A105809 - OEIS

A105809

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

1, 1, 1, 2, 2, 1, 3, 4, 3, 1, 5, 7, 7, 4, 1, 8, 12, 14, 11, 5, 1, 13, 20, 26, 25, 16, 6, 1, 21, 33, 46, 51, 41, 22, 7, 1, 34, 54, 79, 97, 92, 63, 29, 8, 1, 55, 88, 133, 176, 189, 155, 92, 37, 9, 1, 89, 143, 221, 309, 365, 344, 247, 129, 46, 10, 1, 144, 232, 364, 530, 674, 709, 591

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,4

COMMENTS

Previous name was: A Fibonacci-Pascal matrix.

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

From Wolfdieter Lang, Oct 04 2014: (Start)

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

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.

The alternating row sums are A212804. (End)

LINKS

Reinhard Zumkeller, Rows n = 0..120 of table, flattened

H. Belbachir and A. Belkhir, Combinatorial Expressions Involving Fibonacci, Hyperfibonacci, and Incomplete Fibonacci Numbers, Journal of Integer Sequences, Vol. 17 (2014), Article 14.4.3.

Hung Viet Chu, Partial Sums of the Fibonacci Sequence, arXiv:2106.03659 [math.CO], 2021.

Index entries for triangles and arrays related to Pascal's triangle

FORMULA

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

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

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

EXAMPLE

The triangle T(n,k) begins:

n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ...

0: 1

1: 1 1

2: 2 2 1

3: 3 4 3 1

4: 5 7 7 4 1

5: 8 12 14 11 5 1

6: 13 20 26 25 16 6 1

7: 21 33 46 51 41 22 7 1

8: 34 54 79 97 92 63 29 8 1

9: 55 88 133 176 189 155 92 37 9 1

10: 89 143 221 309 365 344 247 129 46 10 1

11: 144 232 364 530 674 709 591 376 175 56 11 1

12: 233 376 596 894 1204 1383 1300 967 551 231 67 12 1

13: 377 609 972 1490 2098 2587 2683 2267 1518 782 298 79 13 1

... reformatted and extended - Wolfdieter Lang, Oct 03 2014

------------------------------------------------------------------

Recurrence from Z-sequence (see a comment above): 8 = T(0,5) = (+1)*5 + (+1)*7 + (-1)*7 + (+1)*4 + (-1)*1 = 8. - Wolfdieter Lang, Oct 04 2014

MAPLE

T := (n, k) -> `if`(n=0, 1, binomial(n, k)*hypergeom([1, k/2-n/2, k/2-n/2+1/2], [k+1, -n], -4)); for n from 0 to 13 do seq(simplify(T(n, k)), k=0..n) od; # Peter Luschny, Oct 10 2014

MATHEMATICA

T[n_, k_] := Sum[Binomial[n-j, k+j], {j, 0, n}]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] (* Jean-François Alcover, Jun 11 2019 *)

PROG

(Haskell)

a105809 n k = a105809_tabl !! n !! k

a105809_row n = a105809_tabl !! n

a105809_tabl = map fst $ iterate

(\(u:_, vs) -> (vs, zipWith (+) ([u] ++ vs) (vs ++ [0]))) ([1], [1, 1])

-- Reinhard Zumkeller, Aug 15 2013

CROSSREFS

Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A109906, A111006, A114197, A162741, A228074.

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