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 1-9 of 9 results.

A102522 Consider A100060 as a binary fraction: this is its decimal fractional equivalent.

Original entry on oeis.org

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

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Author

Keywords

Examples

			0.6617104154252787297744583315053700790798944281724706799325...
		

Crossrefs

Cf. A100060.

Programs

  • Mathematica
    zz = { (* the list of values in the link *) }; yy = Drop[zz, 1] - Drop[zz, -1]; xx = Drop[yy, 1] - Drop[yy, -1]; RealDigits[ FromDigits[ {Join[{1}, Table[If[xx[[n]] > 0, 1, 0], {n, 370}]], 0}, 2], 10, 111][[1]]

A102523 Lengths of runs in A100060.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 3, 1, 2, 1, 1, 1, 1, 2, 1, 3, 2, 1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 3, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 3, 1, 3, 1, 1, 2, 1, 2
Offset: 1

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Author

Keywords

Comments

The first case of exactly n consecutive ones occurs in A100060 at 1, 10, 16, 688, 1246, 29237, 13898, ....
The first case of exactly n consecutive zeros occurs in A100060 at 2, 6, 29, 276, 1529, 34277, ....

Crossrefs

Programs

  • Mathematica
    zz = { (* the list of values in the link *) }; yy = Drop[zz, 1] - Drop[zz, -1]; xx = Drop[yy, 1] - Drop[yy, -1]; Length /@ Split[Join[{1}, Table[ If[xx[[n]] > 0, 1, 0], {n, 145}]]]

A106399 Binomial transform of A100060.

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 31, 58, 107, 200, 387, 782, 1640, 3512, 7542, 16020, 33406, 68257, 136971, 271341, 534302, 1053441, 2092840, 4206655, 8564397, 17631551, 36575711, 76137143, 158427407, 328537434, 677598776, 1388303692, 2824261172, 5704560206, 11443336382
Offset: 1

Views

Author

Gary W. Adamson, May 01 2005

Keywords

Comments

1 or 0 are assigned to increase or decrease in magnitude of a succession of terms in the first difference row of Zeta function heights, as recorded in A100060.

Examples

			a(5) = 8 = 1*1 + 4*0 + 6*1 + 4*0 + 1*1.
		

Crossrefs

Cf. A100060.

Programs

  • Mathematica
    nmax = 40; A100060 = (Sign[Differences[Im[ZetaZero[Range[nmax+2]]], 2]] + 1)/2; Flatten[{1, Table[1 + Sum[Binomial[n, k]*A100060[[k]], {k, 1, n}], {n, 1, nmax}]}] (* Vaclav Kotesovec, Nov 28 2017 *)

Formula

Use the binomial transform operation (bto) on A100060, i.e. (bto): [1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1,

Extensions

a(11) corrected by Vaclav Kotesovec, Nov 28 2017
More terms from Vaclav Kotesovec, Nov 28 2017

A254297 Consider the nontrivial zeros of the Riemann zeta function on the critical line 1/2 + i*t and the gap, or first difference, between two consecutive such zeros; a(n) is the lesser of the two zeros at a place where the gap attains a new minimum.

Original entry on oeis.org

1, 2, 3, 5, 8, 10, 14, 20, 25, 28, 35, 64, 72, 92, 136, 160, 187, 213, 299, 316, 364, 454, 694, 923, 1497, 3778, 4766, 6710, 18860, 44556, 73998, 82553, 87762, 95249, 354770, 415588, 420892, 1115579, 8546951
Offset: 1

Views

Author

Robert G. Wilson v, Jan 27 2015

Keywords

Comments

Since all zeros are assumed to be on the critical line, the gap, or first difference, between two consecutive zeros is measured as the difference between the two imaginary parts.
Inspired by A002410.
No other terms < 10000000. The minimum gap so far is 0.002323...

