OFFSET
1,1
COMMENTS
Suppose that x and y are positive integers and that x <=y. Let c(1) = y and c(2) = greatest k such that H(k) - H(y) < H(y) - H(x); for n > 2, let c(n) = greatest such that H(k) - H(c(n-1)) < H(c(n-1)) - H(c(n-2)). Then 1/x + ... + 1/c(1) > 1/(c(1)+1) + ... + 1/(c(2)) > 1/(c(2)+1) + ... + 1/(c(3)) > ... The decreasing sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1) converge. For what choices of (x,y) is the sequence c(n) linearly recurrent?
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
FORMULA
a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) - a(n-4) (conjectured).
G.f.: (15 - 18 x + 12 x^2 - 8 x^3)/(1 - 3 x + 3 x^2 - 2 x^3 + x^4) (conjectured).
EXAMPLE
The first three values (a(1),a(2),a(3)) = (10,43,179) match the beginning of the following inequality chain (and partition of the harmonic numbers H(n) for n >= 5 ): 1/5 + 1/6 + 1/7 + 1/8 > 1/9 + ... + 1/15 < 1/16 + ... + 1/27 < 1/28 + ... + 1/48 > ...
MATHEMATICA
z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 5; a[1] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}]; m = Map[a, Range[z]] (* A227804, Peter J. C. Moses, Jul 23 2013 *)
CROSSREFS
KEYWORD
nonn,frac,easy
AUTHOR
Clark Kimberling, Jul 31 2013
STATUS
approved