkin polynomials of A097610. - Tom Copeland, Jan 29 2016
Coefficients of the polynomials p(n,x) = hypergeom([(1-n)/2, -n/2], [2], 4x). - Peter Luschny, Jan 23 2018
Sum_{k=1..floor(n/2)} k * T(n,k) = A014531(n-1) for n>1. - Alois P. Heinz, Mar 29 2020

A178252 Triangle T(n,m) read by rows: the numerator of the coefficient [x^m] of the umbral inverse Bernoulli polynomials B^{-1}(n,x), 0 <= m <= n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 1, 5, 10, 5, 1, 1, 1, 3, 5, 5, 3, 1, 1, 1, 7, 7, 35, 7, 7, 1, 1, 1, 4, 28, 14, 14, 28, 4, 1, 1, 1, 9, 12, 21, 126, 21, 12, 9, 1, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 1, 11, 55, 165, 66, 77, 66, 165, 55, 11, 1, 1, 1, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 1

Offset: 0

Paul Curtz, May 24 2010

The fractions A053382(n,m)/A053383(n,m) give the triangle of the coefficients of the Bernoulli polynomials:
    1;
  -1/2,   1;
   1/6,  -1,    1;
    0,   1/2, -3/2,  1;
  -1/30,  0,    1,  -2,    1;
    0,  -1/6,   0,  5/3, -5/2,  1;
   1/42,  0,  -1/2,  0,   5/2, -3, 1;
The matrix inverse of this triangle defines coefficients of the umbral inverse Bernoulli polynomials B^{-1}(n,x) in row n:
   1;
  1/2,  1;
  1/3,  1,  1;
  1/4,  1, 3/2,   1;
  1/5,  1,  2,    2,   1;
  1/6,  1, 5/2, 10/3, 5/2,   1;
  1/7,  1,  3,    5,   5,    3,   1;
  1/8,  1, 7/2,   7, 35/4,   7,  7/2,  1;
  1/9,  1,  4,  28/3, 14,   14, 28/3,  4,  1;
  1/10, 1, 9/2,  12,  21, 126/5, 21,  12, 9/2, 1;
  1/11, 1,  5,   15,  30,   42,  42,  30, 15,  5, 1;
The current triangle T(n,m) is the numerator of the entry in row n and column m.
In the majority of cases, T(n,m) = A050169(n,m), but since we use the numerators of the reduced fractions, an integer factor may be missing in this equation.
Umbral composition (e.g., B(.,x)^k = B(k,x)) gives B^(-1)(n,B(.,x)) = x^n = B(n,B^(-1)(.,x)). - Tom Copeland, Aug 25 2015

			The triangle T(n,k) begins:
n\k 0 1  2  3   4   5   6    7   8   9  10 11 12 13
0:  1
1:  1 1
2:  1 1  1
3:  1 1  3  1
4:  1 1  2  2   1
5:  1 1  5 10   5   1
6:  1 1  3  5   5   3   1
7:  1 1  7  7  35   7   7    1
8:  1 1  4 28  14  14  28    4   1
9:  1 1  9 12  21 126  21   12   9   1
10: 1 1  5 15  30  42  42   30  15   5   1
11: 1 1 11 55 165  66  77   66 165  55  11  1
12: 1 1  6 22  55  99 132  132  99  55  22  6  1
13: 1 1 13 26 143 143 429 1716 429 143 143 26 13  1
... reformatted. - _Wolfdieter Lang_, Aug 25 2015

Cf. A178340 (denominators).

Cf. A074909, A258820, A003989, A053382, A053383.

nm := 15 : eM := Matrix(nm,nm) :
for n from 0 to nm-1 do for m from 0 to n do eM[n+1,m+1] :=coeff(bernoulli(n,x),x,m) ; end do: for m from n+1 to nm-1 do eM[n+1,m+1] := 0 ; end do: end do:
eM := LinearAlgebra[MatrixInverse](eM) :
for n from 1 to nm do for m from 1 to n do printf("%a,", numer(eM[n,m])) ; end do: end do: # R. J. Mathar, Dec 21 2010

max = 13; coes = Table[ PadRight[ CoefficientList[ BernoulliB[n, x], x], max], {n, 0, max-1}]; inv = Inverse[coes]; Table[ Take[inv[[n]], n], {n, 1, max}] // Flatten // Numerator (* Jean-François Alcover, Aug 09 2012 *)

tabl(nn) = {for (n=0, nn, for (k=0, n, print1(numerator(binomial(n+1,k)/(n+1)), ", ");); print(););} \\ after Tom Copeland comment; Michel Marcus, Jul 25 2015

