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In the column k of this triangle (without leading zeros) is the k-fold iterated partial sums of the Fibonacci numbers, starting with 1. A000045(n+1), A000071(n+3), A001924(n+1), A014162(n+1), A014166(n+1), ..., n >= 0. See the Riordan property. - _Wolfdieter Lang_, Oct 03 2014

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Riordan array (1/(1 - x - x^2), x/(1 - x)).

Row sums are A027934, antidiagonal sums are A010049(n+1). Inverse is A105810.

For a combinatorial interpretation of these iterated partial sums see the H. Belbachir and A. Belkhir link. There table 1 shows in the rows these columns. In their notation (with r = k) f^(k)(n) = T(k, n+k).

The A-sequence of this Riordan triangle is [1, 1] (see the recurrence for T(n, k), k >= 1, given in the formula section). The Z-sequence is A165326 = [1, repeat(1, -1)]. See the W. Lang link under A006232 for Riordan A- and Z-sequences. (End)

The alternating row sums are A212804. (End)

Number triangle Triangle T(n, k) = Sum_{j=0..n} binomial(n-j, k+j); T(n, 0) = A000045(n+1);

T(n, m) = T(n-1, m-1) + T(n-1, m).

T(n, k) = Sum_{j=0..n} binomial(j, n+k-j). - Paul Barry, Oct 23 2006

G.f. of row polynomials Sum_{k=0..n} T(n, k)*x^k is (1 - z)/((1 - z - z^2)*(1 - (1 + x)*z)) (Riordan property). - Wolfdieter Lang, Oct 04 2014

T(n, k) = binomial(n, k)*hypergeom([1, k/2 - n/2, k/2 - n/2 + 1/2],[k + 1, -n], -4) for n > 0. - Peter Luschny, Oct 10 2014

A(k, n) = FnF(n+1+2*k) - Sum_{j=0..k-1} F(2*(k-j)-1) * binomial(n+1+j, j), (from iteration of partial sums).

T(n, k) = F(n+1+k) - Sum_{j=0..k-1} F(2*(k-j)-1) * binomial(n - (k-1-j), j). (End)

Cf. A165326 (Z-sequence), A027934 (row sums), A010049(n+1) (antidiagonal sums), A212804 (alternating row sums), inverse is A105810.

Cf.A165326 (Z-sequence), A027934 (row sums, see comment above), A212804 (alternating row sums). - Wolfdieter Lang, Oct 04 2014

T(n, k) = F(n+1+k)-Sum_{j=0..k-1} F(2*(k-j)-1) * binomial(n-(k-1-j)), , j). (End)

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Number triangle T(n, k) = Sum_{j=0..n} binomial(n-j, k+j); T(n, 0) = A000045(n+1);

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From ~~~_Wolfdieter Lang_, Feb 13 2025: (Start)

Triangle T(n, k) = A(k, n-k) = Sum_{j=k..n} F(n-j+1) * binomial(j-1, k-1), 0 <= k <= n.

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