A099573 Reverse of number triangle A054450.
1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 4, 5, 1, 1, 5, 5, 8, 8, 1, 1, 6, 6, 12, 12, 13, 1, 1, 7, 7, 17, 17, 21, 21, 1, 1, 8, 8, 23, 23, 33, 33, 34, 1, 1, 9, 9, 30, 30, 50, 50, 55, 55, 1, 1, 10, 10, 38, 38, 73, 73, 88, 88, 89, 1, 1, 11, 11, 47, 47, 103, 103, 138, 138, 144, 144, 1, 1, 12, 12, 57, 57, 141, 141, 211, 211, 232, 232, 233
Offset: 0
Examples
First few rows of the array: 1, 1, 2, 3, 5, 8, ... (A000045) 1, 1, 3, 4, 8, 12, ... (A052952) 1, 1, 4, 5, 12, 17, ... (A054451) 1, 1, 5, 6, 17, 23, ... (A099571) 1, 1, 6, 7, 23, 30, ... (A099572) ... Triangle begins as: 1; 1, 1; 1, 1, 2; 1, 1, 3, 3; 1, 1, 4, 4, 5; 1, 1, 5, 5, 8, 8; 1, 1, 6, 6, 12, 12, 13; 1, 1, 7, 7, 17, 17, 21, 21; 1, 1, 8, 8, 23, 23, 33, 33, 34; 1, 1, 9, 9, 30, 30, 50, 50, 55, 55;
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Crossrefs
Programs
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Magma
[(&+[Binomial(n-j,j): j in [0..Floor(k/2)]]): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022
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Mathematica
T[n_, k_]:= Sum[Binomial[n-j,j], {j,0,Floor[k/2]}]; Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 25 2022 *)
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SageMath
def A099573(n,k): return sum(binomial(n-j, j) for j in (0..(k//2))) flatten([[A099573(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 25 2022
Formula
Number triangle T(n, k) = Sum_{j=0..floor(k/2)} binomial(n-j, j) if k <= n, 0 otherwise.
T(n, n) = A000045(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A099574(n).
Sum_{k=0..n} T(n, k) = A029907(n+1).
Antidiagonals of the following array: the first row equals the Fibonacci numbers, (1, 1, 2, 3, 5, ...), and the (n+1)-st row is obtained by the matrix-vector product A128174 * n-th row. - Gary W. Adamson, Jan 19 2011
From G. C. Greubel, Jul 25 2022: (Start)
T(n, n-1) = A052952(n-1), n >= 1.
T(n, n-2) = A054451(n-2), n >= 2.
T(n, n-3) = A099571(n-3), n >= 3.
T(n, n-4) = A099572(n-4), n >= 4. (End)