A034827 -id:A034827 - cp's OEIS frontend

A359364 Triangle read by rows. The Motzkin triangle, the coefficients of the Motzkin polynomials. M(n, k) = binomial(n, k) * CatalanNumber(k/2) if k is even, otherwise 0.

1, 1, 0, 1, 0, 1, 1, 0, 3, 0, 1, 0, 6, 0, 2, 1, 0, 10, 0, 10, 0, 1, 0, 15, 0, 30, 0, 5, 1, 0, 21, 0, 70, 0, 35, 0, 1, 0, 28, 0, 140, 0, 140, 0, 14, 1, 0, 36, 0, 252, 0, 420, 0, 126, 0, 1, 0, 45, 0, 420, 0, 1050, 0, 630, 0, 42, 1, 0, 55, 0, 660, 0, 2310, 0, 2310, 0, 462, 0

Offset: 0

Peter Luschny, Jan 09 2023

The generalized Motzkin numbers M(n, k) are a refinement of the Motzkin numbers M(n) (A001006) in the sense that they are coefficients of polynomials M(n, x) = Sum_{n..k} M(n, k) * x^k that take the value M(n) at x = 1. The coefficients of x^n are the aerated Catalan numbers A126120.
Variants are the irregular triangle A055151 with zeros deleted, A097610 with reversed rows, A107131 and A080159.
In the literature the name 'Motzkin triangle' is also used for the triangle A026300, which is generated from the powers of the generating function of the Motzkin numbers.

			Triangle M(n, k) starts:
[0] 1;
[1] 1, 0;
[2] 1, 0,  1;
[3] 1, 0,  3, 0;
[4] 1, 0,  6, 0,   2;
[5] 1, 0, 10, 0,  10, 0;
[6] 1, 0, 15, 0,  30, 0,   5;
[7] 1, 0, 21, 0,  70, 0,  35, 0;
[8] 1, 0, 28, 0, 140, 0, 140, 0,  14;
[9] 1, 0, 36, 0, 252, 0, 420, 0, 126, 0;

M. Artioli, G. Dattoli, S. Licciardi, and S. Pagnutti, Motzkin Numbers: an Operational Point of View, arXiv:1703.07262 [math.CO], 2017, (Table 1 on p. 3).

Variants: A055151, A080159, A097610, A107131.
Columns M(n, 2*k): A000012, A000217, A034827, A000910, A088625, A088626, A213380.
Cf. A001006 (Motzkin numbers), A026300 (Motzkin gf. triangle), A126120 (aerated Catalan), A000108 (Catalan).
Cf. A138364, A107587, A002457, A002522, A025179, A025235, A056107, A014531, A023426, A091147, A189912, A343386, A343773, A359647, A359649.

CatalanNumber := n -> binomial(2*n, n)/(n + 1):
M := (n, k) -> ifelse(irem(k, 2) = 1, 0, CatalanNumber(k/2)*binomial(n, k)):
for n from 0 to 9 do seq(M(n, k), k = 0..n) od;
# Alternative, as coefficients of polynomials:
p := n -> hypergeom([(1 - n)/2, -n/2], [2], (2*x)^2):
seq(print(seq(coeff(simplify(p(n)), x, k), k = 0..n)), n = 0..9);
# Using the exponential generating function:
egf := exp(x)*BesselI(1, 2*x*t)/(x*t): ser:= series(egf, x, 11):
seq(print(seq(coeff(simplify(n!*coeff(ser, x, n)), t, k), k = 0..n)), n = 0..9);

from functools import cache
@cache
def M(n: int, k: int) -> int:
    if k %  2: return 0
    if n <  3: return 1
    if n == k: return (2 * (n - 1) * M(n - 2, n - 2)) // (n // 2 + 1)
    return (M(n - 1, k) * n) // (n - k)
for n in range(10): print([M(n, k) for k in range(n + 1)])

