Susanne Wienand - cp's OEIS frontend

A202409 Triangle read by rows, n>=1, 1<=k<=n, T(n,k) = kbinomial(n,k)^3(n^2+n-kn-k+k^2)/((n-k+1)^2n).

1, 4, 4, 9, 36, 9, 16, 168, 168, 16, 25, 550, 1400, 550, 25, 36, 1440, 7500, 7500, 1440, 36, 49, 3234, 30135, 61250, 30135, 3234, 49, 64, 6496, 98784, 356720, 356720, 98784, 6496, 64, 81, 11988, 278208, 1629936, 2889432, 1629936, 278208, 11988, 81

Offset: 1

Peter Luschny and Susanne Wienand, Dec 19 2011

Let a meander be defined as in the link and m = 3. Then T(n,k) counts the invertible meanders of length m(n+1) built from arcs with central angle 360/m whose binary representation have mk '1's.

			[1]                1
[2]               4, 4
[3]             9, 36, 9
[4]         16, 168, 168, 16
[5]      25, 550, 1400, 550, 25
[6]  36, 1440, 7500, 7500, 1440, 36
T(2,1) = 4 because the invertible meanders of length 9 and central angle 120 degree which have three '1's in their binary representation are {100100100, 100011000, 110001000, 111000000}.

Peter Luschny, Meanders.

Row sums: A201640. Cf. A132812.

A202409 := (n,k) -> k*binomial(n,k)^3*(n^2+n-k*n-k+k^2)/((n-k+1)^2*n);
seq(print(seq(A202409(n,k),k=1..n)),n=1..6);

t[n_, k_] := k*Binomial[n, k]^3*(n^2 + n - k*n - k + k^2)/((n - k + 1)^2*n); Table[t[n, k], {n, 1, 9}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 02 2013 *)

A198256 Row sums of A197653.

Original entry on oeis.org

1, 5, 46, 485, 5626, 69062, 882540, 11614437, 156343330, 2142556130, 29791689148, 419260001030, 5960334608788, 85469709312860, 1234797737654296, 17955907741675749, 262607675818816050, 3860239468267647914, 57002176852356800700, 845159480056345448610

Offset: 0

Susanne Wienand, Oct 22 2011

nonn

Number of meanders of length (n+1)*4 which are composed by arcs of equal length and a central angle of 90 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 = 4.
The terms are proved by brute force for 0 <= n <= 8, but not yet in general. - Susanne Wienand, Oct 29 2011

			Some examples of list S and allocated values of dir if n = 4:
Length(S) = (4+1)*4 = 20.
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L
dir: 1,2,3,0,1,2,3,0,1,2,3,0,1,2,3,0,1,2,3,0
  S: L,L,L,L,R,L,R,R,L,R,R,R,L,R,L,L,R,L,L,L
dir: 1,2,3,0,0,0,0,3,3,3,2,1,1,1,1,2,2,2,3,0
  S: L,L,R,L,L,L,L,R,R,L,R,R,L,R,L,L,L,L,R,R
dir: 1,2,2,2,3,0,1,1,0,0,0,3,3,3,3,0,1,2,2,1
Each value of dir occurs 20/4 = 5 times.

Peter Luschny, Meanders and walks on the circle.

Cf. A197653, A198060, A198257, A198258.
Cf. A005260, A181069.
Cf. A181067, A374614.

A198256[n_] := Sum[Sum[Sum[(-1)^(j + i)* Binomial[i, j]*Binomial[n, k]^4*(n + 1)^j*(k + 1)^(3 - j)/(k + 1)^3, {i, 0, 3}], {j, 0, 3}], {k, 0, n}]; Table[A198256[n], {n, 0, 16}] (* Peter Luschny, Nov 02 2011 *)

