A201641 Triangle read by rows, T(n,k) for 0<=k<=n, generalizes the colored Motzkin paths of A129400.

1, 2, 1, 8, 4, 1, 32, 20, 6, 1, 144, 96, 36, 8, 1, 672, 480, 200, 56, 10, 1, 3264, 2432, 1104, 352, 80, 12, 1, 16256, 12544, 6048, 2128, 560, 108, 14, 1, 82688, 65536, 33152, 12544, 3680, 832, 140, 16, 1, 427520, 346368, 182016, 72960, 23232, 5904, 1176, 176, 18, 1

Offset: 0

Author

Peter Luschny, Sep 20 2012

Keywords

• nonn
• tabl

Examples

			Triangle begins as:
[0] [1]
[1] [2, 1]
[2] [8, 4, 1]
[3] [32, 20, 6, 1]
[4] [144, 96, 36, 8, 1]
[5] [672, 480, 200, 56, 10, 1]
[6] [3264, 2432, 1104, 352, 80, 12, 1]
[7] [16256, 12544, 6048, 2128, 560, 108, 14, 1]
[8] [82688, 65536, 33152, 12544, 3680, 832, 140, 16, 1]
		

Links

• G. C. Greubel, Rows n = 0..100 of triangle, flattened

Crossrefs

Cf. A129400.

Programs

Magma

[[k eq n select 1 else 2^(n-k)*((k+1)/(n+1))*(&+[(-1)^j* Binomial(n+1,j)*Binomial(2*n-k-3*j, n-k-3*j): j in [0..Floor((n-k)/3)]]) :k in [0..n]]: n in [0..10]]; // G. C. Greubel, Apr 05 2019

Maple

T := (n, k) -> 2^n*add(binomial(n,j)*(binomial(n-j,j+k) - binomial(n-j, j+k+2)) *2^(-k), j=0..n); seq(seq(T(n,k), k=0..n), n=0..8); # Peter Luschny, Dec 31 2019

Mathematica

T[n_, k_]:= If[k==n, 1, 2^(n-k)*((k+1)/(n+1))*Sum[(-1)^j*Binomial[n+1,j]* Binomial[2*n-k-3*j, n-k-3*j], {j, 0, Floor[(n-k)/3]}]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* G. C. Greubel, Apr 04 2019 *)

Maxima

T(n,k):=(k+1)/(n+1)*2^(n-k)*sum((-1)^j*binomial(n+1,j)*binomial(2*n-k-3*j,n-k-3*j),j,0,floor((n-k)/3)); /* Vladimir Kruchinin, Apr 06 2019 */

PARI

{T(n,k) = if(k==n, 1, 2^(n-k)*((k+1)/(n+1))*sum(j=0, floor((n-k)/3), (-1)^j*binomial(n+1,j)*binomial(2*n-k-3*j, n-k-3*j)))};
for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Apr 04 2019

Sage

def A201641_triangle(dim):
    M = matrix(ZZ,dim,dim)
    for n in range(dim): M[n,n] = 1
    for n in (1..dim-1):
        for k in (0..n-1):
            M[n,k] = M[n-1,k-1]+2*M[n-1,k]+4*M[n-1,k+1]
    return M
A201641_triangle(9)

Sage

@CachedFunction
def T(n, k):
    if k==n: return 1
    else: return 2^(n-k)*((k+1)/(n+1))*sum((-1)^j*binomial(n+1,j)* binomial(2*n-k-3*j, n-k-3*j) for j in (0..floor((n-k)/3)))
[[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Apr 05 2019

Formula

Recurrence: T(0,0)=1, T(0,k)=0 for k>0 and for n>=1 T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + 4*T(n-1,k+1).

T(n,k) = ((k+1)/(n+1))*2^(n-k)*Sum_{j=0..floor((n-k)/3)} (-1)^j*C(n+1,j) *C(2*n-k-3*j,n-k-3*j). - Vladimir Kruchinin, Apr 06 2019

T(n,k) = 2^n*Sum_{j=0..n} C(n,j)*(C(n-j, j+k) - C(n-j, j+k+2))*2^(-k). - Peter Luschny, Dec 31 2019