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
Offset: 0
Examples
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
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.
- Anton Vladimirovich Shutov, On some analogue of the Gelfond problem for Zeckendorf representations, Chebyshevskii Sbornik (2024) Vol. 25, No. 5, 195-215. See p. 215. (In Russian)
- Index entries for triangles and arrays related to Pascal's triangle
Crossrefs
Programs
-
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
-
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 *)
Formula
Riordan array (1/(1-x-x^2), x/(1-x)).
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
From Wolfdieter Lang, Feb 13 2025: (Start)
Array A(k, n) = Sum_{j=0..n} F(j+1)*binomial(k-1+n-j, k-1), k >= 0, n >= 0, with F = A000045, (from Riordan triangle k-th convolution in columns without leading 0s).
A(k, n) = F(n+1+2*k) - Sum_{j=0..k-1} F(2*(k-j)-1) * binomial(n+1+j, j), (from iteration of partial sums).
Triangle T(n, k) = A(k, n-k) = Sum_{j=k..n} F(n-j+1) * binomial(j-1, k-1), 0 <= k <= n.
T(n, k) = F(n+1+k) - Sum_{j=0..k-1} F(2*(k-j)-1) * binomial(n - (k-1-j), j). (End)
T(n, k) = A027926(n, n+k), for 0 <= k <= n. - Wolfdieter Lang, Mar 08 2025
Extensions
Use first formula as a more descriptive name, Joerg Arndt, Jun 08 2021
Comments