A227804 a(1) = greatest k such that H(k) - H(8) < H(8) - H(4); a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(8), 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.

15, 27, 48, 85, 150, 264, 464, 815, 1431, 2512, 4409, 7738, 13580, 23832, 41823, 73395, 128800, 226029, 396654, 696080, 1221536, 2143647, 3761839, 6601568, 11584945, 20330162, 35676948, 62608680, 109870575, 192809419, 338356944, 593775045, 1042002566

Offset: 1

Author

Clark Kimberling, Jul 31 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 A227804, (x,y) = (5,8); it appears that the sequence a(n) is linearly recurrent with signature (3,-3,2,-1), that H(c(n)) - H(c(n-1)) approaches a limit 0.56239..., and that c(n)/c(n-1) approaches the constant 1.75487... given at A109134.

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 H(n) for n >= 5 ):  1/5 + 1/6 + 1/7 + 1/8 > 1/9 + ... + 1/15 < 1/16 + ... + 1/27 < 1/28 + ... + 1/48 > ...
		

Links

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

Crossrefs

Cf. A001008, A002805 (numerator and denominator of harmonic numbers).

Cf. A224820, A224868, A227728.

Programs

Mathematica

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 5; 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]] (* A227804, Peter J. C. Moses, Jul 23 2013 *)

Formula

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

G.f.: (15 - 18 x + 12 x^2 - 8 x^3)/(1 - 3 x + 3 x^2 - 2 x^3 + x^4) (conjectured).