A228025 a(1) = least k such that 1/2+1/3+1/4+1/5 < H(k) - H(5); a(2) = least k such that H(a(1)) - H(5) < 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.

20, 76, 285, 1065, 3976, 14840, 55385, 206701, 771420, 2878980, 10744501, 40099025, 149651600, 558507376, 2084377905, 7779004245, 29031639076, 108347552060, 404358569165, 1509086724601, 5631988329240, 21018866592360, 78443478040201, 292755045568445

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 A228025, (x,y) = (2,5); it appears that H(a(n)) - H(a(n-1)) approaches log(2 + sqrt(3)) and that and a(n)/a(n-1) approaches sqrt(3).

Examples

			The first two values (a(1),a(2)) = (20,76) match the beginning of the following inequality chain:  1/2+1/3+1/4+1/5 < 1/6+...+1/20 < 1/21+...+1/76 < ...
		

Links

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

Crossrefs

Cf. A224820, A224965, A228016.

Programs

Mathematica

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 2; 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 *)

Formula

a(n) = A061278(n+1) (conjectured).

a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) (conjectured).

G.f.: (-20 + 24 x - 5 x^2)/(-1 + 5 x - 5 x^2 + x^3) (conjectured).