A227728 a(1) = greatest k such that H(k) - H(2) < 1/1 + 1/2; a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(2); 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.

10, 43, 179, 740, 3054, 12599, 51971, 214376, 884278, 3647546, 15045706, 62061794, 255997704, 1055960840, 4355715996, 17966823308, 74111062350, 305699536774, 1260975134078, 5201376179830, 21455073484758, 88499689759294, 365050956038686, 1505792854949114

Offset: 1

Author

Clark Kimberling, Jul 29 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) = (1,2) and a(n) = c(n+1) for n >= 1. It appears that the sequence a(n) is linearly recurrent with signature (5,-4,2,-2,2,-2,2,-2,2,-2). See A227729 and A225815 for limits of the sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1).

Examples

			The first three values (a(1),a(2),a(3)) = (10,43,179) match the beginning of the following inequality chain (and partition of the harmonic numbers): 1/1 + 1/2 > 1/3 + 1/4 + ... + 1/10 > 1/11 + ... + 1/43 > 1/44 + ... + 1/179 > ...
		

Links

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

Crossrefs

Cf. A224820, A224868, A225815, A227729.

Programs

Mathematica

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 1; y = 2; 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]] (* A227728 *)
N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 50]  (* A227729 *)
N[Table[a[n]/a[n - 1], {n, 2, z, 25}], 50]  (* A225815 *)

Formula

a(n) = 5*a(n-1) - 4*a(n-2) + 2*a(n-3) - 2*a(n-4) + 2*a(n-5) - 2*a(n-6) + 2*a(n-7) - 2*a(n-8) + 2*a(n-9) - 2*a(n-10) (conjectured).

Empirical g.f.: -x*(4*x^9-4*x^8+3*x^7-4*x^6+3*x^5-4*x^4+3*x^3-4*x^2+7*x-10) / ((x-1)*(2*x^9+2*x^7+2*x^5+2*x^3+4*x-1)). - Colin Barker, Mar 23 2015

Extensions

Definition corrected by Clark Kimberling, Jan 14 2017