A103371 -id:A103371 - cp's OEIS frontend

A198062 Array read by antidiagonals, m>=0, n>=0, k>=0, A(m, n, k) = sum{j=0..m} sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 4, 2, 1, 0, 1, 1, 8, 3, 2, 1, 0, 1, 1, 16, 4, 4, 3, 1, 0, 1, 1, 32, 5, 8, 9, 3, 1, 0, 1, 1, 64, 6, 16, 27, 7, 3, 1, 0, 1, 1, 128, 7, 32, 81, 15, 7, 3, 1, 0, 1, 1, 256, 8, 64, 243, 31, 15, 9, 4, 1, 0, 1, 1, 512, 9

Offset: 0

Peter Luschny, Nov 02 2011

			   [0] [1] [2]  [3] [4]  [5]  [6]  [7]  [8]  [9]
-------------------------------------------------
[0]  1   1   1    1   1    1    1    1    1    1    A000012
[1]  0   1   1    2   2    2    3    3    3    3    A003056
[2]  0   1   1    4   3    4    9    7    7    9    A073254
[3]  0   1   1    8   4    8   27   15   15   27    A198063
[4]  0   1   1   16   5   16   81   31   31   81    A198064
[5]  0   1   1   32   6   32  243   63   63  243    A198065

Vincenzo Librandi, Table of n, a(n) for n = 0..350

Cf. A198060, A198061.

A198062_RowAsTriangle := proc(m) local pow; pow :=(a,b)->`if`(a=0 and b=0,1,a^b): proc(n, k) local i, j; add(add((-1)^(j + i)*binomial(i, j)*pow(n, j)* pow(k, m-j), i=0..m),j=0..m) end: end:
for m from 0 to 2 do seq(print(seq(A198062_RowAsTriangle(m)(n,k),k=0..n)),n=0..5) od;

max = 9; RowAsTriangle[m_][n_, k_] := Module[{pow}, pow[a_, b_] := If[a == 0 && b == 0, 1, a^b]; Module[{i, j}, Sum[Sum[(-1)^(j+i)*Binomial[i, j]*pow[n, j]*pow[k, m-j], {i, 0, m}], {j, 0, m}]]]; t = Flatten /@ Table[RowAsTriangle[m][n, k], {m, 0, max}, {n, 0, max}, {k, 0, n}]; Table[t[[n-k+1, k+1]], {n, 0, max}, {k, 0, n }] // Flatten (* Jean-François Alcover, Feb 25 2014, after Maple *)

A007318(n,k) = A(0,n+1,k+1)*C(n,k)^1/(k+1)^0,
A103371(n,k) = A(1,n+1,k+1)*C(n,k)^2/(k+1)^1,
A194595(n,k) = A(2,n+1,k+1)*C(n,k)^3/(k+1)^2,
A197653(n,k) = A(3,n+1,k+1)*C(n,k)^4/(k+1)^3,
A197654(n,k) = A(4,n+1,k+1)*C(n,k)^5/(k+1)^4,
A197655(n,k) = A(5,n+1,k+1)*C(n,k)^6/(k+1)^5.

A335340 North-East paths from (0,0) to (n,n) with k cyclic descents.

Original entry on oeis.org

2, 4, 2, 6, 12, 2, 8, 36, 24, 2, 10, 80, 120, 40, 2, 12, 150, 400, 300, 60, 2, 14, 252, 1050, 1400, 630, 84, 2, 16, 392, 2352, 4900, 3920, 1176, 112, 2, 18, 576, 4704, 14112, 17640, 9408, 2016, 144, 2, 20, 810, 8640, 35280, 63504, 52920, 20160, 3240, 180, 2

Offset: 1

Per W. Alexandersson, Jun 02 2020

A North-East path is a path from (0,0) to (n,n) using steps (1,0) and (0,1). A cyclic descent is a North step followed by an East step, where the last and first step is a cyclic descent if the path ends with a North step and starts with an East step.
The sum of the entries in row n is equal to binomial(2n,n).
I conjecture that the polynomial Sum_{k=1...n} T(n,k) t^k is real-rooted for all n.

			The table starts as
2,
4, 2
6, 12, 2
8, 36, 24, 2
10, 80, 120, 40, 2
12, 150, 400, 300, 60, 2

Per Alexandersson, Svante Linusson, Samu Potka, and Joakim Uhlin, Refined Catalan and Narayana cyclic sieving, arXiv:2010.11157 [math.CO], 2020.

Cf. A103371.

T[n_, k_] = 2 Binomial[n, k] Binomial[n - 1, k - 1];

T(n,k) = 2*binomial(n,k)*binomial(n-1,k-1).

