A197654 -id:A197654 - cp's OEIS frontend

A198257 Row sums of A197654.

1, 6, 94, 1700, 35466, 795312, 18848992, 464517468, 11801240050, 307073982116, 8147186436324, 219664321959524, 6003343077661216, 165975724832822400, 4634768975107569024, 130553813782898706908, 3705740233107582161538, 105902829964290241990332

Offset: 0

Susanne Wienand, Oct 22 2011

nonn

Number of meanders of length (n+1)*5 which are composed by arcs of equal length and a central angle of 72 degrees.
Definition of a meander:
A binary curve C is a triple (m, S, dir) such that
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated a value of dir,
(c) consecutive Ls increment the index of dir,
(d) consecutive Rs decrement the index of dir,
(e) the integer m>0 divides the length of S and
(f) C is a meander if each value of dir occurs length(S)/m times.
For this sequence, m = 5.
The terms are proved by brute force for 0 <= n <= 6, but not yet in general. [Susanne Wienand, Oct 29 2011]

			Some examples of list S and allocated values of dir if n = 5:
Length(S) = (5+1)*5 = 30.
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L
dir: 1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,0
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,R,L,R,R,R,R,R,L,R,L,L,L,L,R,R,R,L
dir: 1,2,3,4,0,1,2,3,4,0,1,2,3,3,3,3,2,1,0,4,4,4,4,0,1,2,2,1,0,0
  S: L,L,L,L,L,R,L,L,L,R,R,L,L,L,L,L,R,R,L,R,R,L,R,R,L,L,L,L,L,R
dir: 1,2,3,4,0,0,0,1,2,2,1,1,2,3,4,0,0,4,4,4,3,3,3,2,2,3,4,0,1,1
Each value of dir occurs 30/5 = 6 times.

Peter Luschny, Meanders and walks on the circle.
Project Euler, Robot Walks: Problem 208

Cf. A198060, A198256, A198258.

A198257 := proc(n) local i, j, k, pow;
pow := (a, b) -> if a=0 and b=0 then 1 else a^b fi;
add(add(add((-1)^(j+i)*binomial(i,j)*binomial(n,k)^5*pow(n+1,j)*pow(k+1,4-j)/(k+1)^4, i=0..4),j=0..4),k=0..n) end: seq(A198257(n), n=0..16); # Peter Luschny, Nov 02 2011

Table[Sum[Sum[ Sum[(-1)^(j + i) Binomial[i, j], {i, 0, 4}] Binomial[n, k]^5*(n + 1)^j*(k + 1)^(4 - j), {j, 0, 4}]/(k + 1)^4, {k, 0, n}], {n, 0, 17}] (* Michael De Vlieger, Aug 18 2016 *)

A198257(n) = {sum(k=0,n,if(n == 1+2*k,5,(1+k)*(1-((n-k)/(1+k))^5)/(1+2*k-n))*binomial(n,k)^5)} \\ Peter Luschny, Nov 24 2011

a(n) = Sum{k=0..n} Sum{j=0..4} Sum{i=0..4} (-1)^(j+i)*C(i,j)*C(n,k)^5*(n+1)^j*(k+1)^(4-j)/(k+1)^4. - Peter Luschny, Nov 02 2011
a(n) = Sum_{k=0..n} h(n,k)*binomial(n,k)^5, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^5)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 5. - Peter Luschny, Nov 24 2011
a(n) ~ sqrt(5) * 2^(5*n+2) / (Pi*n)^2. - Vaclav Kotesovec, Apr 17 2023

A103371 Number triangle T(n,k) = C(n,n-k)*C(n+1,n-k).

