A230448 T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = A226205(n+1), n >= 0 and 0 <= k <= n.

1, 1, 0, 1, 1, 3, 1, 2, 4, 5, 1, 3, 6, 9, 16, 1, 4, 9, 15, 25, 39, 1, 5, 13, 24, 40, 64, 105, 1, 6, 18, 37, 64, 104, 169, 272, 1, 7, 24, 55, 101, 168, 273, 441, 715, 1, 8, 31, 79, 156, 269, 441, 714, 1156, 1869, 1, 9, 39, 110, 235, 425, 710, 1155, 1870, 3025, 4896

Offset: 0

Author

Johannes W. Meijer, Oct 19 2013

Keywords

• nonn
• easy
• tabl

Comments

Triangle T(n, k) is related to the Kn1p sums of the ‘Races with Ties’ triangle A035317. See A230447 for the Kn1p sums and A180662 for the definitions of these sums.

The row sums equal ((-1)^n*3*A083581(n) + A022379(2*n+2))/15.

Note that the partial fraction expansion of the G.f. of the terms in the n-th row of the square array Tsq(n, k) = T(n+k, k) is related to A014334, the exponential convolution of the Fibonacci numbers with themselves, and to A000032, the Lucas numbers.

Examples

			The first few rows of triangle T(n, k), n >= 0 and 0 <= k <= n.
n/k 0   1   2    3    4     5     6     7
------------------------------------------------
0|  1
1|  1,  0
2|  1,  1,  3
3|  1,  2,  4,   5
4|  1,  3,  6,   9,  16
5|  1,  4,  9,  15,  25,   39
6|  1,  5, 13,  24,  40,   64,  105
7|  1,  6, 18,  37,  64,  104,  169,   272
The triangle as a square array Tsq(n, k) = T(n+k, k), n >= 0 and k >= 0.
n/k 0   1   2    3    4    5      6     7
------------------------------------------------
0|  1,  0,  3,   5,  16,  39,   105,  272
1|  1,  1,  4,   9,  25,  64,   169,  441
2|  1,  2,  6,  15,  40,  104,  273,  714
3|  1,  3,  9,  24,  64,  168,  441, 1155
4|  1,  4, 13,  37, 101,  269,  710, 1865
5|  1,  5, 18,  55, 156,  425, 1135, 3000
6|  1,  6, 24,  79, 235,  660, 1795, 4795
7|  1,  7, 31, 110, 345, 1005, 2800, 7595
		

Crossrefs

Cf. (Triangle columns) A000012, A001477, A152950, A226205, A007598, A001654, A064831, A080145

Programs

Maple

T := proc(n, k) option remember: if k=0 then return(1) elif k=n then return(combinat[fibonacci](n+2)*combinat[fibonacci](n-1)) else procname(n-1, k-1) + procname(n-1, k) fi: end: seq(seq(T(n, k), k=0..n), n=0..10); # End first program.
T := proc(n, k): add(A035317(n+k-p-2, p), p=0..k) end: A035317 := proc(n, k): add((-1)^(i+k) * binomial(i+n-k+1, i), i=0..k) end: seq(seq(T(n, k), k=0..n), n=0..10); # End second program.

Formula

T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = F(n+2) * F(n-1) = A226205(n+1) with F(n) = A000045(n), the Fibonacci numbers, n >= 0 and 0 <= k <= n.

T(n, k) = sum(A035317(n+k-p-2, p), p=0..k), n >= 0 and 0 <= k <= n.

T(n+p+2, p-2) = A080239(n+2*p-1) - sum(A035317(n-k+p-1, k+p-1), k=0..floor(n/2)), n >= 0 and p >= 2.

The triangle as a square array Tsq(n, k) = T(n+k, k), n >= 0 and k >= 0.

Tsq(n, k) = sum(Tsq(n-1, i), i=0..k), n >= 1 and k >= 0, with Tsq(0, k) = A226205(k+1).

The two G.f.’s given below generate the terms in the n-th row of the square array Tsq(n, k). The remarkable second G.f. is the partial fraction expansion of the first G.f..

G.f.: 1/((1-x)^(n-2)*(1+x)*(x^2-3*x+1)), n >= 0.

G.f.: sum((-1)^(n+k-1)*A014334(k+2)/(2^(k+2)*(x-1)^(n-k-2)), k=0..n-3) + 1/(5*2^(n-2)*(1+x)) + (A000032(n+1) - A000032(n-1)*x)/(5*(x^2-3*x+1)), n >= 0.