A056576 -id:A056576 - cp's OEIS frontend

A206807 Position of 3^n when {2^j} and {3^k} are jointly ranked; complement of A206805.

2, 5, 7, 10, 12, 15, 18, 20, 23, 25, 28, 31, 33, 36, 38, 41, 43, 46, 49, 51, 54, 56, 59, 62, 64, 67, 69, 72, 74, 77, 80, 82, 85, 87, 90, 93, 95, 98, 100, 103, 105, 108, 111, 113, 116, 118, 121, 124, 126, 129, 131, 134, 137, 139, 142, 144, 147, 149, 152, 155

Offset: 1

Clark Kimberling, Feb 16 2012

nonn

The joint ranking is for j >= 1 and k >= 1, so that the sets {2^j} and {3^k} are disjoint.

			The joint ranking begins with 2,3,4,8,9,16,27,32,64,81,128,243,256, so that
A206805 = (1,3,4,6,8,9,11,13,...)
A206807 = (2,5,7,10,12,...)

Cf. A006899, A056576, A098294, A206805, A122437.

f[n_] := 2^n; g[n_] := 3^n; z = 200;
c = Table[f[n], {n, 1, z}]; s = Table[g[n], {n, 1, z}];
j = Sort[Union[c, s]];
p[n_] := Position[j, f[n]]; q[n_] := Position[j, g[n]];
Flatten[Table[p[n], {n, 1, z}]]           (* A206805 *)
Table[n + Floor[n*Log[3, 2]], {n, 1, 50}] (* A206805 *)
Flatten[Table[q[n], {n, 1, z}]]           (* this sequence *)
Table[n + Floor[n*Log[2, 3]], {n, 1, 50}] (* this sequence as a table *)

a(n) = logint(3^n, 2) + n; \\ Ruud H.G. van Tol, Dec 10 2023

a(n) = n + floor(n*log_2(3)).
A206805(n) = n + floor(n*log_3(2)).
a(n) = n + A056576(n). - Michel Marcus, Dec 12 2023
a(n) = A098294(n) + 2*n - 1. - Ruud H.G. van Tol, Jan 22 2024

A258033 Fractal sequence derived from A022328.

Original entry on oeis.org

0, 0, 2, 1, 0, 2, 1, 3, 0, 5, 2, 4, 1, 3, 0, 5, 2, 4, 1, 6, 3, 0, 8, 5, 2, 7, 4, 1, 6, 3, 0, 8, 5, 2, 10, 7, 4, 1, 9, 6, 3, 0, 8, 5, 2, 10, 7, 4, 1, 9, 6, 3, 11, 0, 8, 5, 13, 2, 10, 7, 4, 12, 1, 9, 6, 3, 11, 0, 8, 5, 13, 2, 10, 7, 4, 12, 1, 9, 6, 14, 3

Offset: 1

Clark Kimberling and Reinhard Zumkeller, May 16 2015

nonn

The sequence is constructed as follows: after partitioning A022328 into segments starting with 0, in each segment the greatest term is to be deleted (see example and comment in A022328); length of k-th mentioned segment = A020914(k); respective greatest term = A056576(k);
this sequence is fractal, i.e. if the first occurrence of each n is removed, the resulting sequence is the original sequence;
A258051 is constructed from this sequence, applying the same transform as described above.

			Segments of A022328 starting with 0, deleted maxima in brackets:
.   1:  0 [1]
.   2:  0 2 1 [3]
.   3:  0 2 [4] 1 3
.   4:  0 5 2 4 1 [6] 3
.   5:  0 5 2 [7] 4 1 6 3
.   6:  0 8 5 2 7 4 1 [9] 6 3
.   7:  0 8 5 2 10 7 4 1 9 6 3 [11]
.   8:  0 8 5 2 10 7 4 [12] 1 9 6 3 11
.   9:  0 8 5 13 2 10 7 4 12 1 9 6 [14] 3 11
.  10:  0 8 5 13 2 10 7 [15] 4 12 1 9 6 14 3 11
.  11:  0 8 16 5 13 2 10 7 15 4 12 1 9 [17] 6 14 3 11
.  12:  0 8 16 5 13 2 10 18 7 15 4 12 1 9 17 6 14 3 11 [19]
.  13:  0 8 16 5 13 2 10 18 7 15 4 12 [20] 1 9 17 6 14 3 11 19
.  14:  0 8 16 5 13 21 2 10 18 7 15 4 12 20 1 9 17 6 14 [22] 3 11 19
.  15:  0 8 16 5 13 21 2 10 18 7 15 [23] 4 12 20 1 9 17 6 14 22 3 11 19