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 18, 12, 1, 5, 40, 60, 20, 1, 6, 75, 200, 150, 30, 1, 7, 126, 525, 700, 315, 42, 1, 8, 196, 1176, 2450, 1960, 588, 56, 1, 9, 288, 2352, 7056, 8820, 4704, 1008, 72, 1, 10, 405, 4320, 17640, 31752, 26460, 10080, 1620, 90, 1, 11, 550, 7425, 39600, 97020

Offset: 0

Paul Barry, Feb 03 2005

Columns include A000027, A002411, A004302, A108647, A134287. Row sums are C(2n+1,n+1) or A001700.
T(n-1,k-1) is the number of ways to put n identical objects into k of altogether n distinguishable boxes. See the partition array A035206 from which this triangle arises after summing over all entries related to partitions with fixed part number k.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) of height k (height(alpha) = |Im(alpha)|). - Abdullahi Umar, Oct 02 2008
The o.g.f. of the (n+1)-th diagonal is given by G(n, x) = (n+1)*Sum_{k=1..n} A001263(n, k)*x^(k-1) / (1 - x)^(2*n+1), for n >= 1 and for n = 0 it is G(0, x) = 1/(1-x). - Wolfdieter Lang, Jul 31 2017

			The triangle T(n, k) begins:
n\k  0   1    2     3     4     5     6    7  8 9 ...
0:   1
1:   2   1
2:   3   6    1
3:   4  18   12     1
4:   5  40   60    20     1
5:   6  75  200   150    30     1
6:   7 126  525   700   315    42     1
7:   8 196 1176  2450  1960   588    56    1
8:   9 288 2352  7056  8820  4704  1008   72  1
9:  10 405 4320 17640 31752 26460 10080 1620 90 1
...  reformatted. - _Wolfdieter Lang_, Jul 31 2017
From _R. J. Mathar_, Mar 29 2013: (Start)
The matrix inverse starts
       1;
      -2,       1;
       9,      -6,      1;
     -76,      54,    -12,      1;
    1055,    -760,    180,    -20,   1;
  -21906,   15825,  -3800,    450, -30,   1;
  636447, -460026, 110775, -13300, 945, -42, 1; (End)
O.g.f. of 4th diagonal [4, 40,200, ...] is G(3, x) = 4*(1 + 3*x + x^2)/(1 - x)^7, from the n = 3 row [1, 3, 1] of A001263. See a comment above. - _Wolfdieter Lang_, Jul 31 2017

Reinhard Zumkeller, Rows n = 0..125 of table, flattened
Per Alexandersson, Svante Linusson, Samu Potka, and Joakim Uhlin, Refined Catalan and Narayana cyclic sieving, arXiv:2010.11157 [math.CO], 2020.
Paul Barry and A. Hennessy, A Note on Narayana Triangles and Related Polynomials, Riordan Arrays, and MIMO Capacity Calculations, J. Int. Seq. 14 (2011), Article 11.3.8.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. II. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See p. 16.
R. Cori and G. Hetyei, Counting genus one partitions and permutations, arXiv:1306.4628 [math.CO], 2013.
Wolfdieter Lang, On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving full transformations, Semigroup Forum 72 (2006), 51-62.

Cf. A008459, A122899, A194595, A197653.
Cf. A007318, A000894 (central terms), A132813 (mirrored).
Cf. A000027, A002411, A004302, A108647, A134287, A135278, A001263.

a103371 n k = a103371_tabl !! n !! k
a103371_row n = a103371_tabl !! n
a103371_tabl = map reverse a132813_tabl
-- Reinhard Zumkeller, Apr 04 2014

/* As triangle */ [[Binomial(n,n-k)*Binomial(n+1,n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 01 2017

A103371 := (n,k) -> binomial(n,k)^2*(n+1)/(k+1);
seq(print(seq(A103371(n, k), k=0..n)), n=0..7); # Peter Luschny, Oct 19 2011

Flatten[Table[Binomial[n,n-k]Binomial[n+1,n-k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, May 26 2014 *)
CoefficientList[Series[Series[E^(x(1+y))(BesselI[0,2*x*Sqrt[y]]+BesselI[1,2*x*Sqrt[y]]/Sqrt[y]),{x,0,8}],{y,0,8}],{x,y}]*Range[0,8]! (* Natalia L. Skirrow, Apr 14 2025 *)

create_list(binomial(n,k)*binomial(n+1,k+1),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */

for(n=0,10, for(k=0,n, print1(binomial(n,k)*binomial(n+1,k+1), ", "))) \\ G. C. Greubel, Nov 09 2018

