OFFSET
0,8
COMMENTS
The function F^{k}(x) satisfies: Sum_{k=0..m} 1/F^{k}(1) = F^{m}(1), with F^{0}(x) = 1.
The series expansion used in this sequence has a pole at x = 0, but the definition of the function F^{t}(x) gives us trivially F^{1}(0) = 1 and F^{t}(0) = F^{t-1}(1), such the series is not needed in this point.
F^{1}(x) is also known as the metallic mean function, thus we may name F^{t}(x) the iterated metallic mean function.
Values of t > 1 in F^{t}(x) can be interpreted as generalizations of metallic ratios. The ordinary generating function of 1/P(x), where P(x) is a minimal polynomial with its greatest root in F^{t}(m), is a linear recurrence of order 2^t. Such a generalized Fibonacci sequence corresponding to F^{2}(1) is A112576.
LINKS
Clark Kimberling, Polynomials associated with reciprocation, Journal of Integer Sequences 12 (2009, Article 09.3.4) 1-11.
Mathematics Stack Exchange, What is this function related with continued fractions?.
FORMULA
EXAMPLE
Triangle T(n, k) begins:
1 t t^2 t^3 t^4 t^5 t^6 t^7
[0] (1/x)^-1 1,
[1] (1/x)^ 1 0, 1,
[2] (1/x)^ 3 0, -1, -1,
[3] (1/x)^ 5 0, 3, 6, 3
[4] (1/x)^ 7 0, -14, -45, -46, -15
[5] (1/x)^ 9 0, 80, 400, 655, 440, 105
[6] (1/x)^11 0, -468, -3900, -9585, -10275, -5067, -945
[7] (1/x)^13 0, 2268, 38556, 142905, 225750, 177072, 68334, 10395
PROG
(PARI)
f(r) = (r+sqrt(r^2+4))/2;
c(n, k) = {my(p = 1/('x+O(x^(2*n+6)))); for(m = 1, k, p = f(p)); polcoef(p, 2*n-1); }
row(n) = {my(p = polinterpolate(vector(2*(n+1), k, k-1), vector(2*(n+1), k, c(n, k-1)))); Vecrev(p*n!); }
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Thomas Scheuerle, Nov 20 2025
STATUS
approved
