A103371 -id:A103371 - cp's OEIS frontend

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.

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

A197654 Triangle by rows T(n,k), showing the number of meanders with length 5(n+1) and containing 5(k+1) L's and 5(n-k) R's, where L's and R's denote arcs of equal length and a central angle of 72 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 5, 1, 31, 62, 1, 121, 1215, 363, 1, 341, 13504, 20256, 1364, 1, 781, 96875, 500000, 193750, 3905, 1, 1555, 501066, 7321875, 9762500, 1252665, 9330, 1, 2801, 2033647, 72656661, 262609375, 121094435, 6100941, 19607, 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 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.
Then C is a meander if each value of dir occurs length(S)/m times.
Let T(m,n,k) = number of meanders (m, S, dir) in which S contains m(k+1) L's and m(n-k) R's, so that length(S) = m(n+1).
For this sequence, m = 5, T(n,k) = T(5,n,k).
The values in the triangle were proved by brute force for 0 <= n <= 6. The formulas have not yet been proved in general.
The number triangle can be calculated recursively by the number triangles and A007318, A103371, A194595 and A197653. The first column seems to be A053699.  The diagonal right hand is A000012. The diagonal with k = n-1 seems to be A152031 and to start with the second number of A152031. Row sums are in A198257.
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 29. - Susanne Wienand, Jul 01 2015

			For n = 5 and k = 2, T(n,k) = 500000
Example for recursive formula:
T(1,5,2) = 10
T(4,5,5-1-2) = T(4,5,2) = 40000
T(5,5,2) = 10^5 + 10*40000 = 500000
Example for closed formula:
T(5,2) = A + B + C + D + E
A = 10^5
B = 10^4 * 10
C = 10^3 * 10^2
D = 10^2 * 10^3
E = 10   * 10^4
T(5,2) = 5 * 10^5 = 500000
Some examples of list S and allocated values of dir if n = 5 and k = 2:
Length(S) = (5+1)*5 = 30 and S contains (2+1)*5 = 15 Ls.
  S: 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,R,R,R
dir: 1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,0,4,3,2,1,0,4,3,2,1,0,4,3,2,1
  S: L,L,L,L,L,L,L,R,R,L,L,R,R,R,L,R,R,R,L,R,L,L,L,R,R,L,R,R,R,R
dir: 1,2,3,4,0,1,2,2,1,1,2,2,1,0,0,0,4,3,3,3,3,4,0,0,4,4,4,3,2,1
  S: L,L,L,L,L,R,L,L,L,L,L,R,L,R,R,R,R,R,R,R,R,R,R,L,L,L,L,R,R,R
dir: 1,2,3,4,0,0,0,1,2,3,4,4,4,4,3,2,1,0,4,3,2,1,0,0,1,2,3,3,2,1
Each value of dir occurs 30/5 = 6 times.
The triangle begins:
1,
5, 1,
31, 62, 1,
121, 1215, 363, 1,
341, 13504, 20256, 1364, 1,
781, 96875, 500000, 193750, 3905, 1,
...

Alois P. Heinz, Rows n = 0..140, flattened
Peter Luschny, Meanders and walks on the circle.
Susanne Wienand, Example of a meander counted by A197654

Cf. A000012, A007318, A053699, A103371, A152031, A194595, A197653, A197655, A198257.

A197654 := (n,k)->(k^4+2*k^3*(1-n)+2*k^2*(2+n+2*n^2)+k*(3+n-n^2-3*n^3)+ n^4+n^3+n^2+n+1)*binomial(n,k)^5/(1+k)^4;
seq(print(seq(A197654(n, k), k=0..n)), n=0..7);  # Peter Luschny, Oct 20 2011

T[n_, k_] := (k^4 + 2*k^3*(1-n) + 2*k^2*(2+n+2*n^2) + k*(3+n-n^2-3*n^3) + n^4+n^3+n^2+n+1)*Binomial[n, k]^5/(1+k)^4; Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 20 2017, after Peter Luschny *)

A197654(n,k) = {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

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 A197654(n,k) : return S(4,n,k)
for n in (0..5) : print([A197654(n,k) for k in (0..n)])  # Peter Luschny, Oct 24 2011