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A022328, A020914, A056576, A258051.

import Data.List (delete)
a258033 n = a258033_list !! (n-1)
a258033_list = 0 : f (tail a022328_list) where
   f xs = (0 : (delete (maximum ys) ys)) ++ f zs
          where (ys, (_ : zs)) = span (> 0) xs

A074720 Least k such that floor(3^n/2^k) is prime.

Original entry on oeis.org

2, 1, 4, 5, 6, 1, 11, 6, 7, 4, 5, 1, 9, 6, 8, 21, 8, 4, 25, 12, 20, 13, 30, 17, 6, 13, 10, 13, 19, 5, 12, 34, 33, 37, 16, 39, 35, 13, 38, 30, 28, 20, 53, 16, 60, 24, 40, 43, 34, 19, 23, 32, 63, 59, 19, 22, 27, 56, 86, 14, 29, 5, 53, 13, 15, 63, 19, 7, 88, 1, 87, 46, 22, 51, 25, 30

Offset: 2

Benoit Cloitre, Sep 04 2002

From Robert Israel, Jan 04 2017: (Start)
a(n) <= A056576(n) - 1.
a(n) = 1 for n in A028491. (End)

Robert Israel, Table of n, a(n) for n = 2..2000

f:= proc(n) local t, k;
   t:= 3^n;
   for k from 1 do t:= t/2; if isprime(floor(t)) then return k fi od:
end proc:
map(f, [$2..100]); # Robert Israel, Jan 04 2017

lk[n_]:=Module[{k=1,n3=3^n},While[!PrimeQ[Floor[n3/2^k]],k++];k]; Array[lk,80,2] (* Harvey P. Dale, Feb 24 2013 *)

a(n)=if(n<0,0,k=1; while(isprime(floor(3^n/2^k)) == 0,k++); k)

A098295 ((3/2)^n)/2^a(n) lies in the half-open interval [1,2).

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 8, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 15, 16, 16, 17, 18, 18, 19, 19, 20, 21, 21, 22, 22, 23, 23, 24, 25, 25, 26, 26, 27, 28, 28, 29, 29, 30, 31, 31, 32, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 39, 40, 40, 41, 42, 42, 43, 43

Offset: 1

Wolfdieter Lang, Oct 18 2004

Stacking perfect fifths (the frequency ratio of a fifth is 3/2), a division by 2^a(n) leads the equivalent tone belonging to the first octave interval [1,2). For example, the third fifth, (3/2)^3, falls into the second octave. This means it lies in the interval [2^1,2^2)=[2,4). Hence ((3/2)^3)/2^1 belongs to the first octave, the interval [1,2).

This sequence coincides for the first 93 term with the floor of y(n)= 4*Pi*log(phi)*n/(Pi^2 + (2*log(phi)^2)), with phi:=(1+sqrt(5))/2. a(n) = floor(y(n)), for n=1..93. Note that y(n) is not the imaginary part of the zero of the Fibonacci function because of a different bracket setting. See A214656. - Wolfdieter Lang, Jul 24 2012

			(3/2)^12 lies in the eighth octave [2^7,2^8) and
((3/2)^12)/2^a(12)= ((3/2)^12)/2^7 = 3^12/2^19 = 531441/524288 = 1.01363... belongs to the first octave [1,2). This ratio is called the Pythagorean comma.

Handbook for Acoustic Ecology, Pythagorean Scale.
Eric Weisstein's World of Music, Pythagorean Scale

This sequence differs from A074840 for the first time at entry a(41)=23: A074840(41)=24.

Cf. A020857, A056576, A098294, A214656.

a(n) = logint(3^n, 2) - n; \\ Ruud H.G. van Tol, Jan 26 2024

a(n) = A098294(n)-1, n >= 1.
a(n) = ceiling(tau*n)-1 with tau = log(3)/log(2)-1 = 0.58496250072..., n >= 1.
a(n) = A056576(n) - n. - Ruud H.G. van Tol, Jan 26 2024

A204399 Numbers k such that floor(2^k / 3^n) = 1.