Let p(n, x) = hypergeom([(1 - n)/2, -n/2], [2], (2*x)^2).
p(n, 1) = A001006(n); p(n, sqrt(2)) = A025235(n); p(n, 2) = A091147(n).
p(2, n) = A002522(n); p(3, n) = A056107(n).
p(n, n) = A359649(n); 2^n*p(n, 1/2) = A000108(n+1).
M(n, k) = [x^k] p(n, x).
M(n, k) = [t^k] (n! * [x^n] exp(x) * BesselI(1, 2*t*x) / (t*x)).
M(n, k) = [t^k][x^n] ((1 - x - sqrt((x-1)^2 - (2*t*x)^2)) / (2*(t*x)^2)).
M(n, n) = A126120(n).
M(n, n-1) = A138364(n), the number of Motzkin n-paths with exactly one flat step.
M(2*n, 2*n) = A000108(n), the number of peakless Motzkin paths having a total of n up and level steps.
M(4*n, 2*n) = A359647(n), the central terms without zeros.
M(2*n+2, 2*n) = A002457(n) = (-4)^n * binomial(-3/2, n).
Sum_{k=0..n} M(n - k, k) = A023426(n).
Sum_{k=0..n} k * M(n, k) = 2*A014531(n-1) = 2*GegenbauerC(n - 2, -n, -1/2).
Sum_{k=0..n} i^k*M(n, k) = A343773(n), (i the imaginary unit), is the excess of the number of even Motzkin n-paths (A107587) over the odd ones (A343386).
Sum_{k=0..n} Sum_{j=0..k} M(n, j) = A189912(n).
Sum_{k=0..n} Sum_{j=0..k} M(n, n-j) = modified A025179(n).
For a recursion see the Python program.

A119462 Triangle read by rows: T(n,k) is the number of circular binary words of length n having k occurrences of 01 (0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 2, 2, 2, 2, 6, 2, 12, 2, 2, 20, 10, 2, 30, 30, 2, 2, 42, 70, 14, 2, 56, 140, 56, 2, 2, 72, 252, 168, 18, 2, 90, 420, 420, 90, 2, 2, 110, 660, 924, 330, 22, 2, 132, 990, 1848, 990, 132, 2, 2, 156, 1430, 3432, 2574, 572, 26, 2, 182, 2002, 6006, 6006, 2002, 182, 2, 2, 210, 2730, 10010, 12870, 6006, 910, 30

Offset: 0

Emeric Deutsch, May 21 2006

Row n contains 1 + floor(n/2) terms.
Sum of entries in row n is 2^n (A000079).
2*binomial(n-1,2k) is also the number of permutations avoiding both 123 and 132 with k valleys, i.e., positions with w[i]>w[i+1]Lara Pudwell, Dec 19 2018

			T(3,1) = 6 because we have 001, 010, 011, 100, 101 and 110.
Triangle starts:
  1;
  2;
  2,  2;
  2,  6;
  2, 12,  2;
  2, 20, 10;
  2, 30, 30, 2;
  ...

Muniru A Asiru, Rows n=0..150, flattened
M. Bukata, R. Kulwicki, N. Lewandowski, L. Pudwell, J. Roth, and T. Wheeland, Distributions of Statistics over Pattern-Avoiding Permutations, arXiv preprint arXiv:1812.07112 [math.CO], 2018.
L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254.

Cf. A000079, A002378, A034827, A034839, A057711.

Concatenation([1],Flat(List([1..15],n->List([0..Int(n/2)],k->2*Binomial(n,2*k))))); # Muniru A Asiru, Dec 20 2018

T:=proc(n,k) if n=0 and k=0 then 1 else 2*binomial(n,2*k) fi end: for n from 0 to 15 do seq(T(n,k),k=0..floor(n/2)) od; # yields sequence in triangular form

T[0,0]:=1; T[n_,k_]:= 2*Binomial[n,2k]; Table[T[n,k],{n,0,15},{k,0,Floor[n/2]}]//Flatten (* Stefano Spezia, Apr 19 2025 *)

T(n,k) = 2*binomial(n,2k) for n >= 1; T(0,0) = 1.
T(n,k) = 2*T(n-1,k) - T(n-2,k) + T(n-2,k-1) for n >= 3.
G.f.: (1 - z^2 + t*z^2)/(1 - 2*z + z^2 - t*z^2).
T(n,0) = 2 for n >= 1.
T(n,1) = 2*binomial(n,2) = A002378(n-1).
T(n,2) = 2*binomial(n,4) = A034827(n).
T(n,k) = 2*A034839(n-1,k) for n >= 1. [Corrected by Georg Fischer, May 28 2023]
Sum_{k=0..floor(n/2)} k*T(n,k) = A057711(n).

