OFFSET
1,1
COMMENTS
Suppose that x and y are positive integers and that x <=y. Let a(1) = least k such that H(y) - H(x-1) < H(k) - H(y); let a(2) = least k such that H(a(1)) - H(y) < H(k) - H(a(1)); and for n > 2, let a(n) = least k such that greatest such H(a(n-1)) - H(a(n-2)) < H(k) - H(a(n-1)). The increasing sequences H(a(n)) - H(a(n-1)) and a(n)/a(n-1) converge. For what choices of (x,y) is the sequence a(n) linearly recurrent?
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..100
EXAMPLE
The first two values (a(1),a(2)) = (11,53) match the beginning of the following inequality chain (and partition of the harmonic numbers): 1/1 + 1/2 < 1/3 + ... + 1/11 < 1/12 + ... + 1/53 < ...
MATHEMATICA
z = 300; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 1; y = 2;
a[1] = Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = 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]] (* A227965 *)
t = N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 60]
Last[RealDigits[t, 10]] (* A227966 *)
t = N[Table[a[t]/a[t - 1], {t, 2, z, 50}], 60]
Last[RealDigits[t, 10]] (* A227967 *)
(* A227965, Peter J. C. Moses, Jul 12 2013*)
CROSSREFS
KEYWORD
nonn,frac,easy
AUTHOR
Clark Kimberling, Aug 01 2013
STATUS
approved