Original entry on oeis.org

0, 2, 4, 5, 7, 8, 10, 12, 13, 15, 16, 18, 20, 21, 23, 24, 26, 27, 29, 31, 32, 34, 35, 37, 39, 40, 42, 43, 45, 46, 48, 50, 51, 53, 54, 56, 58, 59, 61, 62, 64, 65, 67, 69, 70, 72, 73, 75, 77, 78, 80, 81, 83, 85, 86, 88, 89, 91, 92, 94, 96, 97, 99, 100, 102, 104

Offset: 0

Michel Lagneau, Jan 15 2012

nonn

Presumably a(n) ~ 3*n - floor(n*sqrt(2)) = A195176(n). In the first hundred, a(n) = A195176(n) except for n = 41, 70, 82, 94 where a(n) = A195176(n) - 1.
The conjecture is false; A195176(n) - a(n) increases without bound (though not monotonically) since log_2(3) < 3 - sqrt(2). - Charles R Greathouse IV, Jan 15 2012
Basically a duplicate of A020914. - R. J. Mathar, Jan 16 2012

Cf. A020914, A056576.

for n from 0 to 120 do : for k from 0 to 100 do: x:=floor(2^k /3^n):if x=1 then printf(`%d, `,k):else fi:od:od:

a(n)=ceil(n*log(3)/log(2)) \\ Charles R Greathouse IV, Jan 15 2012

a(n) = ceiling( n * log_2(3) ). - Charles R Greathouse IV, Jan 15 2012

A337921 a(n) is the sum of (3^n mod 2^k) for k such that 2^k < 3^n.

Original entry on oeis.org

1, 3, 18, 38, 195, 585, 607, 3948, 11976, 42415, 127921, 56067, 666938, 2082798, 10769251, 22610393, 110616780, 315726436, 408228944, 2384863439, 7159829169, 23350950650, 74348867826, 49863537606, 401947783347, 1296027221145, 6159163094580, 13796041908620, 60717334308629, 181812784262527

Offset: 1

J. M. Bergot and Robert Israel, Jan 29 2021

nonn

a(n) == A056576(n) (mod 2).

			a(3) = (3^3 mod 2^1) + (3^3 mod 2^2) + (3^3 mod 2^3) + (3^3 mod 2^4) = 18.

Robert P. P. McKone, Table of n, a(n) for n = 1..1000

Cf. A056576.

f:= proc(n) local k; add(3 &^ n mod 2^k, k = 1 .. ilog2(3^n)) end proc:
map(f, [$1..100]);

A337921[n_] := Sum[Mod[3^n, 2^k], {k, 1, Floor[n*Log[2, 3]]}]; Table[A337921[n], {n, 1, 30}] (* Robert P. P. McKone, Jan 31 2021 *)

a(n) = sum(k=1, logint(3^n, 2), lift(Mod(3, 2^k)^n)); \\ Michel Marcus, Jan 30 2021

A372779 Numbers m such that v^n - u^m < u^(m+1) - v^n, where u=2, v=3, and u^m < v^n < u^(m+1).

Original entry on oeis.org

2, 4, 6, 7, 9, 11, 12, 14, 16, 18, 19, 21, 23, 24, 26, 28, 30, 31, 33, 35, 36, 38, 40, 42, 43, 45, 47, 48, 50, 52, 53, 55, 57, 59, 60, 62, 64, 65, 67, 69, 71, 72, 74, 76, 77, 79, 81, 83, 84, 86, 88, 89, 91, 93, 95, 96, 98, 100, 101, 103, 105, 106, 108, 110

Offset: 1

Clark Kimberling, May 18 2024

nonn

			The condition u^m < v^n < u^(m + 1) implies m = floor (n*log(v)/log(u)). With u=2 and v=3, for n = 1, we have m = 1 and 3 - 2 >= 4 - 3, so 1 is in A372780. For n = 2, we have m = 3 and 9 - 8 < 16 - 9, so 2 is in this sequence.

Cf. A000079, A000244, A056576, A372780 (complement).

z = 200; {u, v} = {2, 3};
m[n_] := Floor[n*Log[v]/Log[u]];
Table[m[n], {n, 0, z}];
s = Select[Range[z], v^# - u^m[#] < u^(m[#] + 1) - v^# &]  (* this sequence *)
Complement[Range[Max[s]], s]   (* A372780 *)

A372780 Numbers m such that v^n - u^m >= u^(m+1) - v^n, where u=2, v=3, and u^m < v^n < u^(m+1).

