A225605 a(1) = least k such that 1/3 < H(k) - 1/3; a(2) = least k such that H(a(1)) - H(3) < H(k) - H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

5, 9, 16, 29, 53, 97, 178, 327, 601, 1105, 2032, 3737, 6873, 12641, 23250, 42763, 78653, 144665, 266080, 489397, 900141, 1655617, 3045154, 5600911, 10301681, 18947745, 34850336, 64099761, 117897841, 216847937, 398845538, 733591315, 1349284789, 2481721641

Offset: 1

Author

Clark Kimberling, Aug 03 2013

Keywords

• nonn
• frac
• easy

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?

For A225605, (x,y) = (3,3); it appears that H(a(n)) - H(a(n-1)) approaches 0.60937786343... ; it is conjectured a(n)/a(n-1) approaches the constant given at A058265..

Examples

			The first two values (a(1),a(2)) = (5,9) match the beginning of the following inequality chain:
  1/3 < 1/4 + 1/5 < 1/6 + 1/7 + 1/8 + 1/9 < ...
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..300

Crossrefs

Cf. A224820, A224965, A228016.

Programs

Mathematica

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 3;
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]] (* A225605, Peter J. C. Moses, Jul 12 2013 *)

Formula

a(n) = A192804(n+4) (conjectured).

a(n) = 2*a(n-1) - a(n-4) (conjectured).

G.f.: (5 - x - 2 x^2 - 3 x^3)/(1 - 2 x + x^4) (conjectured)