0,4

Consider the fractional triangle A180955/A180956, call it triangle A. Consider also a triangle defined by k=1: T(n,1)=1, k>1 and n>=k: T(n,k)= any random number, else 0, call it triangle B. Calculate the matrix inverse of triangle B, call it triangle C. Multiply C with A, call the result triangle D. Calculate the matrix inverse of D, call it triangle E. Then the first column in both matrix A and matrix E will have the fraction A001790/A046161 in the first column.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

From G. C. Greubel, Sep 22 2024: (Start)

T(n, k) = A001790(n-k) = numerator(binomial(2*(n-k), n-k)/4^(n-k)).

T(n, 0) = T(2*n, n) = A001790(n).

Sum_{k=0..n} (-1)^k*T(n, k) = A001790(n+1) + Sum_{j=0..n+1} (-1)^(n+j)*A001790(j).

Sum_{k=0..floor(n/2)} T(n-k, k) = (1/2)*Sum_{j=0..n} (1+(-1)^(n+j))*A001790(j). (End)

Triangle starts:

1;

1, 1;

3, 1, 1;

5, 3, 1, 1;

35, 5, 3, 1, 1;

63, 35, 5, 3, 1, 1;

231, 63, 35, 5, 3, 1, 1;

429, 231, 63, 35, 5, 3, 1, 1;

6435, 429, 231, 63, 35, 5, 3, 1, 1;

12155, 6435, 429, 231, 63, 35, 5, 3, 1, 1;

46189, 12155, 6435, 429, 231, 63, 35, 5, 3, 1, 1;

88179, 46189, 12155, 6435, 429, 231, 63, 35, 5, 3, 1, 1;

A180955[n_, k_]:= Numerator[Binomial[2*(n-k), n-k]/4^(n-k)];

Table[A180955[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Sep 22 2024 *)

(Magma)

A180955:= func< n, k | Numerator((n-k+1)*Catalan(n-k)/4^(n-k)) >;

[A180955(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 22 2024

(SageMath)

def A180955(n, k): return numerator(binomial(2*(n-k), n-k)/4^(n-k))

flatten([[A180955(n, k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Sep 22 2024

Cf. A001790, A046161, A180956.

Sequence in context: A131247 A114278 A134836 * A349182 A191751 A285409

Adjacent sequences: A180952 A180953 A180954 * A180956 A180957 A180958

nonn,tabl

Mats Granvik, Sep 28 2010

Offset changed by G. C. Greubel, Sep 22 2024

approved