A173622 Triangle T(n,m) read by rows: The number of m-Schroeder paths of order n with 2 diagonal steps.

Original entry on oeis.org

1, 6, 21, 30, 180, 546, 140, 1430, 6120, 17710, 630, 10920, 65835, 245700, 695640, 2772, 81396, 690690, 3322704, 11515140, 32212719, 12012, 596904, 7125300, 44170896, 187336380, 619851960, 1721286532, 51480, 4326300, 72624816

Offset: 2

R. J. Mathar, Nov 08 2010

The case with 1 diagonal step is A060543.

			This is the left-lower portion of the array which starts in row n=2, columns m>=1 as:
1, 2, 3, 4, 5, 6,..
6, 21, 45, 78, 120, 171, 231,.. # A081266
30, 180, 546, 1224, 2310, 3900, 6090, 8976,.. # bisection A055112
140, 1430, 6120, 17710, 40950, 81840, 147630, 246820, 389160,.. # 5-section A034827
630, 10920, 65835, 245700, 695640, 1645020, 3426885, 6497400, ...
2772, 81396, 690690, 3322704, 11515140, 32212719, 77481495, ...
12012, 596904, 7125300, 44170896, 187336380, 619851960, ...

Chunwei Song, The Generalized Schroeder Theory, El. J. Combin. 12 (2005) #R53 Theorem 2.1.

T(n,m) = trinomial(m*n+n-2; m*n-2,n-2,2)/(m*n-1) .

A189913 Triangle read by rows: T(n,k) = binomial(n, k) * k! / (floor(k/2)! * floor((k+2)/2)!).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 3, 1, 4, 6, 12, 2, 1, 5, 10, 30, 10, 10, 1, 6, 15, 60, 30, 60, 5, 1, 7, 21, 105, 70, 210, 35, 35, 1, 8, 28, 168, 140, 560, 140, 280, 14, 1, 9, 36, 252, 252, 1260, 420, 1260, 126, 126, 1, 10, 45, 360, 420, 2520, 1050, 4200, 630, 1260, 42

Offset: 0

Peter Luschny, May 24 2011

The triangle may be regarded a generalization of the triangle A097610:
A097610(n,k) = binomial(n,k)*(2*k)$/(k+1);
T(n,k) = binomial(n,k)*(k)$/(floor(k/2)+1).
Here n$ denotes the swinging factorial A056040(n). As A097610 is a decomposition of the Motzkin numbers A001006, a combinatorial interpretation of T(n,k) in terms of lattice paths can be expected.
T(n,n) = A057977(n) which can be seen as extended Catalan numbers.

			[0]  1
[1]  1, 1
[2]  1, 2,  1
[3]  1, 3,  3,   3
[4]  1, 4,  6,  12,  2
[5]  1, 5, 10,  30, 10,  10
[6]  1, 6, 15,  60, 30,  60,  5
[7]  1, 7, 21, 105, 70, 210, 35, 35

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened
Peter Luschny, The lost Catalan numbers.

Row sums are A189912.

Cf. A097610, A057977, A001006.

/* As triangle */ [[Binomial(n,k)*Factorial(k)/(Factorial(Floor(k/2))*Factorial(Floor((k + 2)/2))): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jan 13 2018

A189913 := (n,k) -> binomial(n,k)*(k!/iquo(k,2)!^2)/(iquo(k,2)+1):
seq(print(seq(A189913(n,k),k=0..n)),n=0..7);

T[n_, k_] := Binomial[n, k]*k!/((Floor[k/2])!*(Floor[(k + 2)/2])!); Table[T[n, k], {n, 0, 10}, {k, 0, n}]// Flatten (* G. C. Greubel, Jan 13 2018 *)

{T(n,k) = binomial(n,k)*k!/((floor(k/2))!*(floor((k+2)/2))!) };
for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jan 13 2018