Recursive formula (conjectured):
T(n,k) = 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(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,              k < n
                     A = (C(n,k))^5
                     B = (C(n,k))^4 * C(n,n-1-k)
                     C = (C(n,k))^3 *(C(n,n-1-k))^2
                     D = (C(n,k))^2 *(C(n,n-1-k))^3
                     E =  C(n,k)    *(C(n,n-1-k))^4     [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,4). [Peter Luschny, Oct 20 2011]
T(n,k) = A198064(n+1,k+1)C(n,k)^5/(k+1)^4. [Peter Luschny, Oct 29 2011]
T(n,k) = 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]

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]

A209805 Triangle read by rows: T(n,k) is the number of k-block noncrossing partitions of n-set up to rotations.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 4, 2, 1, 1, 3, 10, 10, 3, 1, 1, 3, 15, 25, 15, 3, 1, 1, 4, 26, 64, 64, 26, 4, 1, 1, 4, 38, 132, 196, 132, 38, 4, 1, 1, 5, 56, 256, 536, 536, 256, 56, 5, 1, 1, 5, 75, 450, 1260, 1764, 1260, 450, 75, 5, 1

Offset: 1

Tilman Piesk, Mar 13 2012

Like the Narayana triangle A001263 (and unlike A152175) this triangle is symmetric.
The diagonal entries are 1, 1, 4, 25, 196, 1764, ... which is probably sequence A001246 - the squares of the Catalan numbers.
The above conjecture about the diagonal entries T(2*n-1, n) is true since gcd(2*n-1, n) = gcd(2*n-1, n-1) = 1 and then T(2*n-1, n) simplifies to A001246(n-1) using the formula given below. - Andrew Howroyd, Nov 15 2017

			Triangle begins:
  1;
  1,   1;
  1,   1,   1;
  1,   2,   2,   1;
  1,   2,   4,   2,   1;
  1,   3,  10,  10,   3,   1;
  1,   3,  15,  25,  15,   3,   1;
  1,   4,  26,  64,  64,  26,   4,   1;
  1,   4,  38, 132, 196, 132,  38,   4,   1;
  1,   5,  56, 256, 536, 536, 256,  56,   5,   1;

Andrew Howroyd, Table of n, a(n) for n = 1..1275
Tilman Piesk, Partition related number triangles (Wikiversity)
Tilman Piesk, Brute force MATLAB code used to calculate the rows (Pastebin)

Cf. A054357 (row sums), A001246 (square Catalan numbers).

Cf. A001263, A103371, A132812, A209612.

b[n_, k_] := Binomial[n-1, n-k] Binomial[n, n-k];
T[n_, k_] := (DivisorSum[GCD[n, k], EulerPhi[#] b[n/#, k/#]&] + DivisorSum[GCD[n, k - 1], EulerPhi[#] b[n/#, (n + 1 - k)/#]&] - k Binomial[n, k]^2/(n - k + 1))/n;
Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 01 2018, after Andrew Howroyd *)

b(n,k)=binomial(n-1,n-k)*binomial(n,n-k);
T(n,k)=(sumdiv(gcd(n,k), d, eulerphi(d)*b(n/d,k/d)) + sumdiv(gcd(n,k-1), d, eulerphi(d)*b(n/d,(n+1-k)/d)) - k*binomial(n,k)^2/(n-k+1))/n; \\ Andrew Howroyd, Nov 15 2017

T(n,k) = (1/n)*((Sum_{d|gcd(n,k)} phi(d)*A103371(n/d-1,k/d-1)) + (Sum_{d|gcd(n,k-1)} phi(d)*A103371(n/d-1,(n+1-k)/d-1)) - A132812(n,k)). - Andrew Howroyd, Nov 15 2017

A113162 a(n) = binomial(6, n) * binomial(7, 6-n).

Original entry on oeis.org

7, 126, 525, 700, 315, 42, 1

Offset: 0

Zerinvary Lajos, Jan 05 2006

The sequence sums to binomial(13, 6).

Cf. A103371, A113163, A113164.

Table[Binomial[6, n] Binomial[7, 6 - n], {n, 0, 6}] (* Wesley Ivan Hurt, Jan 15 2017 *)

Edited by Don Reble, Jan 26 2006

A122899 Triangle with row sums counting directed animals.

