A224868 a(1) = greatest k such that H(k) - H(4) < 1/3 + 1/4; a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(4); and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

7, 11, 17, 26, 39, 58, 86, 127, 187, 275, 404, 593, 870, 1276, 1871, 2743, 4021, 5894, 8639, 12662, 18558, 27199, 39863, 58423, 85624, 125489, 183914, 269540, 395031, 578947, 848489, 1243522, 1822471, 2670962, 3914486, 5736959, 8407923, 12322411, 18059372

Offset: 1

Author

Clark Kimberling, Jul 23 2013

Keywords

• nonn
• frac
• easy

Comments

Suppose that x and y are positive integers and that x <=y. Let c(1) = y and c(2) = greatest k such that H(k) - H(y) < H(y) - H(x); for n > 2, let c(n) = greatest such that H(k) - H(c(n-1)) < H(c(n-1)) - H(c(n-2)). Then 1/x + ... + 1/c(1) > 1/(c(1)+1) + ... + 1/(c(2)) > 1/(c(2)+1) + ... + 1/(c(3)) > ... The decreasing sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1) converge. For what choices of (x,y) is the sequence c(n) linearly recurrent?

For A224868, (x,y) = (3,4); it appears that the sequence a(n) is linearly recurrent with signature (2,-1,1,-1). Possibly the constant at A202537 is the limit of the sequences H(c(n))-H(c(n-1)). Possibly the constant at A092526 is the limit of c(n)/c(n-1).

Examples

			The first three values (a(1),a(2),a(3)) = (7,11,17) match the beginning of the following inequality chain (and partition of {1/m: m>=3}):
1/3+1/4 > 1/5+1/6+1/7 > 1/8+1/9+1/10+1/11 > 1/12+ ... +1/17 > ...
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A224820.

Programs

Mathematica

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 4; a[1] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = -1 + 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]] (* A224868 *)
N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 5]  (* A202537? *)
N[Table[a[n]/a[n - 1], {n, 2, z, 25}], 5]  (* A092526? *)
(* Peter J. C. Moses, Jul 23 2013 *)

Formula

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - a(n-4) (conjectured).

G.f.: (7 - 3 x + 2 x^2 - 4 x^3)/(1 - 2 x + x^2 - x^3 + x^4) (conjectured).