1,1

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?

For A227965, (x,y) = (1,5); H(a(n)) - H(a(n-1)) approaches a limit 2.29243166956117768780078... and a(n)/a(n-1) approaches a limit 0.8989794855663561963945681494117

Clark Kimberling, Table of n, a(n) for n = 1..100

a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) (conjectured).

G.f.: (-54 + 55 x - 5 x^2)/(-1 + 11 x - 11 x^2 + x^3) (conjectured)

The first two values (a(1),a(2)) = (54,539) match the beginning of the following inequality chain: 1/1+1/2+1/3+1/4+1/5 < 1/6+...+1/54 < 1/55+...+1/539 < ...

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 1; y = 5;

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]] (* A227653, Peter J. C. Moses, Jul 12 2013 *)

Cf. A224820, A224965.

Sequence in context: A238908 A233364 A222968 * A245833 A254701 A376260

Adjacent sequences: A228013 A228014 A228015 * A228017 A228018 A228019

nonn,frac,easy

Clark Kimberling, Aug 02 2013

approved