Number triangle T(n, k) = C(n, n-k)*C(n+1, n-k) = C(n, k)*C(n+1, k+1); Column k of this triangle has g.f. Sum_{j=0..k} (C(k, j)*C(k+1, j) * x^(k+j))/(1-x)^(2*k+2); coefficients of the numerators are the rows of the reverse triangle C(n, k)*C(n+1, k).
T(n,k) = C(n, k)*Sum_{j=0..(n-k)} C(n-j, k). - Paul Barry, Jan 12 2006
T(n,k) = (n+1-k)*N(n+1,k+1), with N(n,k):=A001263(n,k), the Narayana triangle (with offset [1,1]).
O.g.f.: ((1-(1-y)*x)/sqrt((1-(1+y)*x)^2-4*x^2*y) -1)/2, (from o.g.f. of A001263, Narayana triangle). - Wolfdieter Lang, Nov 13 2007
From Peter Bala, Jan 24 2008: (Start)
Matrix product of A007318 and A122899.
O.g.f. for row n: (1-x)^n*P(n,1,0,(1+x)/(1-x)) = 1/(2*x)*(1-x)^(n+1)*( Legendre_P(n+1,(1+x)/(1-x)) - Legendre_P(n,(1+x)/(1-x)) ), where P(n,a,b,x) denotes the Jacobi polynomial.
O.g.f. for column k: x^k/(1-x)^(k+2)*P(k,0,1,(1+x)/(1-x)). Compare with A008459. (End)
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,1). (Cf. A194595, A197653, A197654). - Peter Luschny, Oct 20 2011
T(n,k) = A003056(n+1,k+1)*C(n,k)^2/(k+1). - Peter Luschny, Oct 29 2011
T(n,k) = A007318(n, k)*A135278(n, k), n >= k >= 0. - Wolfdieter Lang, Jul 31 2017
From Natalia L. Skirrow, Apr 14 2025: (Start)
T(n,k) = A008459(n,k) + n*N(n,k+1).
E.g.f.: e^(x*(1+y))*(I_0(2*x*sqrt(y)) + I_1(2*x*sqrt(y))/sqrt(y)), where I_n is the modified Bessel function of the first kind. (The I_0 contributes A008459(n,k), the I_1 contributes n*N(n,k+1))
O.g.f. for row n: (n+1)*2F1(-n,-n;2;y) = (n+1)*2F1(2+n,2+n;2;y)*(1-y)^(2*(n+1)) (by Euler's hypergeometric transformation); (n+1)*2F1(2+n,2+n;2;y) is the o.g.f. for row n of (k+n+1)!^2/(k!*(k+1)!*n!*(n+1)!), which is column n+1 of A132812.
O.g.f. for column k: 2F1(1+k,2+k;1;x)*x^k = 2F1(-k,-1-k;1;x)*x^k/(1-x)^(2+2*k). 2F1(-k,-1-k;1;x) is the kth row of A132813, the reflection of the kth row of this triangle.
O.g.f. for diagonal d (beginning at a(d,0)): (d+1)*x^d*2F1(d+1,d+2;2;x*y). 2F1(d+1,d+2;2;x) = 2F1(1-d,-d;2;x)/(1-x)^(2*d+1), numerator being the o.g.f. of row d of the Narayana triangle.
These respectively yield:
T(n,k) = Sum_{i=0..n+k} C(2*(n+1),i)*(-1)^i*A132812(n+1+k-i,n+1),
T(d+k,k) = Sum_{i=0..k} C(d-i+1+2*k,d-i)*T(k,k-i),
T(d+k,k) = Sum_{i=0..d} C(k-i + 2*d,k-i)*N(d,i+1)*(d+1).
E.g.f. for column k: 1F1(2+k;1;x)*x^k/k!.
E.g.f. for diagonal d: (d+1)*x^d*1F1(d+2;2;x*y)/d!. (End)

A194595 Triangle by rows T(n,k), showing the number of meanders with length (n+1)3 and containing (k+1)3 L's and (n-k)*3 R's, where L's and R's denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 3, 1, 7, 14, 1, 13, 81, 39, 1, 21, 304, 456, 84, 1, 31, 875, 3000, 1750, 155, 1, 43, 2106, 13875, 18500, 5265, 258, 1, 57, 4459, 50421, 128625, 84035, 13377, 399, 1, 73, 8576, 153664, 669536, 836920, 307328, 30016, 584, 1, 91, 15309, 409536, 2815344, 6001128, 4223016, 955584, 61236, 819, 1