A198256(n) = {sum(k=0,n,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

from mpmath import mp, hyper
def A198256(n) : return hyper([1-n, 1-n, 1-n, 1-n], [3, 3, 3], 1)*(n^4-n^6)/4 + hyper([-n, -n, -n, -n], [2, 2, 2], 1)*(1+n+n^2+n^3) + hyper([2, 1-n, 1-n, 1-n, 1-n], [1, 3, 3, 3], 1)*(n^4+n^5)/4
mp.dps = 32
for n in (0..19) : print(int(A198256(n)))  # Peter Luschny, Oct 24 2011

a(n) = Sum_{k=0..n} Sum_{j=0..3} Sum_{i=0..3} (-1)^(j+i)*C(i,j)*C(n,k)^4*(n+1)^j*(k+1)^(3-j)/(k+1)^3. - Peter Luschny, Nov 02 2011
a(n) = Sum_{k=0..n} 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 else h(n,k) = 4. - Peter Luschny, Nov 24 2011
From Peter Bala, Mar 21 2023: (Start)
Conjecture 1: a(n) = Sum_{k = 0..n} binomial(n+1,k)^2*binomial(n,k)^2.
If true, then we have the third-order recurrence equation
n^2*(n + 1)^3*P(n-1)*a(n) = 2*n^2*(400*n^8 - 1260*n^7 + 20*n^6 + 3020*n^5 - 1646*n^4 - 1951*n^3 + 1142*n^2 + 465*n - 290)*a(n-1) + 4*(n - 1)*(2800*n^9 - 15420*n^8 + 30620*n^7 - 23710*n^6 + 808*n^5 + 6863*n^4 - 1309*n^3 - 1218*n^2 + 496*n - 60)*a(n-2) + 8*(n - 2)^2*(2*n - 3)*(4*n - 5)*(4*n - 7)*P(n)*a(n-3) with a(0) = 1, a(1) = 5 and a(2) = 46 and where P(n) = 100*n^5 - 65*n^4 - 35*n^3 + 25*n^2 + 6*n - 5.
Conjecture 2: working with offset 1, that is, a(1) = 1, a(2) = 5, ..., then the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 5. (End)
a(n) ~ 2^(4*n + 5/2) / (Pi*n)^(3/2). - Vaclav Kotesovec, Apr 17 2023
Peter Bala's conjecture 1 can equivalently written: a(n) = hypergeom([-n - 1, -n - 1, -n, -n], [1, 1, 1], 1). - Detlef Meya, May 28 2024
a(n) = Sum_{k=0..n+1} (k/(n+1))^2 * binomial(n+1,k)^4. - Seiichi Manyama, Jul 14 2024

A198258 Row sums of A197655.

Original entry on oeis.org

1, 7, 190, 5831, 219626, 8976562, 394800204, 18211045575, 873216720082, 43136486269382, 2183722698469676, 112795257850321202, 5925951358743662260, 315869014732813229716, 17048301919464100932440, 930217336628892162216135, 51244644190683748419791970

Offset: 0

Susanne Wienand, Oct 24 2011

nonn

Number of meanders of length (n+1)*6 which are composed by arcs of equal length and a central angle of 60 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 = 6.
The terms are proved by brute force for 0 <= n <= 5, but not yet in general. - Susanne Wienand, Oct 29 2011

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

Peter Luschny, Meanders and walks on the circle.

Cf. A198060, A198256, A198257.

A198258 := 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)^6*pow(n+1,j)*pow(k+1,5-j)/(k+1)^5, i=0..5),j=0..5),k=0..n) end:
seq(A198258(n), n=0..16); # Peter Luschny, Nov 02 2011

T[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;
a[n_] := Sum[T[n, k], {k, 0, n}];
Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Jun 29 2019 *)

A198258(n) = {sum(k=0,n,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

a(n) = Sum_{k=0..n} Sum_{j=0..5} Sum_{i=0..5} (-1)^(j+i)*C(i,j)*C(n,k)^6*(n+1)^j*(k+1)^(5-j)/(k+1)^5. - Peter Luschny, Nov 02 2011
a(n) = Sum_{k=0..n} 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
Conjecture: working with offset 1, that is, a(1) = 1, a(2) = 7, ..., then the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 3. - Peter Bala, Mar 21 2023
a(n) ~ sqrt(3) * 2^(6*n+3) / (Pi*n)^(5/2). - Vaclav Kotesovec, Apr 17 2023

A198257 Row sums of A197654.

Original entry on oeis.org

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

A197657 Row sums of A194595.

