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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 Ls and (n-k)*3 Rs, where Ls and Rs denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented. 8

%I

%S 1,3,1,7,14,1,13,81,39,1,21,304,456,84,1,31,875,3000,1750,155,1,43,

%T 2106,13875,18500,5265,258,1,57,4459,50421,128625,84035,13377,399,1,

%U 73,8576,153664,669536,836920,307328,30016,584,1,91,15309,409536,2815344,6001128,4223016,955584,61236,819,1

%N Triangle by rows T(n,k), showing the number of meanders with length (n+1)*3 and containing (k+1)*3 Ls and (n-k)*3 Rs, where Ls and Rs denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented.

%C Definition of a meander:

%C A binary curve C is a triple (m, S, dir) such that

%C (a) S is a list with values in {L,R} which starts with an L,

%C (b) dir is a list of m different values, each value of S being allocated

%C a value of dir,

%C (c) consecutive Ls increment the index of dir,

%C (d) consecutive Rs decrement the index of dir,

%C (e) the integer m>0 divides the length of S and

%C (f) C is a meander if each value of dir occurs length(S)/m times.

%C For this sequence, m = 3.

%C The values in the triangle are proved by brute force for 0 <= n <= 11. The formulas are not yet proved in general. - _Susanne Wienand_

%C 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

%C 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

%C The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 62. - _Susanne Wienand_, Jun 24 2015

%H Susanne Wienand, <a href="/A194595/b194595.txt">Table of n, a(n) for n = 0..2015</a>

%H Peter Luschny, <a href="http://oeis.org/wiki/User:Peter_Luschny/Meander">Meanders and walks on the circle</a>.

%H Susanne Wienand, <a href="http://oeis.org/wiki/File:Meander_m%3D3.gif">Animation of a meander</a>

%H Susanne Wienand, <a href="https://oeis.org/wiki/File:Meander,_m%3D3.png">Example of a meander</a>

%F Recursive formula (conjectured):

%F 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

%F T(3,n,n) = 1, k = n

%F T(2,n,k) = T(1,n,k)^2 + T(1,n,k) * T(1,n,n-1-k), 0 <= k < n

%F T(2,n,n) = 1, k = n

%F T(2,n,k) = A103371,

%F T(1,n,k) = A007318 (Pascal's Triangle).

%F Closed formula (conjectured): T(n,k) = (C(n,k))^3 + C(n,k) * C(n,k+1) * C(n+1,k+1). [_Susanne Wienand_]

%F 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]

%F T(n,k) = A073254(n+1,k+1)C(n,k)^3/(k+1)^2. [_Peter Luschny_, Oct 29 2011]

%F 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]

%e For n = 4 and k = 2, T(3,4,2) = 456

%e recursive example:

%e T(1,4,0) = 1

%e T(1,4,1) = 4

%e T(1,4,2) = 6

%e T(1,4,3) = 4

%e T(1,4,4) = 1

%e T(2,4,0) = 5

%e T(2,4,1) = 40

%e T(2,4,2) = 60

%e T(2,4,3) = 20

%e T(2,4,4) = 1

%e T(3,4,0) = T(1,4,0)^3 + T(1,4,0)*T(2,4,4-1-0) = 1^3 + 1*20 = 21

%e T(3,4,1) = T(1,4,1)^3 + T(1,4,1)*T(2,4,4-1-1) = 4^3 + 4*60 = 304

%e T(3,4,2) = T(1,4,2)^3 + T(1,4,2)*T(2,4,4-1-2) = 6^3 + 6*40 = 456

%e T(3,4,3) = T(1,4,3)^3 +(T(1,4,3)*T(2,4,4-1-3) = 4^3 + 4*5 = 84

%e T(3,4,4) = 1

%e example for closed formula:

%e T(4,2) = (C(4,2))^3 + C(4,2) * C(4,3) * C(5,3) = 6^3 + 6 * 4 * 10 = 456

%e Some examples of list S and allocated values of dir if n = 4 and k = 2:

%e Length(S) = (4+1)*3 = 15 and S contains (2+1)*3 = 9 Ls.

%e S: L,L,L,L,L,L,L,L,L,R,R,R,R,R,R

%e dir: 1,2,0,1,2,0,1,2,0,0,2,1,0,2,1

%e S: L,L,R,L,L,L,L,R,R,L,R,R,L,R,L

%e dir: 1,2,2,2,0,1,2,2,1,1,1,0,0,0,0

%e S: L,R,R,R,L,L,L,L,R,R,L,L,L,R,L

%e dir: 1,1,0,2,2,0,1,2,2,1,1,2,0,0,0

%e Each value of dir occurs 15/3 = 5 times.

%p A194595 := (n,k)->binomial(n,k)^3*(k^2+k+1+n^2+n-k*n)/((k+1)^2);

%p seq(print(seq(A194595(n,k),k=0..n)),n=0..7); # _Peter Luschny_, Oct 14 2011

%t T[n_, k_] := Binomial[n, k]^3*(k^2 + k + 1 + n^2 + n - k*n)/((k + 1)^2);

%t Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Jean-Fran├žois Alcover_, Jul 30 2018, after _Peter Luschny_ *)

%o (C#)//code by _Peter Luschny_

%o static int[] GenBinomial(int n, int k)

%o {

%o int[, ,] T = new int[k, n, n ];

%o for (int m = 0; m < n; m++)

%o {

%o T[0, m, 0] = 1; T[0, m, m] = 1;

%o for (int j = 1; j < m; j++)

%o {

%o T[0, m, j] = T[0, m - 1, j] + T[0, m - 1, j - 1];

%o }

%o for (int r = 1; r < k; r++)

%o {

%o T[r, m, m] = 1;

%o for (int j = 0; j < m; j++)

%o {

%o int p = (int)Math.Pow(T[0, m, j], r + 1);

%o T[r, m, j] = p + T[0, m, j] * T[r - 1, m, m - j - 1];

%o }

%o }

%o }

%o int[] R = new int[n];

%o for (int j = 0; j < n; j++)

%o {

%o R[j] = T[k - 1, n - 1, j];

%o }

%o return R;

%o }

%o static int[] A194595_row(int r)

%o {

%o return GenBinomial(r, 3);

%o }

%o // This C#-program causes numerical overflow for results

%o // larger than 2147483647. - _Susanne Wienand_, Jun 25 2015

%o (PARI)

%o 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

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

%K nonn,tabl

%O 0,2

%A _Susanne Wienand_, Oct 10 2011

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Last modified January 20 08:07 EST 2020. Contains 331081 sequences. (Running on oeis4.)