A228016 a(1) = least k such that 1/1+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.

54, 539, 5340, 52865, 523314, 5180279, 51279480, 507614525, 5024865774, 49741043219, 492385566420, 4874114620985, 48248760643434, 477613491813359, 4727886157490160, 46801248083088245, 463284594673392294, 4586044698650834699, 45397162391834954700

Offset: 1

Author

Clark Kimberling, Aug 02 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 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

Examples

			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 < ...
		

Links

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

Crossrefs

Cf. A224820, A224965.

Programs

Mathematica

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 *)

Formula

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)