Examples

			a(1)=1 since the first Riemann zeta zero, 1/2 + i*14.13472514... (A058303) has no previous zero, so its gap is measured from 0.
a(2)=2 since the second Riemann zeta zero, 1/2 + i*21.02203964... (A065434) has a gap of 6.887314497... which is less than the previous gap of ~14.13472514.
a(3)=3 since the third Riemann zeta zero, 1/2 + i*25.01085758... (A065452) has a gap of 3.988817941... which is less than ~6.887314497.
The fourth Riemann zeta zero, 1/2 + i*30.42487613... (A065453) has a gap of 5.414018546... which is not less than ~6.887314497 and therefore is not in the sequence.
a(4)=5 since the fifth Riemann zeta zero, 1/2 + i*32.93506159... (A192492) has a gap of 2.510185462... which is less than ~3.988817941.
a(5)=8 since the eighth Riemann zeta zero, 1/2 + i*43.32707328...  has a gap of 2.408354269... which is less than ~2.510185462.
		

Crossrefs

Programs

  • Mathematica
    k = 1; mn = Infinity; y = 0; lst = {}; While[k < 10001, z = N[ Im@ ZetaZero@ k, 64]; If[z - y < mn, mn = z - y; AppendTo[lst, k]]; y = z; k++]; lst

Formula

a(n) = A326502(n) + 1. - Artur Jasinski, Oct 24 2019

Extensions

a(38) from Arkadiusz Wesolowski, Nov 08 2015
a(39) from Artur Jasinski, Oct 24 2019

A123504 Sequence generated from the first nontrivial zero of the Riemann zeta function.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0
Offset: 2

Views

Author

Gary W. Adamson, Oct 01 2006

Keywords

Comments

A123505 records the lengths of runs. A123506 uses the second zero.

Examples

			a(8) = 1 since (1/8)^z = (0.353553..., angle 115.943... degrees).
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Boole[Arg[1/n^ZetaZero[1]] > 0]; Array[a, 100, 2] (* Amiram Eldar, May 31 2025 *)
  • PARI
    t=1/2+solve(y=14,15,imag(zeta(1/2+y*I)))*I;
    a(n)=arg(n^-t)>0 \\ Charles R Greathouse IV, Mar 10 2016

Formula

Extract argument from (1/n)^z, z = (1/2 + i*14.1347251417...). a(n) = 1 if the argument is between 0 and 180 degrees, and = 0 if otherwise (n = 2, 3, 4, ...).

Extensions

More terms from Amiram Eldar, May 31 2025

A123505 Lengths of bit runs in A123504.

Original entry on oeis.org

1, 2, 1, 2, 2, 2, 3, 3, 5, 6, 7, 8, 11, 14, 17, 21, 26, 34, 41, 51, 65, 80, 101, 125, 157, 196, 245, 305, 381, 477, 595, 743, 927, 1159, 1448, 1807, 2258, 2819, 3521, 4397, 5492, 6859, 8565, 10698, 13361, 16685, 20839, 26026, 32503, 40593, 50697, 63315, 79074
Offset: 2

Views

Author

Gary W. Adamson, Oct 01 2006

Keywords

Comments

A123507 records the bit runs of A123506 and uses the second zero in an analogous operation.
Record the numbers of consecutive bit runs of A123504, see example.

Examples

			A123504 = 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1...; therefore the numbers of bit runs are 1, 2, 1, 2, 2, 2, 3, ...
		

Crossrefs

Programs

  • Mathematica
    Length /@ Split[Table[Boole[Arg[1/n^ZetaZero[1]] > 0], {n, 2, 10^6}]] (* Amiram Eldar, May 31 2025 *)

Extensions

More terms from Amiram Eldar, May 31 2025

A123507 Lengths of bit runs in A123506.