Original entry on oeis.org

1, 4, 22, 134, 866, 5812, 40048, 281374, 2006698, 14482064, 105527060, 775113440, 5731756720, 42628923040, 318621793472, 2391808860446, 18023208400634, 136271601087352, 1033449449559724, 7858699302115444, 59906766929537116, 457685157123172664

Offset: 0

Susanne Wienand, Oct 17 2011

Number of meanders of length (n+1)*3 which are composed by arcs of equal length and a central angle of 120 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 = 3.
For 0 <= n <= 16, a(n) = the hypergraph Fuss-Catalan number FC_1^(2,n+1) in the notation of Chavan et al. - see 7.1 in the Appendix. - Peter Bala, Apr 11 2023

			Some examples of list S and allocated values of dir if n = 4:
Length(S) = (4+1)*3 = 15.
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L
dir: 1,2,0,1,2,0,1,2,0,1,2,0,1,2,0
  S: L,L,L,L,R,L,L,R,L,L,R,L,L,L,L
dir: 1,2,0,1,1,1,2,2,2,0,0,0,1,2,0
  S: L,R,L,L,L,L,L,R,L,L,R,L,R,R,R
dir: 1,1,1,2,0,1,2,2,2,0,0,0,0,2,1
Each value of dir occurs 15/3 = 5 times.

Parth Chavan, Andrew Lee and Karthik Seetharaman, Hypergraph Fuss-Catalan numbers, arXiv:2202.01111 [math.CO]
Peter Luschny, Meanders and walks on the circle.
Susanne Wienand, Animation of a meander
Susanne Wienand, Example of a meander

Cf. A198060, A198256, A198257, A198258.

A197657 := proc(n)
    (A000172(n) + A000172(n+1)) / 3 ;
end proc; # R. J. Mathar, Jul 26 2014
a := n -> 2^n*hypergeom([n + 1, -n/2, -n/2 - 1/2], [1, 1], 1):
seq(simplify(a(n)), n = 0..21); # Peter Luschny, Mar 26 2023

A197657[n_] := Sum[Sum[Sum[(-1)^(j + i)* Binomial[i, j]*Binomial[n, k]^3*(n + 1)^j*(k + 1)^(2 - j)/(k + 1)^2, {i, 0, 2}], {j, 0, 2}], {k, 0, n}]; Table[A197657[n], {n, 0, 16}]  (* Peter Luschny, Nov 02 2011 *)

A197657(n) = {sum(k=0,n,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

def A197657(n):
    return 2^n*hypergeometric([n + 1, -n/2, -n/2 - 1/2], [1, 1], 1).simplify_hypergeometric()
for n in (0..21): print(A197657(n)) # Peter Luschny, Mar 26 2023

a(n) = Sum{k=0..n} Sum{j=0..2} Sum{i=0..2} (-1)^(j+i)*C(i,j)*C(n,k)^3*(n+1)^j*(k+1)^(2-j)/(k+1)^2. - Peter Luschny, Nov 02 2011
a(n) = Sum_{k=0..n} 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
a(n) = A141147(n+1)/2 = A110707(n+1)/6 = (A000172(n)+A000172(n+1))/3. - Max Alekseyev, Jul 15 2014
Conjecture: (n+1)^2*a(n) -3*(n+1)*(2*n+1)*a(n-1) -3*n*(5*n-7)*a(n-2) -8*(n-2)^2*a(n-3)=0. - R. J. Mathar, Jul 26 2014
a(n) = 2^n*hypergeom([n + 1, -n/2, -n/2 - 1/2], [1, 1], 1). - Peter Luschny, Mar 26 2023
a(n) ~ sqrt(3) * 2^(3*n+1) / (Pi*n). - Vaclav Kotesovec, Apr 17 2023

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]

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]

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

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

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User: Susanne Wienand

A202409 Triangle read by rows, n>=1, 1<=k<=n, T(n,k) = k*binomial(n,k)^3*(n^2+n-k*n-k+k^2)/((n-k+1)^2*n).

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A198256 Row sums of A197653.

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A198258 Row sums of A197655.

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A198257 Row sums of A197654.

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A197657 Row sums of A194595.

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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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A202409 Triangle read by rows, n>=1, 1<=k<=n, T(n,k) = kbinomial(n,k)^3(n^2+n-kn-k+k^2)/((n-k+1)^2n).

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.

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.

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.