A062153 -id:A062153 - cp's OEIS frontend

A365458 The largest power of 3 that is less than or equal to n.

1, 1, 3, 3, 3, 3, 3, 3, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 81, 81, 81, 81, 81

Offset: 1

Antti Karttunen, Sep 17 2023

nonn

			a(2) = 1 because 3^0 = 1 <= 2.
a(3) = 3 because 3^1 = 3 <= 3.
a(4) = 3 because 3^1 = 3 <= 4.

Michael De Vlieger, Table of n, a(n) for n = 1..19683 (3^9 = 19683)

Cf. A000244, A365459.

Cf. also A062153, A064235, A080342.

Array[3^Floor@ Log[3, #] &, 90] (* Michael De Vlieger, Sep 17 2023 *)

A365458(n) = if(1==n,n,my(k=0); while((3^k) < n, k++); if((3^k) > n,k--); (3^k));

a(n) = 3^logint(n, 3); \\ Michel Marcus, Sep 17 2023

def A365458(n):
    kmin, kmax = 0, 1
    while 3**kmax <= n:
        kmax <<= 1
    while True:
        kmid = kmax+kmin>>1
        if 3**kmid > n:
            kmax = kmid
        else:
            kmin = kmid
        if kmax-kmin <= 1:
            break
    return 3**kmin # Chai Wah Wu, Sep 17 2023

a(n) = 3^floor((log n) / (log 3)). - Michael De Vlieger, Sep 17 2023

a(n) = A000244(A062153(n)). - Michel Marcus, Sep 17 2023

A081063 Number of numbers <= n that are 3-smooth or prime powers.

Original entry on oeis.org

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

Offset: 1

Reinhard Zumkeller, Mar 04 2003

nonn

a(n) = #{A081061(k): 1<=k<=n}.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A003586, A000961.

Accumulate[Table[If[PrimePowerQ[n]||Max[FactorInteger[n][[All,1]]]<5,1,0],{n,80}]] (* Harvey P. Dale, Nov 29 2020 *)

from sympy import integer_log, primepi, integer_nthroot
def A081063(n): return int(1-(a:=n.bit_length())-(b:=integer_log(n,3)[0])+sum((n//3**i).bit_length() for i in range(b+1))+sum(primepi(integer_nthroot(n, k)[0]) for k in range(1, a))) # Chai Wah Wu, Sep 16 2024

a(n)=A071521(n)+A065515(n)-A000523(n)-A062153(n)+1.

A374054 a(n) = max_{i=0..n} S_3(i) + S_3(n-i) where S_3(x) = A053735(x) is the base-3 digit sum of x.

Original entry on oeis.org

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

Offset: 0

Hermann Gruber, Jun 26 2024

As shown in the proof of [Gruber and Holzer, lemma 9], the maximum is attained by choosing i as the largest number not exceeding n whose ternary representation is (22...2)_3. By [Gruber and Holzer, lemma 6], for this choice of i we have S_3(i) = 2*floor(log_3(n+1)) and S_3(n-i) = S_3(n+1)-1, giving the formula below.

			For n=31, the maximum is attained by 26 + 5 = (222)_3 + (12)_3. Using 32=(1021)_3, comparing with the formula above, S_3(26) = 2*floor(log_3(n+1)) = 6 and S_3(5) = S_3(31+1)-1 = 3. Notice that other pairs attain the maximum as well, namely 23 + 8 = (22)_3 + (212)_3, as well as 20 + 11 = (202)_3 + (102)_3.

Hermann Gruber and Markus Holzer, Optimal Regular Expressions for Palindromes of Given Length. Extended journal version, in preparation, 2024.

Paolo Xausa, Table of n, a(n) for n = 0..10000
Hermann Gruber and Markus Holzer, Optimal Regular Expressions for Palindromes of Given Length, Proceedings of the 46th International Symposium on Mathematical Foundations of Computer Science, Article No. 53, pp. 53:1-53:15, 2021.

Cf. A053735, A062153, A014701, A374056.

f:= n -> 2 * ilog[3](n+1) + convert(convert(n+1,base,3),`+`) - 1:
map(f, [$0..100]); # Robert Israel, Jun 27 2024

Table[2*Floor[Log[3, k]] + DigitSum[k, 3] - 1, {k, 100}] (* Paolo Xausa, Aug 01 2024 *)

a(n) = 2*logint(n+1, 3) + sumdigits(n+1, 3) - 1; \\ Michel Marcus, Jul 05 2024

a(n) = 2*floor(log_3(n+1)) + A053735(n+1) - 1 [Gruber and Holzer, lemma 9].

