A227816 - OEIS

%I #16 Dec 25 2023 17:52:46

%S 16,41,103,257,640,1592,3958,9839,24457,60792,151107,375596,933591,

%T 2320556,5768028,14337143,35636731,88579473,220175161,547272407,

%U 1360312788,3381224518,8404448844,20890289891,51925381404,129066913288,320811665802,797416799492

%N a(1) = greatest k such that H(k) - H(6) < H(6) - H(3); a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(6), 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.

%C 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.  It appears that for many choices of (x,y), the sequence c(n) is linearly recurrent with signature of length less than 10; (3,6) is not one of them.  H(c(n)) - H(c(n-1)) and c(n)/c(n-1) approach limits given by A227817 and A227818.

%H Clark Kimberling, <a href="/A227816/b227816.txt">Table of n, a(n) for n = 1..1000</a>

%e The first two values (a(1),a(2)) = (16,41) match the beginning of the following inequality chain (and partition of the harmonic numbers H(n) for n >= 3 ):

%e 1/3 + 1/4 + 1/5 + 1/6 > 1/7 + ... + 1/16 < 1/17 + ... + 1/41 <  ...

%t z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 6; 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 *)

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

%Y Cf. A224820, A224868, A227728, A227817, A227818.

%K nonn,frac,easy

%O 1,1

%A _Clark Kimberling_, Jul 31 2013