Offset: 0

Susanne Wienand, Oct 10 2011

Definition of a meander:
A binary curve C is a triple (m, S, dir) such that
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated
    a value of dir,
(c) consecutive L's increment the index of dir,
(d) consecutive R's decrement the index of dir,
(e) the integer m > 0 divides the length of S and
(f) C is a meander if each value of dir occurs length(S)/m times.
For this sequence, m = 3.
The values in the triangle are proved by brute force for 0 <= n <= 11. The formulas are not yet proved in general. - Susanne Wienand
Let S(N,n,k) = C(n,k)^(N+1)*Sum_{j=0..N} Sum_{i=0..N} (-1)^(N-j+i)*C(N-i,j)*((n+1)/(k+1))^j. Then S(0,n,k) = A007318(n,k), S(1,n,k) = A103371(n,k), S(2,n,k) = T(n,k), S(3,n,k) = A197653(n,k), S(4,n,k) = A197654(n,k), S(5,n,k) = A197655(n,k). - Peter Luschny, Oct 21 2011
The number triangle can be calculated recursively by the number triangles A103371 and A007318. The first column of the triangle contains the central polygonal numbers A002061. The diagonal right hand is A000012. The diagonal with k = n-1 seems to be A027444. Row sums are in A197657. - Susanne Wienand, Nov 24 2011
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 62. - Susanne Wienand, Jun 24 2015

			For n = 4 and k = 2, T(3,4,2) = 456.
Recursive example:
T(1,4,0) = 1
T(1,4,1) = 4
T(1,4,2) = 6
T(1,4,3) = 4
T(1,4,4) = 1
T(2,4,0) = 5
T(2,4,1) = 40
T(2,4,2) = 60
T(2,4,3) = 20
T(2,4,4) = 1
T(3,4,0) = T(1,4,0)^3 + T(1,4,0)*T(2,4,4-1-0) = 1^3 + 1*20 = 21
T(3,4,1) = T(1,4,1)^3 + T(1,4,1)*T(2,4,4-1-1) = 4^3 + 4*60 = 304
T(3,4,2) = T(1,4,2)^3 + T(1,4,2)*T(2,4,4-1-2) = 6^3 + 6*40 = 456
T(3,4,3) = T(1,4,3)^3 + T(1,4,3)*T(2,4,4-1-3) = 4^3 + 4*5  = 84
T(3,4,4) = 1.
Example for closed formula:
T(4,2) = (C(4,2))^3 + C(4,2) * C(4,3) * C(5,3) = 6^3 + 6 * 4 * 10 = 456.
Some examples of list S and allocated values of dir if n = 4 and k = 2:
Length(S) = (4+1)*3 = 15 and S contains (2+1)*3 = 9 L's.
  S: L,L,L,L,L,L,L,L,L,R,R,R,R,R,R
dir: 1,2,0,1,2,0,1,2,0,0,2,1,0,2,1
  S: L,L,R,L,L,L,L,R,R,L,R,R,L,R,L
dir: 1,2,2,2,0,1,2,2,1,1,1,0,0,0,0
  S: L,R,R,R,L,L,L,L,R,R,L,L,L,R,L
dir: 1,1,0,2,2,0,1,2,2,1,1,2,0,0,0
Each value of dir occurs 15/3 = 5 times.

Susanne Wienand, Table of n, a(n) for n = 0..2015
Peter Luschny, Meanders and walks on the circle.
Susanne Wienand, Animation of a meander.
Susanne Wienand, Example of a meander.

Cf. A103371, A197653, A197654, A197655, A197657, A007318, A002061, A000012, A027444, A073254.

