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A194595
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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.
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8
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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
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OFFSET
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0,2
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COMMENTS
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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.
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.
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LINKS
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Table of n, a(n) for n=0..48.
Peter Luschny, Meanders and walks on the circle.
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FORMULA
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recursive formula:
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: 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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EXAMPLE
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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 Ls.
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.
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MAPLE
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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
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PROG
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(C#)//code by Peter Luschny
static int[] GenBinomial(int n, int k)
{
int[, , ] T = new int[k, n, n ];
for (int m = 0; m < n; m++)
{
T[0, m, 0] = 1; T[0, m, m] = 1;
for (int j = 1; j < m; j++)
{
T[0, m, j] = T[0, m - 1, j] + T[0, m - 1, j - 1];
}
for (int r = 1; r < k; r++)
{
T[r, m, m] = 1;
for (int j = 0; j < m; j++)
{
int p = (int)Math.Pow(T[0, m, j], r + 1);
T[r, m, j] = p + T[0, m, j] * T[r - 1, m, m - j - 1];
}
}
}
int[] R = new int[n];
for (int j = 0; j < n; j++)
{
R[j] = T[k - 1, n - 1, j];
}
return R;
}
static int[] A194595_row(int r)
{
return GenBinomial(r, 3);
}
(PARI)
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
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CROSSREFS
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Cf. A103371, A197653, A197654, A197655.
Sequence in context: A113647 A161380 A051927 * A219063 A218810 A218638
Adjacent sequences: A194592 A194593 A194594 * A194596 A194597 A194598
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KEYWORD
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nonn,tabl
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AUTHOR
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Susanne Wienand, Oct 10 2011
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STATUS
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approved
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