Original entry on oeis.org

1, 1, 1, 0, 4, 1, 0, 3, 9, 1, 0, 0, 18, 16, 1, 0, 0, 10, 60, 25, 1, 0, 0, 0, 80, 150, 36, 1, 0, 0, 0, 35, 350, 315, 49, 1, 0, 0, 0, 0, 350, 1120, 588, 64, 1, 0, 0, 0, 0, 126, 1890, 2940, 1008

Offset: 0

Paul Barry, Sep 18 2006

Row sums are A005773(n+1). Product of A007318 and A122899 is A103371.

			Triangle begins
1,
1, 1,
0, 4, 1,
0, 3, 9, 1,
0, 0, 18, 16, 1,
0, 0, 10, 60, 25, 1,
0, 0, 0, 80, 150, 36, 1,
0, 0, 0, 35, 350, 315, 49, 1,
0, 0, 0, 0, 350, 1120, 588, 64, 1,
0, 0, 0, 0, 126, 1890, 2940, 1008, 81, 1,
0, 0, 0, 0, 0, 1512, 7350, 6720, 1620, 100, 1

Cf. A123160.

Number triangle T(n,k)=sum{j=0..n, (-1)^(n-j)C(n,j)C(j+1,k+1)C(j,k)}

T(n,k) = C(n,k)*C(k+1,n-k). The columns of this triangle (ignoring leading zeros) give the rows of A123160. - Peter Bala, Jan 24 2008

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

Peter Bala, Jan 18 2008

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.

			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

Fredrik Jansson, Variations on the excedance statistic in permutations

Cf. A000142 (row sums), A008292, A008459, A103371, A120434, A134434, A136716, A136717.

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.

A155497 Triangle T(n, k) = binomial(n, k)binomial(n+1, k+1)binomial(2n, 2k)/(n-k+1), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 18, 1, 1, 90, 90, 1, 1, 280, 1400, 280, 1, 1, 675, 10500, 10500, 675, 1, 1, 1386, 51975, 161700, 51975, 1386, 1, 1, 2548, 196196, 1471470, 1471470, 196196, 2548, 1, 1, 4320, 611520, 9417408, 22702680, 9417408, 611520, 4320, 1, 1, 6885, 1652400, 46781280, 231567336, 231567336, 46781280, 1652400, 6885, 1

Offset: 0

Roger L. Bagula, Jan 23 2009

			Triangle begins as:
  1;
  1,    1;
  1,   18,       1;
  1,   90,      90,        1;
  1,  280,    1400,      280,         1;
  1,  675,   10500,    10500,       675,         1;
  1, 1386,   51975,   161700,     51975,      1386,        1;
  1, 2548,  196196,  1471470,   1471470,    196196,     2548,       1;
  1, 4320,  611520,  9417408,  22702680,   9417408,   611520,    4320,    1;
  1, 6885, 1652400, 46781280, 231567336, 231567336, 46781280, 1652400, 6885, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A103371, A155495.

[Binomial(n, k)*Binomial(n+1, k+1)*Binomial(2*n, 2*k)/(n-k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, May 29 2021

T[n_, k_]:= Binomial[n,k]*Binomial[n+1,k+1]*Binomial[2*n,2*k]/(n-k+1);
Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* modified by G. C. Greubel, May 29 2021 *)

flatten([[binomial(n, k)*binomial(n+1, k+1)*binomial(2*n, 2*k)/(n-k+1) for k in (0..12)] for n in (0..12)]) # G. C. Greubel, May 29 2021

T(n, k) = binomial(n, k)*binomial(2*n, 2*k)*f(n)/(f(k)*f(n-k)), where f(n) = (n+1)!.
T(n, k) = binomial(n, k)*binomial(n+1, k+1)*binomial(2*n, 2*k)/(n-k+1).
From G. C. Greubel, May 29 2021: (Start)
Sum_{k=0..n} T(n, k) = Hypergeometric4F3([-n, -n, -n-1, -n-1/2], [1/2, 1, 2], 1).
T(n, k) = binomial(n, k)*A155495(n, k)/(n-k+1).
T(n, k) = binomial(2*n, 2*k)*A103371(n, k)/(n-k+1). (End)

Edited by G. C. Greubel, May 29 2021

A156136 A triangle of polynomial coefficients related to Mittag-Leffler polynomials: p(x,n)=Sum[Binomial[n, k]Binomial[n - 1, n - k]2^kx^k, {k, 0, n}]/(2x).

