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 1-8 of 8 results.

A246046 [Pi((n + Pi/2)/(Pi -1) - 1/2)]; complement of A062389.

Original entry on oeis.org

2, 3, 5, 6, 8, 9, 11, 12, 13, 15, 16, 18, 19, 21, 22, 24, 25, 27, 28, 30, 31, 33, 34, 35, 37, 38, 40, 41, 43, 44, 46, 47, 49, 50, 52, 53, 55, 56, 57, 59, 60, 62, 63, 65, 66, 68, 69, 71, 72, 74, 75, 77, 78, 79, 81, 82, 84, 85, 87, 88, 90, 91, 93, 94, 96
Offset: 1

Views

Author

Clark Kimberling, Aug 24 2014

Keywords

Comments

In general, the complement of a nonhomogenous Beatty sequence [n*r + h] is given by [n*s + h - h*s], where s = r/(r - 1).
A246046 also gives the nonnegative integers k such that tan(k) < tan(k + 1). The complementary sequence, A062389, gives the nonnegative integers k such that tan(k) > tan(k + 1).

Crossrefs

Cf. A062389.

Programs

  • Mathematica
    r = Pi; s = Pi/(Pi - 1); h = -Pi/2; z = 120;
u = Table[Floor[n*r + h], {n, 1, z}] (* A062389 *)
v = Table[Floor[n*s + h - h*s], {n, 1, z}]  (* A246046 *)

A246393 Nonnegative integers k satisfying cos(k) >= 0 and cos(k+1) <= 0.

Original entry on oeis.org

1, 7, 14, 20, 26, 32, 39, 45, 51, 58, 64, 70, 76, 83, 89, 95, 102, 108, 114, 120, 127, 133, 139, 146, 152, 158, 164, 171, 177, 183, 190, 196, 202, 208, 215, 221, 227, 234, 240, 246, 252, 259, 265, 271, 278, 284, 290, 296, 303, 309, 315, 322, 328, 334, 340
Offset: 0

Views

Author

Clark Kimberling, Aug 24 2014

Keywords

Comments

A246393 and A246394 partition A062389 (the nonhomogeneous Beatty sequence {floor((n-1/2)*Pi)}). Likewise, A246046, the complement of A062389, is partitioned by A246395 and A246396. (See the Mathematica program.)

Links

Crossrefs

Cf. A062389, A246394, A246046, A246395, A246396, A246388.

Programs

  • Mathematica
    z = 400; f[x_] := Cos[x]
Select[Range[0, z], f[#]*f[# + 1] <= 0 &]  (* A062389 *)
Select[Range[0, z], f[#] >= 0 && f[# + 1] <= 0 &]  (* A246393 *)
Select[Range[0, z], f[#] <= 0 && f[# + 1] >= 0 &]  (* A246394 *)
Select[Range[0, z], f[#]*f[# + 1] > 0 &]  (* A246046 *)
Select[Range[0, z], f[#] >= 0 && f[# + 1] >= 0 &]  (* A246395 *)
Select[Range[0, z], f[#] <= 0 && f[# + 1] <= 0 &]  (* A246396 *)

A246394 Nonnegative integers k satisfying cos(k) <= 0 and cos(k+1) >= 0.

Original entry on oeis.org

4, 10, 17, 23, 29, 36, 42, 48, 54, 61, 67, 73, 80, 86, 92, 98, 105, 111, 117, 124, 130, 136, 142, 149, 155, 161, 168, 174, 180, 186, 193, 199, 205, 212, 218, 224, 230, 237, 243, 249, 256, 262, 268, 274, 281, 287, 293, 300, 306, 312, 318, 325, 331, 337, 344
Offset: 0

Views

Author

Clark Kimberling, Aug 24 2014

Keywords

Comments

A246393 and A246394 partition A062389 (the nonhomogeneous Beatty sequence {floor((n-1/2)*Pi)}). Likewise, A246046, the complement of A062389, is partitioned by A246395 and A246396. (See the Mathematica program.)

Links

Crossrefs

Cf. A062389, A246393, A246046, A246395, A246396.