A069924 Number of k, 1<=k<=n, such that phi(k) divides k.

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15

Offset: 1

Benoit Cloitre, May 05 2002

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000
Vaclav Kotesovec, Plot of a(n)/log(n)^2 for n = 1..10000

Cf. A007694, A071521, A062153.

Table[Length[Select[Range[n], Divisible[#, EulerPhi[#]] &]], {n, 1, 100}] (* Vaclav Kotesovec, Feb 16 2019 *)
Accumulate[Table[If[Divisible[n,EulerPhi[n]],1,0],{n,80}]] (* Harvey P. Dale, Jul 04 2021 *)

for(n=1,150,print1(sum(i=1,n,if(i%eulerphi(i),0,1)),","))

a(n) = Card(k: 1<=k<=n : k==0 (mod phi(k))) asymptotically: a(n) = C*log(n)^2 + o(log(n)^2) with C=0.6....

a(n) = A071521(n) - A062153(n). - Ridouane Oudra, Mar 05 2025

A206722 Parameters of Chebyshev function psi.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 3, 2, 1, 1, 3, 2, 1, 1, 3, 2, 1, 1, 1, 3, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 4, 2, 1, 1, 1, 1, 4, 2, 1, 1, 1, 1, 1, 4, 2, 1, 1, 1, 1, 1

Offset: 2

John W. Nicholson, Feb 11 2012

a(x,n) is the exponent k such that prime(n)^k <= x and x < prime(n)^(k+1).
psi(x) = Sum_{p_n <= x} k*log(p_n), where a(x,n) = k is the unique integer such that p_n^k <= x but p_n^(k+1) > x.
Related to Firoozbakht's Conjecture (1982): p_n^(1/n) > p_(n+1)^(1/(n+1)) for all n >= 1.

			If x = 7, then 2^2, 3^1, 5^1, 7^1 <= x < 2^3, 3^2, 5^2, 7^2, respectively so k = 2, 1, 1, 1, respectively.
The table starts in row x=2 with columns n >= 1 as:
  1;
  1, 1;
  2, 1;
  2, 1, 1;
  2, 1, 1;
  2, 1, 1, 1;
  3, 1, 1, 1;
  3, 2, 1, 1;
  3, 2, 1, 1, 1;

N. Kanti Sinha, On a new property of primes that leads to a generalization of Cramer's conjecture, arXiv:1010.1399 [math.NT], 2010.
Wikipedia, Chebyshev function

Columns: A000523 (n=1), A062153 (n=2).

A206722[x_, n_] := Module[{p = Prime[n]}, For[k = 0, True, k++, If[p^(k+1) > x && p^k <= x, Return[k]]]];
Table[DeleteCases[Table[A206722[x, n], {n, 1, 17}], 0], {x, 2, 20}] // Flatten (* Jean-François Alcover, Sep 15 2018, after R. J. Mathar *)

prime(n) := block(
    if n = 1 then
       return(2)
    else
    return(next_prime(prime(n-1)))
)$ /* very slow recursive definition of A000040 */
A206722(x,n) := block(
    local(p),
    p : prime ( n ),
    for k : 0 do (
       if p^(k+1) > x and p^k <= x then
          return(k)
       )
)$
for x : 2 thru 20 do (
    for n : 1 thru 17 do
      sprint(A206722(x,n)),
    newline()
)$ /* R. J. Mathar, Feb 14 2012 */

A329195 a(n) = floor(log_5(n^2)) = floor(2 log_5(n)).

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5

Offset: 1

M. F. Hasler, Nov 07 2019

nonn

Cf. A000290 (n^2), A062153 (log_3), A329202 (log_2(n^2)), A329193 (log_2(n^3)), A329194 (log_3(n^2)).

Table[Floor[2Log[5,n]],{n,100}] (* Harvey P. Dale, Dec 14 2021 *)

apply( A329195(n)=logint(n^2,5), [1..99])

A329199 a(n) = round(log_3(n)).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 1

M. F. Hasler, Nov 07 2019

nonn

Cf. A062153 (floor log_3), A000523 (floor log_2), A004257 (round log_2), A029837 (ceiling log_2), A329194 (log_3(n^2)).

Round[Log[3,Range[100]]] (* Harvey P. Dale, Sep 06 2022 *)

apply( A329199(n)=log(n)\/log(3), [1..130])

A280724 Expansion of 1/(1 - x) + (1/(1 - x)^2)*Sum_{k>=0} x^(3^k).

