

A228025


a(1) = least k such that 1/2+1/3+1/4+1/5 < H(k)  H(5); a(2) = least k such that H(a(1))  H(5) < H(k) H(a(1)), and for n > 2, a(n) = least k such that H(a(n1))  H(a(n2)) > H(k)  H(a(n1)), where H = harmonic number.


1



20, 76, 285, 1065, 3976, 14840, 55385, 206701, 771420, 2878980, 10744501, 40099025, 149651600, 558507376, 2084377905, 7779004245, 29031639076, 108347552060, 404358569165, 1509086724601, 5631988329240, 21018866592360, 78443478040201, 292755045568445
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OFFSET

1,1


COMMENTS

Suppose that x and y are positive integers and that x <=y. Let a(1) = least k such that H(y)  H(x1) < H(k)  H(y); let a(2) = least k such that H(a(1))  H(y) < H(k)  H(a(1)); and for n > 2, let a(n) = least k such that greatest such H(a(n1))  H(a(n2)) < H(k)  H(a(n1)). The increasing sequences H(a(n))  H(a(n1)) and a(n)/a(n1) converge. For what choices of (x,y) is the sequence a(n) linearly recurrent?
For A228025, (x,y) = (2,5); it appears that H(a(n))  H(a(n1)) approaches log(2 + sqrt(3)) and that and a(n)/a(n1) approaches sqrt(3).


LINKS



FORMULA

a(n) = 5*a(n1)  5*a(n2) + a(n3) (conjectured).
G.f.: (20 + 24 x  5 x^2)/(1 + 5 x  5 x^2 + x^3) (conjectured)


EXAMPLE

The first two values (a(1),a(2)) = (20,76) match the beginning of the following inequality chain: 1/2+1/3+1/4+1/5 < 1/6+...+1/20 < 1/21+...+1/76 < ...


MATHEMATICA

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 2; y = 5;
a[1] = Ceiling[w /. FindRoot[h[w] == 2 h[y]  h[x  1], {w, 1}, WorkingPrecision > 400]]; a[2] = Ceiling[w /. FindRoot[h[w] == 2 h[a[1]]  h[y], {w, a[1]}, WorkingPrecision > 400]]; Do[s = 0; a[t] = Ceiling[w /. FindRoot[h[w] == 2 h[a[t  1]]  h[a[t  2]], {w, a[t  1]}, WorkingPrecision > 400]], {t, 3, z}];


CROSSREFS



KEYWORD

nonn,frac,easy


AUTHOR



STATUS

approved