A194595 := (n,k)->binomial(n,k)^3*(k^2+k+1+n^2+n-k*n)/((k+1)^2);
seq(print(seq(A194595(n,k),k=0..n)),n=0..7); # Peter Luschny, Oct 14 2011

T[n_, k_] := Binomial[n, k]^3*(k^2 + k + 1 + n^2 + n - k*n)/((k + 1)^2);
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Peter Luschny *)

A194595(n,k) = {if(n == 1+2*k,3,(1+k)*(1-((n-k)/(1+k))^3)/(1+2*k-n))*binomial(n,k)^3} \\ Peter Luschny, Nov 24 2011

Recursive formula (conjectured):
T(n,k) = T(3,n,k) = T(1,n,k)^3 + T(1,n,k)*T(2,n,n-1-k),  0 <= k < n
T(3,n,n) = 1,                                            k = n
T(2,n,k) = T(1,n,k)^2 + T(1,n,k) * T(1,n,n-1-k),         0 <= k < n
T(2,n,n) = 1,                                            k = n
T(2,n,k) = A103371,
T(1,n,k) = A007318 (Pascal's Triangle).
Closed formula (conjectured): T(n,k) = (C(n,k))^3 + C(n,k) * C(n,k+1) * C(n+1,k+1). - Susanne Wienand
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,2). - Peter Luschny, Oct 20 2011
T(n,k) = A073254(n+1,k+1)C(n,k)^3/(k+1)^2. - Peter Luschny, Oct 29 2011
T(n,k) = h(n,k)*binomial(n,k)^3, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^3)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 3. - Peter Luschny, Nov 24 2011

A197653 Triangle by rows T(n,k), showing the number of meanders with length (n+1)4 and containing (k+1)4 Ls and (n-k)*4 Rs, where Ls and Rs denote arcs of equal length and a central angle of 90 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 4, 1, 15, 30, 1, 40, 324, 120, 1, 85, 2080, 3120, 340, 1, 156, 9375, 40000, 18750, 780, 1, 259, 32886, 328125, 437500, 82215, 1554, 1, 400, 96040, 1959216, 6002500, 3265360, 288120, 2800, 1

Offset: 0

Susanne Wienand, Oct 17 2011

Definition of a meander:
A binary curve C is a triple (m, S, dir) such that
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated a value of dir,
(c) consecutive Ls increment the index of dir,
(d) consecutive Rs decrement the index of dir,
(e) the integer m > 0 divides the length of S and
(f) C is a meander if each value of dir occurs length(S)/m times.
For this sequence, m = 4.
The values in the triangle are proved by brute force for 0 <= n <= 8. The formulas are not yet proved in general.
The number triangle can be calculated recursively by the number triangles A007318, A103371 and A194595. The first column of the triangle seems to be A053698. The diagonal right hand is A000012. The diagonal with k = n-1 seems to be A027445. Row sums are in A198256.
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 43. - Susanne Wienand, Jun 29 2015

			For n = 4 and k = 2, T(4,4,2) = 3120.
Recursive example:
T(1,4,0) = 1,
T(1,4,1) = 4,
T(1,4,2) = 6,
T(1,4,3) = 4,
T(1,4,4) = 1,
T(3,4,0) = 21,
T(3,4,1) = 304,
T(3,4,2) = 456,
T(3,4,3) = 84,
T(3,4,1) = 1,
T(4,4,2) = 6^4 + 6*304 = 3120.
Example for closed formula:
T(4,2) = 6^4 + 6^3 * 4 + 6^2 * 4^2 + 6 * 4^3 = 3120.
Some examples of list S and allocated values of dir if n = 4 and k = 2:
Length(S) = (4+1)*4 = 20 and S contains (2+1)*4 = 12 Ls.
  S: L,L,L,L,L,L,L,L,L,L,L,L,R,R,R,R,R,R,R,R
dir: 1,2,3,0,1,2,3,0,1,2,3,0,0,3,2,1,0,3,2,1
  S: L,L,L,R,L,L,R,L,L,R,R,L,L,L,R,L,L,R,R,R
dir: 1,2,3,3,3,0,0,0,1,1,0,0,1,2,2,2,3,3,2,1
  S: L,R,L,L,L,L,R,R,R,L,L,R,R,L,L,L,R,L,L,R
dir: 1,1,1,2,3,0,0,3,2,2,3,3,2,2,3,0,0,0,1,1
Each value of dir occurs 20/4 = 5 times.
Triangle begins:
   1;
   4,    1;
  15,   30,    1;
  40,  324,  120,   1;
  85, 2080, 3120, 340, 1;
  ...

Susanne Wienand, Table of n, a(n) for n = 0..989
Peter Luschny, Meanders and walks on the circle.

Cf. A000012, A007318, A027445, A053698, A103371, A194595, A197654, A197655, A198063, A198256.

A197653 := (n,k) -> binomial(n,k)^4*(n+1)*(n^2-2*n*k+1+2*k+2*k^2)/((1+k)^3);
seq(print(seq(A197653(n, k), k=0..n)), n=0..7); # Peter Luschny, Oct 19 2011

T[n_, k_] := Binomial[n, k]^4 (n+1)(n^2 - 2n*k + 1 + 2k + 2k^2)/((1+k)^3);
Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Peter Luschny *)

A197653(n,k) = {if(n==1+2*k,4,(1+k)*(1-((n-k)/(1+k))^4)/(1+2*k-n))*binomial(n,k)^4} \\ Peter Luschny, Nov 24 2011

def S(N,n,k) : return binomial(n,k)^(N+1)*sum(sum((-1)^(N-j+i)*binomial(N-i,j)*((n+1)/(k+1))^j for i in (0..N) for j in (0..N)))
def A197653(n,k) : return S(3,n,k)
for n in (0..5) : print([A197653(n,k) for k in (0..n)])  ## Peter Luschny, Oct 24 2011

Recursive formula (conjectured):
T(n,k)   = T(4,n,k)
T(4,n,k) = T(1,n,k)^4 + T(1,n,k)*T(3,n,n-1-k), 0 <= k < n
T(4,n,n) = 1                                        k = n
T(3,n,k) = T(1,n,k)^3 + T(1,n,k)*T(2,n,n-1-k), 0 <= k < n
T(3,n,n) = 1                                        k = n
T(2,n,k) = T(1,n,k)^2 + T(1,n,k)*T(1,n,n-1-k), 0 <= k < n
T(2,n,n) = 1                                        k = n
T(3,n,k) = A194595
T(2,n,k) = A103371
T(1,n,k) = A007318 (Pascal's Triangle)
closed formula (conjectured): T(n,n) = 1,                         k = n
                T(n,k) = A + B + C + D,             k < n
                     A = (C(n,k))^4
                     B = (C(n,k))^3 * C(n,n-1-k)
                     C = (C(n,k))^2 *(C(n,n-1-k))^2
                     D =  C(n,k)    *(C(n,n-1-k))^3
[Susanne Wienand]
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,3). - Peter Luschny, Oct 20 2011
T(n,k) = A198063(n+1,k+1)*C(n,k)^4/(k+1)^3. - Peter Luschny, Oct 29 2011
T(n,k) = h(n,k)*binomial(n,k)^4, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^4)/(1+2*k-n) if 1+2*k-n <> 0, otherwise h(n,k) = 4. - Peter Luschny, Nov 24 2011

A197655 Triangle by rows T(n,k), showing the number of meanders with length (n+1)6 and containing (k+1)6 Ls and (n-k)*6 Rs, where Ls and Rs denote arcs of equal length and a central angle of 60 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 6, 1, 63, 126, 1, 364, 4374, 1092, 1, 1365, 85120, 127680, 5460, 1, 3906, 984375, 6000000, 1968750, 19530, 1, 9331, 7562646, 157828125, 210437500, 18906615, 55986, 1, 19608, 42824236, 2628749256, 11029593750, 4381248760, 128472708, 137256, 1

Offset: 0

Susanne Wienand, Oct 19 2011

Definition of a meander:
A binary curve C is a triple (m, S, dir) such that
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated a value of dir,
(c) consecutive Ls increment the index of dir,
(d) consecutive Rs decrement the index of dir,
(e) the integer m>0 divides the length of S and
(f) C is a meander if each value of dir occurs length(S)/m times.
For this sequence, m = 6.
The values in the triangle are proved by brute force for 0 <= n <= 5. The formulas are not yet proved in general.
The number triangle can be calculated recursively by the number triangles and A007318, A103371, A194595, A197653 and A197654. For n > 0, the first column seems to be A053700. The diagonal right hand is A000012. Row sums are in A198258.
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 19. - Susanne Wienand, Jul 04 2015

