cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A225815 Decimal expansion of limit of c(n)/c(n-1) where c = A227728.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Jul 29 2013

Keywords

Comments

See A227728.

Examples

			4.12488401972446016179218972867267683862221288...

Crossrefs

Cf. A227728, A227729.

Programs

A227729 Decimal expansion of limit of H(c(n)) - H(c(n-1)) where c = A227728, H = harmonic number.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Jul 29 2013

Keywords

Comments

See A227728.

Examples

			1.41703790296093616771059989527013922...

Crossrefs

Cf. A227728, A225815.

Programs

A227965 a(1) = least k such that 1 + 1/2 < H(k) - H(2); a(2) = least k such that H(a(1)) - 1/2 < 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

11, 53, 249, 1164, 5435, 25371, 118428, 552798, 2580343, 12044484, 56221045, 262427666, 1224955522, 5717827134, 26689578960, 124581175389, 581517950673, 2714399875409, 12670230858892, 59141894115145, 276061555506087, 1288595564424512, 6014885070144844
Offset: 1

Views

Author

Clark Kimberling, Aug 01 2013

Keywords

Comments

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) = (1,2); H(a(n)) - H(a(n-1)) approaches a limit 1.540684... given by A227966, and a(n)/a(n-1) approaches a limit 4.6677834... given by A227967. It is unknown whether the sequence a(n) is linearly recurrent.

Examples

			The first two values (a(1),a(2)) = (11,53) match the beginning of the following inequality chain (and partition of the harmonic numbers):  1/1 + 1/2 < 1/3 + ... + 1/11 < 1/12 + ... + 1/53 < ...

Links

Crossrefs

Cf. A224820, A224868, A227728, A227966, A227967.

Programs

  • Mathematica
    z = 300; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 1; y = 2;
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]] (* A227965 *)
t = N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 60]
Last[RealDigits[t, 10]]  (* A227966 *)
t = N[Table[a[t]/a[t - 1], {t, 2, z, 50}], 60]
Last[RealDigits[t, 10]]  (* A227967 *)
(* A227965,  Peter J. C. Moses, Jul 12 2013*)

A227816 a(1) = greatest k such that H(k) - H(6) < H(6) - H(3); a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(6), 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.

Original entry on oeis.org

16, 41, 103, 257, 640, 1592, 3958, 9839, 24457, 60792, 151107, 375596, 933591, 2320556, 5768028, 14337143, 35636731, 88579473, 220175161, 547272407, 1360312788, 3381224518, 8404448844, 20890289891, 51925381404, 129066913288, 320811665802, 797416799492
Offset: 1

Views

Author

Clark Kimberling, Jul 31 2013

Keywords

Comments

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. It appears that for many choices of (x,y), the sequence c(n) is linearly recurrent with signature of length less than 10; (3,6) is not one of them. H(c(n)) - H(c(n-1)) and c(n)/c(n-1) approach limits given by A227817 and A227818.

Examples

			The first two values (a(1),a(2)) = (16,41) match the beginning of the following inequality chain (and partition of the harmonic numbers H(n) for n >= 3 ):
1/3 + 1/4 + 1/5 + 1/6 > 1/7 + ... + 1/16 < 1/17 + ... + 1/41 <  ...

Links

Crossrefs

Cf. A001008, A002805 (numerator and denominator of harmonic numbers).
Cf. A224820, A224868, A227728, A227817, A227818.

Programs

  • Mathematica
    z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 6; 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 *)

A227804 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.

Original entry on oeis.org

15, 27, 48, 85, 150, 264, 464, 815, 1431, 2512, 4409, 7738, 13580, 23832, 41823, 73395, 128800, 226029, 396654, 696080, 1221536, 2143647, 3761839, 6601568, 11584945, 20330162, 35676948, 62608680, 109870575, 192809419, 338356944, 593775045, 1042002566
Offset: 1

Views

Author

Clark Kimberling, Jul 31 2013

Keywords

Comments

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?
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.

Examples

			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 > ...

Links

Crossrefs

Cf. A001008, A002805 (numerator and denominator of harmonic numbers).
Cf. A224820, A224868, A227728.

Programs

  • Mathematica
    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 *)

Formula

a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) - a(n-4) (conjectured).
G.f.: (15 - 18 x + 12 x^2 - 8 x^3)/(1 - 3 x + 3 x^2 - 2 x^3 + x^4) (conjectured).
Showing 1-5 of 5 results.

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