A227653 a(1) = least k such that 1/2 + 1/3 < H(k) - H(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.

8, 21, 54, 138, 352, 897, 2285, 5820, 14823, 37752, 96148, 244872, 623645, 1588311, 4045140, 10302237, 26237926, 66823230, 170186624, 433434405, 1103878665, 2811378360, 7160069791, 18235396608, 46442241368, 118279949136, 301237536249, 767197263003

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) = (2,3); H(a(n)) - H(a(n-1)) approaches a limit 0.9348448455..., and a(n)/a(n-1) approaches a limit 2.546818276...

Examples

			The first two values (a(1),a(2)) = (8,21) match the beginning of the following inequality chain:  1/2 + 1/3 < 1/4 + ... + 1/8 < 1/9 + ... + 1/21 < ...
		

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 = 2; 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]] (* A227653, Peter J. C. Moses, Jul 12 2013 *)

Formula

a(n) = A077849(n+1) (conjectured).

a(n) = 3*a(n-1) - a(n-2) - a(n-4) (conjectured).G.f.: (8 - 3 x - x^2 - 3 x^3)/(1 - 3 x + x^2 + x^4) (conjectured).