A227965 -id:A227965 - cp's OEIS frontend

A227966 Decimal expansion of limit of H(c(n)) - H(c(n-1)), where c = A227965 and H = harmonic number.

1, 5, 4, 0, 6, 8, 4, 3, 2, 6, 4, 8, 6, 0, 7, 1, 6, 4, 5, 0, 6, 1, 5, 6, 1, 5, 0, 7, 2, 9, 3, 9, 2, 0, 5, 2, 0, 1, 8, 7, 5, 2, 5, 9, 6, 7, 2, 6, 2, 2, 8, 6, 5, 7, 3, 5, 4, 4, 0, 0

Offset: 1

Clark Kimberling, Aug 01 2013

See A227965.

			1.54068432648607164506156150729392052018...

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

Cf. A227965, A227967.

Mathematica
```
(* See A227965. *)
```

A227967 Decimal expansion of limit of c(n)/c(n-1), where c = A227965.

Original entry on oeis.org

4, 6, 6, 7, 7, 8, 3, 4, 6, 6, 1, 2, 6, 6, 2, 4, 2, 5, 6, 1, 1, 8, 8, 1, 6, 9, 9, 5, 1, 7, 1, 1, 5, 5, 8, 3, 1, 6, 0, 3, 8, 3, 4, 9, 7, 2, 4, 8, 6, 8, 7, 3, 3, 3, 9, 4, 5, 7, 0, 4

Offset: 1

Clark Kimberling, Aug 01 2013

See A227965.

			4.6677834661266242561188169951711558316038349724...

Cf. A227965, A227966.

Mathematica
```
(* See A227965. *)
```

A227653 a(1) = least k such that 1/2 + 1/3 < H(k) - H(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.

Original entry on oeis.org

8, 21, 54, 138, 352, 897, 2285, 5820, 14823, 37752, 96148, 244872, 623645, 1588311, 4045140, 10302237, 26237926, 66823230, 170186624, 433434405, 1103878665, 2811378360, 7160069791, 18235396608, 46442241368, 118279949136, 301237536249, 767197263003

Offset: 1

Clark Kimberling, Aug 02 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 A227965, (x,y) = (2,3); H(a(n)) - H(a(n-1)) approaches a limit 0.9348448455..., and a(n)/a(n-1) approaches a limit 2.546818276...

			The first two values (a(1),a(2)) = (8,21) match the beginning of the following inequality chain:  1/2 + 1/3 < 1/4 + ... + 1/8 < 1/9 + ... + 1/21 < ...

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

Cf. A224820, A224965.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 2; 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]] (* A227653, Peter J. C. Moses, Jul 12 2013 *)

a(n) = A077849(n+1) (conjectured).

a(n) = 3*a(n-1) - a(n-2) - a(n-4) (conjectured).G.f.: (8 - 3 x - x^2 - 3 x^3)/(1 - 3 x + x^2 + x^4) (conjectured).

A228016 a(1) = least k such that 1/1+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

54, 539, 5340, 52865, 523314, 5180279, 51279480, 507614525, 5024865774, 49741043219, 492385566420, 4874114620985, 48248760643434, 477613491813359, 4727886157490160, 46801248083088245, 463284594673392294, 4586044698650834699, 45397162391834954700

Offset: 1

Clark Kimberling, Aug 02 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 A227965, (x,y) = (1,5); H(a(n)) - H(a(n-1)) approaches a limit 2.29243166956117768780078... and a(n)/a(n-1) approaches a limit 0.8989794855663561963945681494117

			The first two values (a(1),a(2)) = (54,539) match the beginning of the following inequality chain:  1/1+1/2+1/3+1/4+1/5 < 1/6+...+1/54 < 1/55+...+1/539 < ...

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

Cf. A224820, A224965.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 1; 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) = 11*a(n-1) - 11*a(n-2) + a(n-3) (conjectured).

G.f.: (-54 + 55 x - 5 x^2)/(-1 + 11 x - 11 x^2 + x^3) (conjectured)

cp's OEIS Frontend

A227966 Decimal expansion of limit of H(c(n)) - H(c(n-1)), where c = A227965 and H = harmonic number.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A227967 Decimal expansion of limit of c(n)/c(n-1), where c = A227965.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A227653 a(1) = least k such that 1/2 + 1/3 < H(k) - H(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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A228016 a(1) = least k such that 1/1+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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula