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A099573
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Reverse of number triangle A054450.
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3
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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
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OFFSET
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0,6
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LINKS
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FORMULA
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Number triangle T(n, k) = Sum_{j=0..floor(k/2)} binomial(n-j, j) if k <= n, 0 otherwise.
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
T(n, n-4) = A099572(n-4), n >= 4. (End)
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EXAMPLE
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First few rows of the array:
...
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;
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MATHEMATICA
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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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PROG
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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
(SageMath)
def A099573(n, k): return sum(binomial(n-j, j) for j in (0..(k//2)))
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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