A237124 Triangle of numbers related to Catalan numbers (A000108).
1, 1, 1, 1, 2, 1, 3, 4, 3, 1, 9, 11, 8, 4, 1, 28, 33, 24, 13, 5, 1, 90, 104, 76, 43, 19, 6, 1, 297, 339, 249, 145, 69, 26, 7, 1, 1001, 1133, 836, 497, 248, 103, 34, 8, 1, 3432, 3861, 2860, 1727, 891, 394, 146, 43, 9, 1, 11934, 13364, 9932, 6071, 3211, 1484, 593, 199, 53, 10, 1
Offset: 0
Examples
Triangle begins:
1;
1, 1;
1, 2, 1;
3, 4, 3, 1;
9, 11, 8, 4, 1;
28, 33, 24, 13, 5, 1;
90, 104, 76, 43, 19, 6, 1;
297, 339, 249, 145, 69, 26, 7, 1;
1001, 1133, 836, 497, 248, 103, 34, 8, 1;
3432, 3861, 2860, 1727, 891, 394, 146, 43, 9, 1;
11934, 13364, 9932, 6071, 3211, 1484, 593, 199, 53, 10, 1;
...
Links
- G. C. Greubel, Rows n = 0..30 of the triangle, flattened
Crossrefs
Cf. A000108.
Programs
-
Mathematica
b[n_, k_]:= Binomial[2*n-k+1, n-k]; T[n_, k_]:= If[n<3, Binomial[n, k], b[n, k] -2*b[n, k+1] -b[n, k+2] +3*b[n, k+3] - 2*b[n, k+4]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, May 08 2021 *) -
Sage
def b(n,k): return binomial(2*n-k+1, n-k) def T(n,k): return binomial(n,k) if (n<3) else b(n,k) -2*b(n, k+1) -b(n, k+2) +3*b(n, k+3) -2*b(n, k+4) flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 08 2021
Formula
From Peter Bala, Feb 18 2018: (Start)
T(n,k) = C(2*n+1-k, n-k) - 2*C(2*n-k, n-k-1) - C(2*n-1-k, n-k-2) + 3*C(2*n-2-k, n-k-3) - 2*C(2*n-3-k, n-k-4), for n > 2, otherwise C(n, k).
The n-th row polynomial of the row reverse triangle equals the n-th degree Taylor polynomial of the function (1 - x^2 + x^3)*(1 - 2*x)/(1 - x)^2 * 1/(1 - x)^n about 0. For example, for n = 4, (1 - x^2 + x^3)*(1 - 2*x)/(1 - x)^2 * 1/(1 - x)^4 = 1 + 4*x + 8*x^2 + 11*x^3 + 9*x^4 + O(x^5), giving (9, 11, 8, 4, 1) as row 4. (End)
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