Original entry on oeis.org

1, 2, 3, 5, 7, 9, 11, 13, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 73, 77, 81, 85, 89, 93, 97, 101, 105, 109, 113, 117, 121, 125, 129, 133, 137, 141, 145, 149, 153, 157, 161, 165, 169, 173, 177, 181, 185, 189, 193, 197, 201, 205, 209, 213, 217, 221, 225, 229, 233, 237, 241, 245

Offset: 0

Ilya Gutkovskiy, Jan 07 2017

Sums of lengths of ternary numbers (A007089).

			-----------------------
n  base 3 length  a(n)
-----------------------
0 |  0   |  1   |  1
1 |  1   |  1   |  2
2 |  2   |  1   |  3
3 |  10  |  2   |  5
4 |  11  |  2   |  7
5 |  12  |  2   |  9
6 |  20  |  2   |  11
7 |  21  |  2   |  13
8 |  22  |  2   |  15
9 |  100 |  3   |  18
-----------------------

Eric Weisstein's World of Mathematics, Ternary

Cf. A007089, A062153, A081604, A083652, A117804.

CoefficientList[Series[1/(1 - x) + (1/(1 - x)^2) Sum[x^3^k, {k, 0, 15}], {x, 0, 70}], x]
Table[1 + Sum[Floor[Log[3, k]] + 1, {k, 1, n}], {n, 0, 70}]

G.f.: 1/(1 - x) + (1/(1 - x)^2)*Sum_{k>=0} x^(3^k).

a(n) = 1 + Sum_{k=1..n} floor(log_3(k)) + 1.

A333355 Number of bits in binary expansion of n minus the number of digits of n when written in base 3.

Original entry on oeis.org

0, 1, 0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2

Offset: 1

Frank Ellermann, Mar 15 2020

Record highs are at n = 2^A054414. All n=2^k >= 2 are increases, all n=3^j are decreases, and there is either one or none 3^j between 2^(k-1) and 2^k. When one, a(2^k) = a(2^(k-1)) so not a record high. When none, a(2^k) = a(2^(k-1)) + 1 which is a record high. If 2^k and 2^(k-1) are the same length in ternary then there is no 3^j between them. This is when 2^k has most significant ternary digit 2 since 2^(k-1) >= 3^j is 2^k >= 2*3^j. These k are A054414. Non-record increases are at its complement n = 2^A020914 >= 2. - Kevin Ryde, Apr 04 2020

			a(8) = 2 = 4 - 2 for binary 1000 and ternary 22.
a(64) = 3 = 7 - 4 for binary 1000000 and ternary 2101.

Cf. A007088 ( binary), A000523 (floor(log_2(n))), A029837.

Cf. A007089 (ternary), A062153 (floor(log_3(n))), A117966.

a:= n-> ilog[2](n)-ilog[3](n):
seq(a(n), n=1..100);  # Alois P. Heinz, Mar 15 2020

a[n_]: = Floor @ Log[2, n] - Floor @ Log[3, n]; Array[a, 100] (* Amiram Eldar, Mar 16 2020 *)

a(n) = logint(n,2) - logint(n,3); \\ Kevin Ryde, May 15 2020

L = 1 ;  M = 1 ;  B = 2 ;  T = 3       ;  S = 0
do N = 2 while length( S ) < 258
   if B = N then  do    ;  B = B * 2   ;  L = L + 1   ;  end
   if T = N then  do    ;  T = T * 3   ;  M = M + 1   ;  end
   S = S || ',' L - M
end N
say S                   ;  return S

a(n) = A000523(n) - A062153(n) = floor(log_2(n)) - floor(log_3(n)).

a(n) = length(A007088(n)) - length(A007089(n)).

cp's OEIS Frontend

A365458 The largest power of 3 that is less than or equal to n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Python

Formula

A081063 Number of numbers <= n that are 3-smooth or prime powers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Python

Formula

A374054 a(n) = max_{i=0..n} S_3(i) + S_3(n-i) where S_3(x) = A053735(x) is the base-3 digit sum of x.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A069924 Number of k, 1<=k<=n, such that phi(k) divides k.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A206722 Parameters of Chebyshev function psi.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Maxima

A329195 a(n) = floor(log_5(n^2)) = floor(2 log_5(n)).

Views

Author

Keywords

Crossrefs

Programs

Mathematica

PARI

A329199 a(n) = round(log_3(n)).

Views

Author

Keywords

Crossrefs

Programs

Mathematica

PARI

A280724 Expansion of 1/(1 - x) + (1/(1 - x)^2)*Sum_{k>=0} x^(3^k).

Views

Author

Keywords