From R. J. Mathar, Jun 07 2011: (Start)
T(n,1) = n.
T(n,2) = A000217(n-1).
T(n,3) = A027480(n-2).
T(n,4) = A034827(n). (End)

A213380 a(n) = 132*binomial(n,12).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 132, 1716, 12012, 60060, 240240, 816816, 2450448, 6651216, 16628040, 38798760, 85357272, 178474296, 356948592, 686439600, 1274816400, 2294669520, 4015671660, 6850263420, 11417105700, 18627909300, 29804654880, 46835886240, 72382733280, 110147637600, 165221456400, 244527755472, 357386719536, 516225261552, 737464659360

Offset: 0

N. J. A. Sloane, Jun 10 2012. This sequence was in the 1973 "Handbook", but was later dropped from the database. I have now reinstated it. - N. J. A. Sloane, Jun 10 2012

Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 450.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Cf. A010965, A000217, A034827, A000910, A088625, A088626. A column of A088617.

[132*Binomial(n,12):n in [0..40]]; // Vincenzo Librandi, Feb 17 2020

132*Binomial[Range[0,40],12] (* Harvey P. Dale, Feb 16 2020 *)

G.f.: -132*x^12 / (x-1)^13. - Colin Barker, Jul 22 2013

a(n) = 132*A010965(n). - R. J. Mathar, Jul 15 2015

A360665 Square array T(n, k) = k((2n-1)*k+1)/2 read by rising antidiagonals.

Original entry on oeis.org

0, 0, 0, 0, 1, -1, 0, 2, 3, -3, 0, 3, 7, 6, -6, 0, 4, 11, 15, 10, -10, 0, 5, 15, 24, 26, 15, -15, 0, 6, 19, 33, 42, 40, 21, -21, 0, 7, 23, 42, 58, 65, 57, 28, -28, 0, 8, 27, 51, 74, 90, 93, 77, 36, -36, 0, 9, 31, 60, 90, 115, 129, 126, 100, 45, -45

Offset: 0

Paul Curtz, Mar 17 2023

			By rows:
   0,   0,  -1,  -3,  -6,  -10,  -15,  -21,  -28, ...   = -A161680
   0,   1,   3,   6,  10,   15,   21,   28,   36, ...   =  A000217
   0,   2,   7,  15,  26,   40,   57,   77,  100, ...   =  A005449
   0,   3,  11,  24,  42,   65,   93,  126,  164, ...   =  A005475
   0,   4,  15,  33,  58,   90,  129,  175,  228, ...   =  A022265
   0,   5,  19,  42,  74,  115,  165,  224,  292, ...   =  A022267
   0,   6,  23,  51,  90,  140,  201,  273,  356, ...   =  A022269
   ... .

Cf. Antidiagonal sums: A034827(n+1).
Cf. A000217, A000290, A005449, A005475, A022265.
Cf. A022267, A022269, A059270, A081436, A101986.
Cf. A161680, A162254, A179297, A360962, A361226.

T[n_, k_] := ((2*n - 1)*k^2 + k)/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Mar 31 2023 *)

T(n, k) = ((2*n-1)*k^2+k)/2 \\ Thomas Scheuerle, Mar 17 2023

T(n,k) = T(n,k-1)+k^2.
T(n,n) = A081436(n-1).
T(n,n+1) = A059270(n).
T(n,n+4) = -3*A179297(n+4).
T(n+3,n) = A162254(n).
T(n+5,n) = 3*A101986(n).
From Stefano Spezia, Mar 31 2023: (Start)
O.g.f.: (x*y - y^2 + 2*x*y^2)/((1 - x)^2*(1 - y)^3).
E.g.f.: exp(x+y)*y*(2*x - y + 2*x*y)/2. (End)

A361226 Square array T(n,k) = k((1+2n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 0, 2, 5, 3, 0, 3, 9, 12, 6, 0, 4, 13, 21, 22, 10, 0, 5, 17, 30, 38, 35, 15, 0, 6, 21, 39, 54, 60, 51, 21, 0, 7, 25, 48, 70, 85, 87, 70, 28, 0, 8, 29, 57, 86, 110, 123, 119, 92, 36, 0, 9, 33, 66, 102, 135, 159, 168, 156, 117, 45

