A228016 -id:A228016 - cp's OEIS frontend

A225605 a(1) = least k such that 1/3 < H(k) - 1/3; a(2) = least k such that H(a(1)) - H(3) < H(k) - H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

5, 9, 16, 29, 53, 97, 178, 327, 601, 1105, 2032, 3737, 6873, 12641, 23250, 42763, 78653, 144665, 266080, 489397, 900141, 1655617, 3045154, 5600911, 10301681, 18947745, 34850336, 64099761, 117897841, 216847937, 398845538, 733591315, 1349284789, 2481721641

Offset: 1

Clark Kimberling, Aug 03 2013

Suppose that x and y are positive integers and that x <=y. Let a(1) = least k such that H(y) - H(x-1) < 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(n-1)) - H(a(n-2)) < H(k) - H(a(n-1)). The increasing sequences H(a(n)) - H(a(n-1)) and a(n)/a(n-1) converge. For what choices of (x,y) is the sequence a(n) linearly recurrent?

For A225605, (x,y) = (3,3); it appears that H(a(n)) - H(a(n-1)) approaches 0.60937786343... ; it is conjectured a(n)/a(n-1) approaches the constant given at A058265..

			The first two values (a(1),a(2)) = (5,9) match the beginning of the following inequality chain:
  1/3 < 1/4 + 1/5 < 1/6 + 1/7 + 1/8 + 1/9 < ...

Clark Kimberling, Table of n, a(n) for n = 1..300

Cf. A224820, A224965, A228016.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 3;
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}];
m = Map[a, Range[z]] (* A225605, Peter J. C. Moses, Jul 12 2013 *)

a(n) = A192804(n+4)  (conjectured).
a(n) = 2*a(n-1) - a(n-4)  (conjectured).
G.f.: (5 - x - 2 x^2 - 3 x^3)/(1 - 2 x + x^4)  (conjectured)

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(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, Aug 03 2013

For A228025, (x,y) = (2,5); it appears that H(a(n)) - H(a(n-1)) approaches log(2 + sqrt(3)) and that and a(n)/a(n-1) approaches sqrt(3).

			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 < ...

Clark Kimberling, Table of n, a(n) for n = 1..100

Cf. A224820, A224965, A228016.

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}];
m = Map[a, Range[z]] (* A227653, Peter J. C. Moses, Jul 12 2013 *)

a(n) = A061278(n+1) (conjectured).
a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) (conjectured).
G.f.: (-20 + 24 x - 5 x^2)/(-1 + 5 x - 5 x^2 + x^3) (conjectured).

cp's OEIS Frontend

A225605 a(1) = least k such that 1/3 < H(k) - 1/3; a(2) = least k such that H(a(1)) - H(3) < H(k) - H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

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