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%I #14 Dec 25 2023 17:58:02
%S 7,11,17,26,39,58,86,127,187,275,404,593,870,1276,1871,2743,4021,5894,
%T 8639,12662,18558,27199,39863,58423,85624,125489,183914,269540,395031,
%U 578947,848489,1243522,1822471,2670962,3914486,5736959,8407923,12322411,18059372
%N 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.
%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. For what choices of (x,y) is the sequence c(n) linearly recurrent?
%C 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).
%H Clark Kimberling, <a href="/A224868/b224868.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = 2*a(n-1) - a(n-2) + a(n-3) - a(n-4) (conjectured).
%F G.f.: (7 - 3 x + 2 x^2 - 4 x^3)/(1 - 2 x + x^2 - x^3 + x^4) (conjectured).
%e 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}):
%e 1/3+1/4 > 1/5+1/6+1/7 > 1/8+1/9+1/10+1/11 > 1/12+ ... +1/17 > ...
%t 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 *)
%t N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 5] (* A202537? *)
%t N[Table[a[n]/a[n - 1], {n, 2, z, 25}], 5] (* A092526? *)
%t (* _Peter J. C. Moses_, Jul 23 2013 *)
%Y Cf. A224820.
%K nonn,frac,easy
%O 1,1
%A _Clark Kimberling_, Jul 23 2013
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