T(n,k) = 2 * A103371(n-1,k-1). - Alois P. Heinz, Jun 02 2020

A344563 T(n, k) = binomial(n - 1, k - 1) * binomial(n, k) * 2^k, T(0, 0) = 1. Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 2, 0, 4, 4, 0, 6, 24, 8, 0, 8, 72, 96, 16, 0, 10, 160, 480, 320, 32, 0, 12, 300, 1600, 2400, 960, 64, 0, 14, 504, 4200, 11200, 10080, 2688, 128, 0, 16, 784, 9408, 39200, 62720, 37632, 7168, 256, 0, 18, 1152, 18816, 112896, 282240, 301056, 129024, 18432, 512

Offset: 0

Peter Luschny, May 30 2021

			[0] 1;
[1] 0,  2;
[2] 0,  4,    4;
[3] 0,  6,   24,     8;
[4] 0,  8,   72,    96,     16;
[5] 0, 10,  160,   480,    320,     32;
[6] 0, 12,  300,  1600,   2400,    960,     64;
[7] 0, 14,  504,  4200,  11200,  10080,   2688,    128;
[8] 0, 16,  784,  9408,  39200,  62720,  37632,   7168,   256;
[9] 0, 18, 1152, 18816, 112896, 282240, 301056, 129024, 18432, 512.

T. Amdeberhan, Power of 2 dividing a specialized Mittag-Leffler polynomial, MathOverflow.

Row sums are A002003 with a(0) = 1, cf. also A047781.

The coefficients of the associated polynomials are in A103371.

aRow := n -> seq(binomial(n-1, k-1)*binomial(n,k)*2^k, k=0..n):
seq(print(aRow(n)), n=0..9);

T[n_, k_] := Binomial[n-1, k-1] * Binomial[n, k] * 2^k;
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten

from math import comb
def T(n, k):
    return comb(n-1, k-1)*comb(n, k)*2**k if k > 0 else k**n
print([T(n, k) for n in range(10) for k in range(n+1)]) # Michael S. Branicky, May 30 2021

A361894 Triangle read by rows. T(n, k) is the number of Fibonacci meanders with a central angle of 360/m degrees that make mk left turns and whose length is mn, where m = 2.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 6, 2, 1, 5, 16, 6, 2, 1, 6, 35, 20, 6, 2, 1, 7, 66, 65, 20, 6, 2, 1, 8, 112, 186, 70, 20, 6, 2, 1, 9, 176, 462, 246, 70, 20, 6, 2, 1, 10, 261, 1016, 812, 252, 70, 20, 6, 2, 1, 11, 370, 2025, 2416, 917, 252, 70, 20, 6, 2, 1, 12, 506, 3730, 6435, 3256, 924, 252, 70, 20, 6, 2, 1

Offset: 1

Peter Luschny, Mar 31 2023

For an overview of the terms used see A361574. A201631 gives the row sums of this triangle.
The corresponding sequence counting meanders without the requirement of being Fibonacci is A103371 (for which in turn A103327 is a termwise majorant counting permutations of the same type).
The diagonals, starting from the main diagonal, apparently converge to A000984.

			Triangle T(n, k) starts:
  [ 1]  1;
  [ 2]  2,   1;
  [ 3]  3,   2,    1;
  [ 4]  4,   6,    2,    1;
  [ 5]  5,  16,    6,    2,    1;
  [ 6]  6,  35,   20,    6,    2,   1;
  [ 7]  7,  66,   65,   20,    6,   2,   1;
  [ 8]  8, 112,  186,   70,   20,   6,   2,  1;
  [ 9]  9, 176,  462,  246,   70,  20,   6,  2,  1;
  [10] 10, 261, 1016,  812,  252,  70,  20,  6,  2, 1;
  [11] 11, 370, 2025, 2416,  917, 252,  70, 20,  6, 2, 1;
  [12] 12, 506, 3730, 6435, 3256, 924, 252, 70, 20, 6, 2, 1.
.
T(4, k) counts Fibonacci meanders with central angle 180 degrees and length 8 that make k left turns. Written as binary strings (L = 1, R = 0):
k = 1: 11000000, 10010000, 10000100, 10000001;
k = 2: 11110000, 11100100, 11100001, 11010010, 11001001, 10100101;
k = 3: 11111100, 11111001;
k = 4: 11111111.

Jean-Luc Baril, Sergey Kirgizov, Rémi Maréchal, and Vincent Vajnovszki, Enumeration of Dyck paths with air pockets, arXiv:2202.06893 [cs.DM], 2022-2023.
Peter Luschny, Fibonacci meanders.

Cf. A201631 (row sums), A361681 (m=3), A132812, A361574, A103371, A000984.

# using function 'FibonacciMeandersByLeftTurns' from A361681.
for n in range(1, 12):
    print(FibonacciMeandersByLeftTurns(2, n))

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A198062 Array read by antidiagonals, m>=0, n>=0, k>=0, A(m, n, k) = sum{j=0..m} sum{i=0..m} (-1)^(j+i)*C(i,j)*n^j*k^(m-j).

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A335340 North-East paths from (0,0) to (n,n) with k cyclic descents.

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A344563 T(n, k) = binomial(n - 1, k - 1) * binomial(n, k) * 2^k, T(0, 0) = 1. Triangle read by rows, T(n, k) for 0 <= k <= n.

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A361894 Triangle read by rows. T(n, k) is the number of Fibonacci meanders with a central angle of 360/m degrees that make m*k left turns and whose length is m*n, where m = 2.

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A198062 Array read by antidiagonals, m>=0, n>=0, k>=0, A(m, n, k) = sum{j=0..m} sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

A361894 Triangle read by rows. T(n, k) is the number of Fibonacci meanders with a central angle of 360/m degrees that make mk left turns and whose length is mn, where m = 2.