Original entry on oeis.org

1, 2, 1, 3, 1, 1, 2, 2, 3, 2, 4, 3, 5, 5, 5, 7, 8, 9, 11, 12, 14, 17, 19, 22, 26, 31, 34, 41, 47, 55, 64, 73, 86, 100, 115, 135, 156, 181, 210, 244, 283, 329, 383, 443, 516, 598, 695, 807, 936, 1088, 1263, 1467, 1703, 1978, 2297, 2666, 3097, 3595, 4176, 4848, 5630
Offset: 2

Views

Author

Gary W. Adamson, Oct 01 2006

Keywords

Comments

The sequence uses operations based on the second nontrivial Riemann zero: (1/2 + i*t), t = 21.022039639... A123504 and A123505 use the first nontrivial zero.
Record the numbers of consecutive bit runs of A123506, see example.

Examples

			a(4) = 3 since A123506 = 0, 1, 1, 0, 1, 1, 1, ...
		

References

  • John Derbyshire, Prime Obsession, Bernhard Riemann and the Greatest Unsolved Problem in Mathematics, Plume - a Penguin Group, NY, 2003, pp. 198-199.

Crossrefs

Programs

  • Mathematica
    Length /@ Split[Table[Boole[Arg[1/n^ZetaZero[2]] > 0], {n, 2, 10^6}]] (* Amiram Eldar, May 31 2025 *)

Extensions

More terms from Amiram Eldar, May 31 2025

A102524 Continued fraction expansion of A102522.

Original entry on oeis.org

0, 1, 1, 1, 21, 1, 3, 36, 3, 1, 1, 6, 2, 1, 1, 1, 1, 3, 1, 44, 1, 3, 3, 4, 13, 1, 4, 7, 1, 1, 3, 1, 3, 1, 4, 2, 1, 1, 1, 17, 5, 1, 9, 1, 1, 6, 1, 6, 12, 3, 5, 1, 1, 8, 1, 3, 30, 1, 4, 4, 1, 2, 2, 7, 1, 7, 1, 16, 7, 8, 7, 1, 51, 1, 1, 1, 240, 7, 1, 2, 1, 1, 1, 7, 4, 1, 10, 19, 3, 1, 6, 1, 22, 1, 6, 1, 1, 2, 5
Offset: 0

Views

Author

Keywords

Comments

Increasing PQ's are: 1,21,36,44,51,240,298,999,2004,18156,130055, ...

Crossrefs

Cf. A100060, A102522 (decimal expansion).

Programs

  • Mathematica
    ContinuedFraction[ FromDigits[ {Join[{1}, Table[ If[ xx[[n]] > 0, 1, 0], {n, 1000}]], 0}, 2]]

Extensions

Offset changed by Andrew Howroyd, Aug 04 2024

A125642 Divide the circle into ten "decants" (each of 36 degrees). Let z = 1/2 + i*14.134725142... be the first nontrivial zero of the Riemann zeta function. Then a(n) is the decant containing the argument of 1/n^z.

Original entry on oeis.org

1, 5, -5, -2, 4, -1, -4, 4, 1, -2, -4, 5, 3, 1, -1, -3, -4, 5, 4, 3, 2, 1, -1, -2, -3, -4, -5, -5, 5, 4, 3, 3, 2, 1, 1, -1, -2, -2, -3, -3, -4, -5, -5, 5, 5, 4, 4, 3, 3, 2
Offset: 1

Views

Author

Gary W. Adamson, Nov 28 2006

Keywords

Examples

			a(5) = 4 since 1/4^z = has angle 136.58045... and the argument is between 108 and 144 deg., which is the 4th decant.
		

Crossrefs

Cf. A100060.

Formula

Given the first Riemann nontrivial zero, z = (1/2 + i*14.134725142...), extract the argument of 1/n^z (in polar coordinates) and map it on a unit circle by decants: (0 to 36 deg. = 1), (36 to 72 deg. = 2), (72 to 108 deg. = 3), (108 to 144 deg. = 4), (144 to 180 deg. = 5), (0 to -36 deg. = -1), (-36 to -72 deg. = -2), (-72 to -108 deg. = -3), (-108 to -144 deg. = -4), (-144 to -180 deg. = -5).

Extensions

Edited by N. J. A. Sloane, Aug 10 2019
Showing 1-9 of 9 results.