Programs

  • Mathematica
    z = 400; f[x_] := Cos[x]
Select[Range[0, z], f[#]*f[# + 1] <= 0 &]  (* A062389 *)
Select[Range[0, z], f[#] >= 0 && f[# + 1] <= 0 &]  (* A246393 *)
Select[Range[0, z], f[#] <= 0 && f[# + 1] >= 0 &]  (* A246394 *)
Select[Range[0, z], f[#]*f[# + 1] > 0 &]  (* A246046 *)
Select[Range[0, z], f[#] >= 0 && f[# + 1] >= 0 &]  (* A246395 *)
Select[Range[0, z], f[#] <= 0 && f[# + 1] <= 0 &]  (* A246396 *)

A246395 Nonnegative integers k satisfying cos(k) >= 0 and cos(k+1) >= 0.

Original entry on oeis.org

0, 5, 6, 11, 12, 13, 18, 19, 24, 25, 30, 31, 37, 38, 43, 44, 49, 50, 55, 56, 57, 62, 63, 68, 69, 74, 75, 81, 82, 87, 88, 93, 94, 99, 100, 101, 106, 107, 112, 113, 118, 119, 125, 126, 131, 132, 137, 138, 143, 144, 145, 150, 151, 156, 157, 162, 163, 169, 170
Offset: 0

Views

Author

Clark Kimberling, Aug 24 2014

Keywords

Comments

A246393 and A246394 partition A062389 (the nonhomogeneous Beatty sequence {floor((n-1/2)*Pi)}). Likewise, A246046, the complement of A062389, is partitioned by A246395 and A246396. (See the Mathematica program.)

Links

Crossrefs

Cf. A062389, A246393, A246046, A246394, A246396.

Programs

  • Mathematica
    z = 400; f[x_] := Cos[x]
Select[Range[0, z], f[#]*f[# + 1] <= 0 &]  (* A062389 *)
Select[Range[0, z], f[#] >= 0 && f[# + 1] <= 0 &]  (* A246393 *)
Select[Range[0, z], f[#] <= 0 && f[# + 1] >= 0 &]  (* A246394 *)
Select[Range[0, z], f[#]*f[# + 1] > 0 &]  (* A246046 *)
Select[Range[0, z], f[#] >= 0 && f[# + 1] >= 0 &]  (* A246395 *)
Select[Range[0, z], f[#] <= 0 && f[# + 1] <= 0 &]  (* A246396 *)

A246396 Nonnegative integers k satisfying cos(k) <= 0 and cos(k+1) <= 0.

Original entry on oeis.org

2, 3, 8, 9, 15, 16, 21, 22, 27, 28, 33, 34, 35, 40, 41, 46, 47, 52, 53, 59, 60, 65, 66, 71, 72, 77, 78, 79, 84, 85, 90, 91, 96, 97, 103, 104, 109, 110, 115, 116, 121, 122, 123, 128, 129, 134, 135, 140, 141, 147, 148, 153, 154, 159, 160, 165, 166, 167, 172
Offset: 0

Views

Author

Clark Kimberling, Aug 24 2014

Keywords

Comments

A246393 and A246394 partition A062389 (the nonhomogeneous Beatty sequence {floor((n-1/2)*Pi)}). Likewise, A246046, the complement of A062389, is partitioned by A246395 and A246396. (See the Mathematica program.)
Conjecture: every term t has at least one neighbor which is equal to t plus or minus one. - Harvey P. Dale, Jul 11 2023

Links

Crossrefs

Cf. A062389, A246393, A246046, A246394, A246395.

Programs

  • Mathematica
    z = 400; f[x_] := Cos[x]
Select[Range[0, z], f[#]*f[# + 1] <= 0 &]  (* A062389 *)
Select[Range[0, z], f[#] >= 0 && f[# + 1] <= 0 &]  (* A246393 *)
Select[Range[0, z], f[#] <= 0 && f[# + 1] >= 0 &]  (* A246394 *)
Select[Range[0, z], f[#]*f[# + 1] > 0 &]  (* A246046 *)
Select[Range[0, z], f[#] >= 0 && f[# + 1] >= 0 &]  (* A246395 *)
Select[Range[0, z], f[#] <= 0 && f[# + 1] <= 0 &]  (* A246396 *)
SequencePosition[Table[If[Cos[k]<=0,1,0],{k,200}],{1,1}][[;;,1]] (* Harvey P. Dale, Jul 11 2023 *)

A257984 Nonhomogeneous Beatty sequence: ceiling((n - 1/2)*Pi).

