OFFSET
1,2
COMMENTS
Arrange the quotients F(i)/F(j) of Fibonacci numbers, for 2<=i<j<=n, in increasing order. Then the positions of the F(i)/F(n-k) are the first n-k-2 terms of the diagonal T(i,i-k), for k=0,1,2,...,n-3.
LINKS
FORMULA
From Werner Schulte, Sep 09 2024: (Start)
T(n, k) = (n^2 + (-1)^k * (n - k) + (3 + (-1)^k) / 2) / 2.
T(n, 1) = (n^2 - n + 2) / 2 = A000124(n).
T(n, 2) = (n^2 + n) / 2 = A000217(n) for n >= 2.
T(n, k) = T(n, k-2) - (-1)^k for 3 <= k <= n. (End)
G.f.: x*y*(1 + x*(y - 1) - x^4*(y - 1)*y^2 + x^5*y^3 + x^3*y*(y^2 - y - 1) - x^2*(y^2 + y - 1))/((1 - x)^3*(1 - x*y)^3*(1 + x*y)). - Stefano Spezia, Sep 10 2024
EXAMPLE
For n=6, the ordered quotients are 1/8, 1/5, 2/8, 1/3, 3/8, 2/5, 1/2, 3/5, 5/8, 2/3; the positions of 1/5, 2/5, 3/5 are 2, 6, 8 (first terms of diagonal T(i, i-1)).
Triangle starts:
1;
2, 3;
4, 6, 5;
7,10, 8, 9;
...
MATHEMATICA
nn= 12; t = Table[Range[Binomial[n, 2] + 1, Binomial[n + 1, 2]], {n, nn}]; Table[t[[n, If[OddQ@ k, Ceiling[k/2], -k/2] ]], {n, nn}, {k, n}] // Flatten (* Michael De Vlieger, Jul 02 2016 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Jul 28 2000
EXTENSIONS
Corrected and extended by Vladeta Jovovic, Oct 18 2001
STATUS
approved