			For n = 4 and k = 2, T(n,k) = 127680
Example for recursive formula:
T(1,4,2) = 6
T(5,4,4-1-2) = T(5,4,1) = 13504
T(6,4,2) = 6^6 + 6*13504 = 127680
Example for closed formula:
T(4,2) = A + B + C + D + E + F
A = 6^6       =  46656
B = 6^5 * 4   =  31104
C = 6^4 * 4^2 =  20736
D = 6^3 * 4^3 =  13824
E = 6^2 * 4^4 =   9216
F = 6   * 4^5 =   6144
T(4,2)        = 127680
Some examples of list S and allocated values of dir if n = 4 and k = 2:
Length(S) = (4+1)*6 = 30 and S contains (2+1)*6 = 18 Ls.
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,R,R,R,R,R,R,R,R,R,R,R,R
dir: 1,2,3,4,5,0,1,2,3,4,5,0,1,2,3,4,5,0,0,5,4,3,2,1,0,5,4,3,2,1
  S: L,L,L,L,L,L,L,L,L,L,R,L,L,R,L,R,R,R,L,R,R,L,R,R,R,L,L,R,R,L
dir: 1,2,3,4,5,0,1,2,3,4,4,4,5,5,5,5,4,3,3,3,2,2,2,1,0,0,1,1,0,0
  S: L,L,L,L,R,L,L,R,L,L,R,L,R,R,L,L,L,L,L,R,R,L,L,L,R,R,R,R,L,R
dir: 1,2,3,4,4,4,5,5,5,0,0,0,0,5,5,0,1,2,3,3,2,2,3,4,4,3,2,1,1,1
Each value of dir occurs 30/6 = 5 times.

Susanne Wienand, Table of n, a(n) for n = 0..209
Peter Luschny, Meanders and walks on the circle.

Cf. A000012, A007318, A053700, A103371, A194595, A197653, A197654, A198258.

A197655 := (n,k) -> (1+n)*(1+3*k+3*k^2-n-3*k*n+n^2)*(1+k+k^2+n-k*n+n^2)* binomial(n,k)^6/(1+k)^5; seq(print(seq(A197655(n, k), k=0..n)), n=0..7); # Peter Luschny, Oct 21 2011

T[n_, k_] := (1 + n)(1 + 3k + 3k^2 - n - 3k*n + n^2)(1 + k + k^2 + n - k*n + n^2) Binomial[n, k]^6/(1 + k)^5;
Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Peter Luschny *)

A197655(n,k) = {if(n==1+2*k,6,(1+k)*(1-((n-k)/(1+k))^6)/(1+2*k-n))*binomial(n,k)^6} \\ Peter Luschny, Nov 24 2011

def S(N,n,k) : return binomial(n,k)^(N+1)*sum(sum((-1)^(N-j+i)*binomial(N-i,j)*((n+1)/(k+1))^j for i in (0..N) for j in (0..N)))
def A197655(n,k) : return S(5,n,k)
for n in (0..5) : print([A197655(n,k) for k in (0..n)])  # Peter Luschny, Oct 24 2011

recursive formula (conjectured):
T(n,k) = T(6,n,k)
T(6,n,k) = T(1,n,k)^6 + T(1,n,k)*T(5,n,n-1-k), 0 <= k < n
T(6,n,n) = 1                                        k = n
T(5,n,k) = T(1,n,k)^5 + T(1,n,k)*T(4,n,n-1-k), 0 <= k < n
T(5,n,n) = 1                                        k = n
T(4,n,k) = T(1,n,k)^4 + T(1,n,k)*T(3,n,n-1-k), 0 <= k < n
T(4,n,n) = 1                                        k = n
T(3,n,k) = T(1,n,k)^3 + T(1,n,k)*T(2,n,n-1-k), 0 <= k < n
T(3,n,n) = 1                                        k = n
T(2,n,k) = T(1,n,k)^2 + T(1,n,k)*T(1,n,n-1-k), 0 <= k < n
T(2,n,n) = 1                                        k = n
T(5,n,k) = A197654
T(4,n,k) = A197653
T(3,n,k) = A194595
T(2,n,k) = A103371
T(1,n,k) = A007318 (Pascal's Triangle)
closed formula (conjectured): T(n,n) = 1,                         k = n
                T(n,k) = A + B + C + D + E + F,     k < n
                     A = (C(n,k))^6
                     B = (C(n,k))^5 * C(n,n-1-k)
                     C = (C(n,k))^4 *(C(n,n-1-k))^2
                     D = (C(n,k))^3 *(C(n,n-1-k))^3
                     E = (C(n,k))^2 *(C(n,n-1-k))^4
                     F =  C(n,k)    *(C(n,n-1-k))^5
[Susanne Wienand]
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,5). (Cf. A103371, A194595, A197653). [Peter Luschny, Oct 21 2011]
T(n,k) = A198065(n+1,k+1)C(n,k)^6/(k+1)^5. [Peter Luschny, Oct 29 2011]
T(n,k) = h(n,k)*binomial(n,k)^6, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^6)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 6. [Peter Luschny, Nov 24 2011]

