

A197655


Triangle by rows T(n,k), showing the number of meanders with length (n+1)*6 and containing (k+1)*6 Ls and (nk)*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
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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,n1k), 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,n1k), 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,n1k), 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,n1k), 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,n1k), 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,n1k)
C = (C(n,k))^4 *(C(n,n1k))^2
D = (C(n,k))^3 *(C(n,n1k))^3
E = (C(n,k))^2 *(C(n,n1k))^4
F = C(n,k) *(C(n,n1k))^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((nk)/(1+k))^6)/(1+2*kn) if 1+2*kn <> 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(5,4,412) = 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^2n3*k*n+n^2)*(1+k+k^2+nk*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 (* JeanFrançois 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)^(Nj+i)*binomial(Ni, 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((nk)/(1+k))^6)/(1+2*kn))*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