"Palindromic:" T(n,m+1) = T(n,n-m). T(n,0)=1.
From Tom Copeland, Jun 18 2015: (Start)
The umbral inverse Bernoulli polynomials are Binv(n,x) = [(1+x)^(n+1)-x^(n+1)]/(n+1) with the e.g.f. e^(t*x) * (e^t-1)/t. (See A074909 for more details.) Therefore, T(n,k) is the numerator of the reduced fraction C(n+1,k)/(n+1) for 0 <= k < (n+1).
The reversed rows are presented as the diagonals of A258820.
T(n,k) = A258820(2n-k,n-k) = A003989(n+1,n+1-k) * n! / [ k! (n+1-k)! ], where A003989(j,k) = gcd(j,k). (End)
From Wolfdieter Lang, Aug 26 2015: (Start)
The following refers to the rational triangle TBinv with entries T(n,k)/A178340(n, m), n >= m >= 0.
The inverse of the Bernoulli triangle TB(n, m) with entries A196838(n,m)/A196839(n,m), n >= m >= 0, is the Sheffer triangle (z/(exp(z)-1),z). Therefore, the inverse triangle TBinv is the Sheffer triangle ((exp(z)-1)/z, z). This means that the e.g.f. of the sequence of column m of TBinv ((exp(x)-1)/x)*x^m/m! for m >= 0.
The e.g.f. of the row polynomials of TBinv, called Binv(n, x) = Sum_{m=0..n} TBinv(n,m)*x^m, is gBinv(z,x) = ((exp(z)-1)/z)*exp(x*z) (of the so-called Appell type).
The e.g.f. of the row sums is gBinv(x,1).
The e.g.f. of the alternating row sums is gBinv(x,-1) = (1 - exp(-x))/x.
The e.g.f. of the a-sequence of this Sheffer triangle is 1, and the e.g.f. of the z-sequence is (exp(x) - x -1)/((exp(x) -1)*x). This is the sequence 1/2, -1/12, 0, 1/120, 0, -1/252, 0, 1/240, 0, -1/132, .... For a- and z-sequences of Sheffer triangles and the corresponding recurrences see A006232.
The convolution property of the row polynomials Binv(n, x) is Binv(n, x+y) = Sum_{k=0..n} binomial(n, k)*Binv(n-k, x)*y^n (or with x and y exchanged).
The row polynomials satisfy (d/dx)Binv(n, x) = n*Binv(n-1, x), with Binv(0, x) = 1 (from Meixner's identity).
(End)

Redefined based on reduced fractions by R. J. Mathar, Dec 21 2010

The term umbral was added by Tom Copeland, Aug 25 2015

A161642 Triangle (by rows): T(n,k) = A007318(n,k) / A003989(n+1,k+1).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 2, 2, 2, 1, 1, 5, 10, 10, 5, 1, 1, 3, 15, 5, 15, 3, 1, 1, 7, 7, 35, 35, 7, 7, 1, 1, 4, 28, 28, 14, 28, 28, 4, 1, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 1, 5, 15, 30, 210, 42, 210, 30, 15, 5, 1

Offset: 0

Jason Richardson-White, Jun 15 2009

Taking each row polynomial listed on p. 12 of the Alexeev et al. link and listing the GCD of each sub-polynomial in the indeterminate q gives the left half of this entry's symmetric/palindromic triangle. E.g., for k=6, q*s^6 + (6*q + 9*q^2) s^4 + (15*q + 15*q^2) s^2 + 5 = q*s^6 + 3*(2*q + 3*q^2)*s^4 + 15*(q + q^2)*s^2 + 5 generates (1,3,15,5). See also A055151. - Tom Copeland, Jun 18 2015

			The triangle T(n,k) begins:
n\k 0 1  2  3   4   5   6  7  8 9 10 ...
0:  1
1:  1 1
2:  1 1  1
3:  1 3  3  1
4:  1 2  2  2   1
5:  1 5 10 10   5   1
6:  1 3 15  5  15   3   1
7:  1 7  7 35  35   7   7  1
8:  1 4 28 28  14  28  28  4  1
9:  1 9 36 84 126 126  84 36  9 1
10: 1 5 15 30 210  42 210 30 15 5  1
... reformatted. - _Wolfdieter Lang_, Aug 24 2015

N. Alexeev, J. Andersen, R. Penner, P. Zograf, Enumeration of chord diagrams on many intervals and their non-orientable analogs, arXiv:1307.0967 [math.CO], 2013-2014 [From Tom Copeland, Jun 18 2015]

Cf. A000108, A003989, A055151, A258820, A007318.

T[n_, k_] := Binomial[n, k]/GCD[n-k+1, k+1]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 06 2015, after R. J. Mathar *)

T(2n,n) = A000108(n).
T(n,k) = binomial(n,k)/A003989(n+1,k+1), 0<=k<=n. - R. J. Mathar, Sep 04 2013
For first half (k <= floor(n/2)) of each palindromic row, T(n,k) = A055151(n,k) / A258820(n,k) = A007318(n,2k) * A000108(k) / A258820(n,k) = n! / [(n-2k)! k! (k+1)! A258820(n,k)]. - Tom Copeland, Jun 18 2015

Name changed, and R. J. Mathar's formula corrected, by Wolfdieter Lang, Aug 24 2015

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A055151 Triangular array of Motzkin polynomial coefficients.

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A178252 Triangle T(n,m) read by rows: the numerator of the coefficient [x^m] of the umbral inverse Bernoulli polynomials B^{-1}(n,x), 0 <= m <= n.

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A161642 Triangle (by rows): T(n,k) = A007318(n,k) / A003989(n+1,k+1).

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