A198064 Triangle read by rows (n >= 0, 0 <= k <= n, m = 4); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

Original entry on oeis.org

0, 1, 1, 16, 5, 16, 81, 31, 31, 81, 256, 121, 80, 121, 256, 625, 341, 211, 211, 341, 625, 1296, 781, 496, 405, 496, 781, 1296, 2401, 1555, 1031, 781, 781, 1031, 1555, 2401, 4096, 2801, 1936, 1441, 1280, 1441, 1936, 2801, 4096, 6561, 4681, 3355, 2511, 2101

Offset: 0

Peter Luschny, Oct 26 2011

			[0]                      0
[1]                     1, 1
[2]                  16, 5, 16
[3]                81, 31, 31, 81
[4]            256, 121, 80, 121, 256
[5]         625, 341, 211, 211, 341, 625
[6]     1296, 781, 496, 405, 496, 781, 1296
[7] 2401, 1555, 1031, 781, 781, 1031, 1555, 2401

Cf. A057427, A003056, A073254, A198063, A198065.

A198064 := (n,k) -> k^4-2*k^3*n+4*k^2*n^2-3*k*n^3+n^4:

T(n,k) = k^4-2*k^3*n+4*k^2*n^2-3*k*n^3+n^4.
T(n,0) = T(n,n) = n^m = n^4 = A000583(n).
T(2n,n) = (m+1)n^m = 5n^4.
T(2n+1,n+1) = (n+1)^(m+1)-n^(m+1) = (n+1)^5-n^5 = A022521(n).
Sum{k=0..n} T(n,k) = (16n^5+30n^4+15n^3-n)/30.
T(n+1,k+1)C(n,k)^5/(k+1)^4 = A197654(n,k).

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).

Original entry on oeis.org

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.

cp's OEIS Frontend

A198257 Row sums of A197654.

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A103371 Number triangle T(n,k) = C(n,n-k)*C(n+1,n-k).

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A194595 Triangle by rows T(n,k), showing the number of meanders with length (n+1)*3 and containing (k+1)*3 L's and (n-k)*3 R's, where L's and R's denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented.

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A197653 Triangle by rows T(n,k), showing the number of meanders with length (n+1)*4 and containing (k+1)*4 Ls and (n-k)*4 Rs, where Ls and Rs denote arcs of equal length and a central angle of 90 degrees which are positively or negatively oriented.

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A197655 Triangle by rows T(n,k), showing the number of meanders with length (n+1)*6 and containing (k+1)*6 Ls and (n-k)*6 Rs, where Ls and Rs denote arcs of equal length and a central angle of 60 degrees which are positively or negatively oriented.

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A198064 Triangle read by rows (n >= 0, 0 <= k <= n, m = 4); T(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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A194595 Triangle by rows T(n,k), showing the number of meanders with length (n+1)3 and containing (k+1)3 L's and (n-k)*3 R's, where L's and R's denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented.

A197653 Triangle by rows T(n,k), showing the number of meanders with length (n+1)4 and containing (k+1)4 Ls and (n-k)*4 Rs, where Ls and Rs denote arcs of equal length and a central angle of 90 degrees which are positively or negatively oriented.

A197655 Triangle by rows T(n,k), showing the number of meanders with length (n+1)6 and containing (k+1)6 Ls and (n-k)*6 Rs, where Ls and Rs denote arcs of equal length and a central angle of 60 degrees which are positively or negatively oriented.

A198064 Triangle read by rows (n >= 0, 0 <= k <= n, m = 4); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

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).