Original entry on oeis.org

2, 5, 8, 11, 15, 18, 21, 24, 27, 30, 33, 37, 40, 43, 46, 49, 52, 55, 59, 62, 65, 68, 71, 74, 77, 81, 84, 87, 90, 93, 96, 99, 103, 106, 109, 112, 115, 118, 121, 125, 128, 131, 134, 137, 140, 143, 147, 150, 153, 156, 159, 162, 165, 169, 172, 175, 178, 181, 184
Offset: 1

Views

Author

Clark Kimberling, Jun 15 2015

Keywords

Comments

Let r = Pi, s = r/(r-1), and t = 1/2. Let R be the ordered set {floor[(n + t)*r] : n is an integer} and let S be the ordered set {floor[(n - t)*s : n is an integer}; thus,
R = (..., -10, -9, -7, -6, -4, -3, -1, 0, 2, 3, 5, 6, 8, ...);
S = (..., -15, -11, -8, -5, -2, 1, 4, 7, 10, 14, 17, 20, ...).
By Fraenkel's theorem (Theorem XI in the cited paper); R and S partition the integers.
R is the set of integers n such that (cos n)*(cos(n + 1)) < 0;
S is the set of integers n such that (cos n)*(cos(n + 1)) > 0.
A246046 = (2,3,5,6,8,...), positive terms of R;
A062389 = (1,4,7,10,14,17,...), positive terms of S;
A258048 = (1,3,4,6,7,9,10,...), - (negative terms of R);
A257984 = (2,5,8,11,15,...), - (negative terms of S).
A062389 and A246046 partition the positive integers, and A258048 and A257984 partition the positive integers.

Links

Crossrefs

Cf. A258048 (complement), A246046, A062380, A258833.

Programs

  • Mathematica
    Table[Ceiling[(n - 1/2) Pi], {n, 1, 120}] (* A257984 *)
Table[Ceiling[(n + 1/2) Pi/(Pi - 1)], {n, 0, 120}]  (* A258048 *)

Formula

a(n) = ceiling((n - 1/2)*Pi).

A163584 Number of singularities of tan(x) in integer intervals starting with (0,1).

Original entry on oeis.org

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

Views

Author

A. Timothy Royappa, Jul 31 2009

Keywords

Examples

			a(0) = 0 because tan(x) has no singularities in the interval (0,1).
a(1) = 1 because tan(x) has one singularity in the interval (1,2).

Crossrefs

Cf. A163581.

Formula

For n>0, a(A062389(n)) = 1. - Michel Marcus, Aug 07 2013

Extensions

More terms from Michel Marcus, Aug 07 2013

A214628 Intersections of radii with the cycloid.

Original entry on oeis.org

2, 2, 2, 2, 4, 4, 4, 6, 6, 6, 8, 8, 8, 8, 10, 10, 10, 12, 12, 12, 14, 14, 14, 16, 16, 16, 18, 18, 18, 20, 20, 20, 22, 22, 22, 22, 24, 24, 24, 26, 26, 26, 28, 28, 28, 30, 30, 30, 32, 32, 32, 34, 34, 34, 36, 36, 36, 36, 38, 38, 38, 40, 40, 40
Offset: 1

Views

Author

Gordon Roesler, Jul 23 2012

Keywords

Comments

Number of times the line y=x/n intersects the cycloid specified by x=t-sin(t), y=1-cos(t) or, by symmetry, number of times the line y=n*x intersects the cycloid specified by x=1-cos(t), y=t-sin(t). It is equal to twice the number of arches that are intersected by the lines (2 intersection points by arch).
To find this sequence one can look for the slopes of the tangents to the n-th arch when these tangents pass through the origin (see PARI script). If one consider the indices where a(n) change value, one gets: 1, 4, 7, 10, 14, 17, 20, 23, 26, ... that may well be A062389, as this is the slope of the line joining the origin to the summit of the n-th arch. Will this be true for all n? - Michel Marcus, Aug 29 2013

Examples

			For n=1..4, a(n)=2; for n=5..7, a(n)=4.

Programs

  • PARI
    slop(n) = {ang = 2*n*Pi; val = solve(x=ang + Pi/100, ang + Pi, 2 - 2*cos(x) - x*sin(x)); vinvn = floor((1 - cos(val))/sin(val));}
lista(nn) = {nbc = 0; nbi = 1; for (i=1, nn, nnbc = slop(i); for (j = 1, nnbc - nbc, print1(2*nbi, ", ")); nbi++; nbc = nnbc;);} \\ Michel Marcus, Aug 29 2013

Extensions

More terms from Michel Marcus, Aug 29 2013
Showing 1-8 of 8 results.

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