Offset: 0

Paul Curtz, Mar 05 2023

The main diagonal is A002414.
The first upper diagonal is A160378(n+1).
The antidiagonals sums are A034827(n+2).
b(n) = (A034827(n+3) = 0, 2, 10, 30, 70, ...) - (A002414(n) = 0, 1, 9, 30, 70, ...) = 0, 1, 1, 0, 0, 5, 21, 56, ... .
b(n+2) = A299120(n). b(n+4) = A033275(n). b(n+4) - b(n) = A002492(n).

			The rows are
  0, 0,  1,  3,  6,  10,  15,  21, ...   = A161680
  0, 1,  5, 12, 22,  35,  51,  70, ...   = A000326
  0, 2,  9, 21, 38,  60,  87, 119, ...   = A005476
  0, 3, 13, 30, 54,  85, 123, 168, ...   = A022264
  0, 4, 17, 39, 70, 110, 159, 217, ...   = A022266
  ... .
Columns: A000004, A001477, A016813, A017197=3*A016777, 2*A017101, 5*A016873, 3*A017581, 7*A017017, ... (coefficients from A026741).
Difference between two consecutive rows: A000290. Hence A143844.
This square array read by antidiagonals leads to the triangle
  0
  0   0
  0   1   1
  0   2   5   3
  0   3   9  12   6
  0   4  13  21  22  10
  0   5  17  30  38  35  15
  ... .

Cf. A000004, A000290, A000326, A001477, A002414, A005476, A016777, A016813, A016873, A017017, A017101, A017197, A017581, A022264, A022266, A026741, A034827, A160378, A161680, A360962.

Cf. A002492, A033275, A143844, A299120.

T[n_, k_] := k*((2*n + 1)*k - 1)/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Mar 05 2023 *)

a(n) = { my(row = (sqrtint(8*n+1)-1)\2, column = n - binomial(row + 1, 2)); binomial(column, 2) + column^2 * (row - column) } \\ David A. Corneth, Mar 05 2023

# Seen as a triangle:
from functools import cache
@cache
def Trow(n: int) -> list[int]:
    if n == 0: return [0]
    r = Trow(n - 1)
    return [r[k] + k * k if k < n else r[n - 1] + n - 1 for k in range(n + 1)]
for n in range(7): print(Trow(n)) # Peter Luschny, Mar 05 2023

Take successively sequences n*(n-1)/2, n*(3*n-1)/2, n*(5*n-1)/2, ... listed in the EXAMPLE section.
G.f.: y*(x + y)/((1 - y)^3*(1 - x)^2). - Stefano Spezia, Mar 06 2023
E.g.f.: exp(x+y)*y*(2*x + y + 2*x*y)/2. - Stefano Spezia, Feb 21 2024

cp's OEIS Frontend

A359364 Triangle read by rows. The Motzkin triangle, the coefficients of the Motzkin polynomials. M(n, k) = binomial(n, k) * CatalanNumber(k/2) if k is even, otherwise 0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Python

Formula

A119462 Triangle read by rows: T(n,k) is the number of circular binary words of length n having k occurrences of 01 (0 <= k <= floor(n/2)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Maple

Mathematica

Formula

A173622 Triangle T(n,m) read by rows: The number of m-Schroeder paths of order n with 2 diagonal steps.

Views

Author

Keywords

Comments

Examples

References

Formula

A189913 Triangle read by rows: T(n,k) = binomial(n, k) * k! / (floor(k/2)! * floor((k+2)/2)!).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A213380 a(n) = 132*binomial(n,12).

Views

Author

Keywords

References

Links

Crossrefs

Programs

Magma

Mathematica

Formula

A360665 Square array T(n, k) = k*((2*n-1)*k+1)/2 read by rising antidiagonals.

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

PARI

Formula

A361226 Square array T(n,k) = k*((1+2*n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

Python

A360665 Square array T(n, k) = k((2n-1)*k+1)/2 read by rising antidiagonals.

A361226 Square array T(n,k) = k((1+2n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.