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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.

%I A194595 #121 Jun 22 2024 22:38:03
%S A194595 1,3,1,7,14,1,13,81,39,1,21,304,456,84,1,31,875,3000,1750,155,1,43,
%T A194595 2106,13875,18500,5265,258,1,57,4459,50421,128625,84035,13377,399,1,
%U A194595 73,8576,153664,669536,836920,307328,30016,584,1,91,15309,409536,2815344,6001128,4223016,955584,61236,819,1
%N 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.
%C A194595 Definition of a meander:
%C A194595 A binary curve C is a triple (m, S, dir) such that
%C A194595 (a) S is a list with values in {L,R} which starts with an L,
%C A194595 (b) dir is a list of m different values, each value of S being allocated
%C A194595     a value of dir,
%C A194595 (c) consecutive L's increment the index of dir,
%C A194595 (d) consecutive R's decrement the index of dir,
%C A194595 (e) the integer m > 0 divides the length of S and
%C A194595 (f) C is a meander if each value of dir occurs length(S)/m times.
%C A194595 For this sequence, m = 3.
%C A194595 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 A194595 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 A194595 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 A194595 The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 62. - _Susanne Wienand_, Jun 24 2015
%H A194595 Susanne Wienand, <a href="/A194595/b194595.txt">Table of n, a(n) for n = 0..2015</a>
%H A194595 Peter Luschny, <a href="http://oeis.org/wiki/User:Peter_Luschny/Meander">Meanders and walks on the circle</a>.
%H A194595 Susanne Wienand, <a href="http://oeis.org/wiki/File:Meander_m%3D3.gif">Animation of a meander</a>.
%H A194595 Susanne Wienand, <a href="https://oeis.org/wiki/File:Meander,_m%3D3.png">Example of a meander</a>.
%F A194595 Recursive formula (conjectured):
%F A194595 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 A194595 T(3,n,n) = 1,                                            k = n
%F A194595 T(2,n,k) = T(1,n,k)^2 + T(1,n,k) * T(1,n,n-1-k),         0 <= k < n
%F A194595 T(2,n,n) = 1,                                            k = n
%F A194595 T(2,n,k) = A103371,
%F A194595 T(1,n,k) = A007318 (Pascal's Triangle).
%F A194595 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 A194595 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 A194595 T(n,k) = A073254(n+1,k+1)C(n,k)^3/(k+1)^2. - _Peter Luschny_, Oct 29 2011
%F A194595 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 A194595 For n = 4 and k = 2, T(3,4,2) = 456.
%e A194595 Recursive example:
%e A194595 T(1,4,0) = 1
%e A194595 T(1,4,1) = 4
%e A194595 T(1,4,2) = 6
%e A194595 T(1,4,3) = 4
%e A194595 T(1,4,4) = 1
%e A194595 T(2,4,0) = 5
%e A194595 T(2,4,1) = 40
%e A194595 T(2,4,2) = 60
%e A194595 T(2,4,3) = 20
%e A194595 T(2,4,4) = 1
%e A194595 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 A194595 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 A194595 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 A194595 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 A194595 T(3,4,4) = 1.
%e A194595 Example for closed formula:
%e A194595 T(4,2) = (C(4,2))^3 + C(4,2) * C(4,3) * C(5,3) = 6^3 + 6 * 4 * 10 = 456.
%e A194595 Some examples of list S and allocated values of dir if n = 4 and k = 2:
%e A194595 Length(S) = (4+1)*3 = 15 and S contains (2+1)*3 = 9 L's.
%e A194595   S: L,L,L,L,L,L,L,L,L,R,R,R,R,R,R
%e A194595 dir: 1,2,0,1,2,0,1,2,0,0,2,1,0,2,1
%e A194595   S: L,L,R,L,L,L,L,R,R,L,R,R,L,R,L
%e A194595 dir: 1,2,2,2,0,1,2,2,1,1,1,0,0,0,0
%e A194595   S: L,R,R,R,L,L,L,L,R,R,L,L,L,R,L
%e A194595 dir: 1,1,0,2,2,0,1,2,2,1,1,2,0,0,0
%e A194595 Each value of dir occurs 15/3 = 5 times.
%p A194595 A194595 := (n,k)->binomial(n,k)^3*(k^2+k+1+n^2+n-k*n)/((k+1)^2);
%p A194595 seq(print(seq(A194595(n,k),k=0..n)),n=0..7); # _Peter Luschny_, Oct 14 2011
%t A194595 T[n_, k_] := Binomial[n, k]^3*(k^2 + k + 1 + n^2 + n - k*n)/((k + 1)^2);
%t A194595 Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 30 2018, after _Peter Luschny_ *)
%o A194595 (C#)//code by _Peter Luschny_
%o A194595 static int[] GenBinomial(int n, int k)
%o A194595 {
%o A194595     int[, ,] T = new int[k, n, n ];
%o A194595     for (int m = 0; m < n; m++)
%o A194595     {
%o A194595         T[0, m, 0] = 1; T[0, m, m] = 1;
%o A194595         for (int j = 1; j < m; j++)
%o A194595         {
%o A194595             T[0, m, j] = T[0, m - 1, j] + T[0, m - 1, j - 1];
%o A194595         }
%o A194595         for (int r = 1; r < k; r++)
%o A194595         {
%o A194595             T[r, m, m] = 1;
%o A194595             for (int j = 0; j < m; j++)
%o A194595             {
%o A194595                 int p = (int)Math.Pow(T[0, m, j], r + 1);
%o A194595                 T[r, m, j] = p + T[0, m, j] * T[r - 1, m, m - j - 1];
%o A194595             }
%o A194595         }
%o A194595     }
%o A194595     int[] R = new int[n];
%o A194595     for (int j = 0; j < n; j++)
%o A194595     {
%o A194595         R[j] = T[k - 1, n - 1, j];
%o A194595     }
%o A194595     return R;
%o A194595 }
%o A194595 static int[] A194595_row(int r)
%o A194595 {
%o A194595      return GenBinomial(r, 3);
%o A194595 }
%o A194595 // This C#-program causes numerical overflow for results
%o A194595 // larger than 2147483647. - _Susanne Wienand_, Jun 25 2015
%o A194595 (PARI)
%o A194595 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 A194595 Cf. A103371, A197653, A197654, A197655, A197657, A007318, A002061, A000012, A027444, A073254.
%K A194595 nonn,tabl
%O A194595 0,2
%A A194595 _Susanne Wienand_, Oct 10 2011