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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. 7
 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 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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 LINKS Susanne Wienand, Table of n, a(n) for n = 0..209 Peter Luschny, Meanders and walks on the circle. FORMULA 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] EXAMPLE 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. MAPLE 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 MATHEMATICA 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 *) PROG (C#) static int[] A197655_row(int r) { return GenBinomial(r, 6); } // The function GenBinomial(r, s) is defined in A194595. // This C#-program causes numerical overflow for results // larger than 2147483647. - Susanne Wienand, Jul 04 2015 (Sage) 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 (PARI) 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 CROSSREFS Cf. A000012, A007318, A053700, A103371, A194595, A197653, A197654, A198258. Sequence in context: A174502 A056218 A292219 * A134279 A134280 A134278 Adjacent sequences:  A197652 A197653 A197654 * A197656 A197657 A197658 KEYWORD nonn,tabl AUTHOR Susanne Wienand, Oct 19 2011 STATUS approved

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Last modified December 16 00:33 EST 2019. Contains 330013 sequences. (Running on oeis4.)