Original entry on oeis.org

1, 3, 5, 8, 10, 13, 15, 17, 20, 22, 25, 27, 29, 32, 34, 37, 39, 41, 44, 46, 49, 51, 54, 56, 58, 61, 63, 66, 68, 70, 73, 75, 78, 80, 82, 85, 87, 90, 92, 94, 97, 99, 102, 104, 107, 109, 111, 114, 116, 119, 121, 123, 126, 128, 131, 133, 135, 138, 140, 143, 145

Offset: 1

Clark Kimberling, May 18 2024

nonn

			The condition u^m < v^n < u^(m + 1) implies m = floor(n*log(v)/log(u)). With u=2 and v=3, for n = 1, we have m = 1 and 3 - 2 >= 4 - 3, so 1 is in this sequence. For n = 2, we have m = 3 and 9 - 8 < 16 - 9, so 2 is in A372779.

Cf. A000079, A000244, A056576, A372779 (complement).

z = 200; {u, v} = {2, 3};
m[n_] := Floor[n*Log[v]/Log[u]];
Table[m[n], {n, 0, z}];
s = Select[Range[z], v^# - u^m[#] < u^(m[#] + 1) - v^# &]  (* A372779 *)
Complement[Range[Max[s]], s]   (* this sequence *)

A375619 a(n) is the largest integer such that there exists a simple graph with n vertices, a(n) edges, and no cycles of length 0 mod 4.

Original entry on oeis.org

0, 1, 3, 4, 6, 7, 9, 11, 12, 14, 15, 17, 19, 20, 22, 23, 25, 26, 28, 30, 31, 33, 34, 36, 38, 39, 41, 42, 44, 45, 47, 49, 50, 52, 53, 55, 57, 58, 60, 61, 63, 64, 66, 68, 69, 71, 72, 74, 76, 77, 79, 80, 82, 83, 85, 87, 88, 90, 91, 93, 95, 96, 98, 99, 101, 102

Offset: 1

Luc Ta, Aug 21 2024

In the parlance of extremal graph theory, a(n) is the extremal number ex(n, C_(0 mod 4)).

			For n = 4, any simple graph with 4 vertices and 5 edges contains a cycle of length 4 == 0 (mod 4), so a(4) < 5. There are exactly two nonisomorphic graphs with 4 vertices and 4 edges. One of them has no cycles of any length other than 3, so a(4) = 4.

Luc Ta, Table of n, a(n) for n = 1..10000
Ervin Győri, Binlong Li, Nika Salia, Casey Tompkins, Kitti Varga, and Manran Zhu, On graphs without cycles of length 0 modulo 4, arXiv: 2312.09999 [math.CO], 2023.
Index entries for sequences related to graphs

Cf. A172272, A056576, A006855, A000066, A345248, A006184.

Mathematica
```
Table[Floor[19/12 * (n - 1)], {n, 100}]
```

a(n) = floor(19/12(n-1)). See Győri et al. in Links.
a(n) = A172272(n-1) for all n <= 77; then a(78) = 121 != 122 = A172272(77).
a(n) = A056576(n-1) for all n <= 53; then a(54) = 83 != 84 = A056576(53).

cp's OEIS Frontend

A206807 Position of 3^n when {2^j} and {3^k} are jointly ranked; complement of A206805.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

Formula

A258033 Fractal sequence derived from A022328.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

A074720 Least k such that floor(3^n/2^k) is prime.

Views

Author

Keywords

Comments

Links

Programs

Maple

Mathematica

PARI

A098295 ((3/2)^n)/2^a(n) lies in the half-open interval [1,2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A204399 Numbers k such that floor(2^k / 3^n) = 1.

Views

Author

Keywords

Comments

Crossrefs

Programs

Maple

PARI

Formula

A337921 a(n) is the sum of (3^n mod 2^k) for k such that 2^k < 3^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

A372779 Numbers m such that v^n - u^m < u^(m+1) - v^n, where u=2, v=3, and u^m < v^n < u^(m+1).

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

A372780 Numbers m such that v^n - u^m >= u^(m+1) - v^n, where u=2, v=3, and u^m < v^n < u^(m+1).

Views

Author

Keywords

Examples

Crossrefs

Programs