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%I #12 Dec 25 2023 17:53:24
%S 15,27,48,85,150,264,464,815,1431,2512,4409,7738,13580,23832,41823,
%T 73395,128800,226029,396654,696080,1221536,2143647,3761839,6601568,
%U 11584945,20330162,35676948,62608680,109870575,192809419,338356944,593775045,1042002566
%N 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.
%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 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.
%H Clark Kimberling, <a href="/A227804/b227804.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) - a(n-4) (conjectured).
%F G.f.: (15 - 18 x + 12 x^2 - 8 x^3)/(1 - 3 x + 3 x^2 - 2 x^3 + x^4) (conjectured).
%e 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 > ...
%t 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 *)
%Y Cf. A001008, A002805 (numerator and denominator of harmonic numbers).
%Y Cf. A224820, A224868, A227728.
%K nonn,frac,easy
%O 1,1
%A _Clark Kimberling_, Jul 31 2013
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