Original entry on oeis.org

1, 2, 2, 3, 12, 4, 4, 36, 48, 8, 5, 80, 240, 160, 16, 6, 150, 800, 1200, 480, 32, 7, 252, 2100, 5600, 5040, 1344, 64, 8, 392, 4704, 19600, 31360, 18816, 3584, 128, 9, 576, 9408, 56448, 141120, 150528, 64512, 9216, 256, 10, 810, 17280, 141120, 508032

Offset: 0

Roger L. Bagula, Feb 04 2009

			1;
2, 2;
3, 12, 4;
4, 36, 48, 8;
5, 80, 240, 160, 16;
6, 150, 800, 1200, 480, 32;
7, 252, 2100, 5600, 5040, 1344, 64;
8, 392, 4704, 19600, 31360, 18816, 3584, 128;
9, 576, 9408, 56448, 141120, 150528, 64512, 9216, 256;
10, 810, 17280, 141120, 508032, 846720, 645120, 207360, 23040, 512;

Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 75-76

A142983, A142978, A047781 (row sums).

Clear[t0, p, x, n, m];
p[x_, n_] = Sum[Binomial[n, k]*Binomial[n - 1, n - k]*2^k*x^k, {k, 0, n}]/(2*x);
Table[FullSimplify[ExpandAll[p[x, n]]], {n, 1, 10}];
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 1, 10}];
Flatten[%]

p(x,n)=Sum[Binomial[n, k]*Binomial[n - 1, n - k]*2^k*x^k, {k, 0, n}]/(2*x);
p(x,n)=n Hypergeometric2F1[1 - n, 1 - n, 2, 2 x];
t(n,m)=coefficiemts(p(x,n))
T(n,m) = 2^m*A103371(n,m). - R. J. Mathar, Dec 05 2017

A187552 Triangle a(n,k) = binomial(n,k)binomial(n+1,k+1)binomial(n+2,k+2) read by rows.

Original entry on oeis.org

1, 6, 1, 18, 24, 1, 40, 180, 60, 1, 75, 800, 900, 120, 1, 126, 2625, 7000, 3150, 210, 1, 196, 7056, 36750, 39200, 8820, 336, 1, 288, 16464, 148176, 308700, 164640, 21168, 504, 1, 405, 34560, 493920, 1778112, 1852200, 564480, 45360, 720, 1, 550, 66825, 1425600, 8149680, 14669424, 8731800, 1663200, 89100, 990, 1, 726, 121000, 3675375, 31363200, 89646480, 92207808, 34303500, 4356000, 163350, 1320, 1

Offset: 0

Emanuele Munarini, Mar 11 2011

Row sums are 1, 7, 43, 281, 1896, 13112, 92359,...

			Triangle begins:
1
6,1
18,24,1
40,180,60,1
75,800,900,120,1
126,2625,7000,3150,210,1
196,7056,36750,39200,8820,336,1
288,16464,148176,308700,164640,21168,504,1
405,34560,493920,1778112,1852200,564480,45360,720,1

Cf. A103371, A002411 (column k=0), A165187 (column k=1), A007531 (subdiagonal)

A187552 := proc(n,k) binomial(n,k)*binomial(n+1,k+1)*binomial(n+2,k+2) ; end proc:

Table[Binomial[n, k]Binomial[n + 1, k + 1]Binomial[n + 2, k + 2], {n, 0, 8}, {k, 0, 8}]//MatrixForm

create_list(binomial(n,k)*binomial(n+1,k+1)*binomial(n+2,k+2),n,0,8,k,0,n);

cp's OEIS Frontend

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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A197654 Triangle by rows T(n,k), showing the number of meanders with length 5(n+1) and containing 5(k+1) L's and 5(n-k) R's, where L's and R's denote arcs of equal length and a central angle of 72 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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A209805 Triangle read by rows: T(n,k) is the number of k-block noncrossing partitions of n-set up to rotations.

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A113162 a(n) = binomial(6, n) * binomial(7, 6-n).

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A122899 Triangle with row sums counting directed animals.

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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.

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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.

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.

A155497 Triangle T(n, k) = binomial(n, k)binomial(n+1, k+1)binomial(2n, 2k)/(n-k+1), read by rows.

A156136 A triangle of polynomial coefficients related to Mittag-Leffler polynomials: p(x,n)=Sum[Binomial[n, k]Binomial[n - 1, n - k]2^kx^k, {k, 0, n}]/(2x).

A187552 Triangle a(n,k) = binomial(n,k)binomial(n+1,k+1)binomial(n+2,k+2) read by rows.