A090387 -id:A090387 - cp's OEIS frontend

A090395 Denominator of d(n)/n, where d(n) is the number of divisors of n (A000005).

1, 1, 3, 4, 5, 3, 7, 2, 3, 5, 11, 2, 13, 7, 15, 16, 17, 3, 19, 10, 21, 11, 23, 3, 25, 13, 27, 14, 29, 15, 31, 16, 33, 17, 35, 4, 37, 19, 39, 5, 41, 21, 43, 22, 15, 23, 47, 24, 49, 25, 51, 26, 53, 27, 55, 7, 57, 29, 59, 5, 61, 31, 21, 64, 65, 33, 67, 34, 69, 35, 71, 6, 73, 37, 25, 38

Offset: 1

Ivan_E_Mayle(AT)a_provider.com, Jan 31 2004

The first occurrence of k (if it exists) is studied in A091895.

Sequence A353011 gives indices of "late birds": n such that a(k) > a(n) for all k > n. - M. F. Hasler, Apr 15 2022

			a(6) = 3 because the number of divisors of 6 is 4 and 4 divided by 6 equals 2/3, which has 3 as its denominator.

Antti Karttunen, Table of n, a(n) for n = 1..65537

Cf. A000005, A090387 (numerators), A091896 (numbers not in this sequence), A353011 (indices of terms such that all subsequent terms are larger).

with(numtheory): seq(denom(tau(n)/n), n=1..75) ; # Zerinvary Lajos, Jun 04 2008

Table[ Denominator[ DivisorSigma[0, n]/n], {n, 1, 80}] (* Robert G. Wilson v, Feb 04 2004 *)

A090395(n) = denominator(numdiv(n)/n); \\ Antti Karttunen, Sep 25 2018

from math import gcd
from sympy import divisor_count
def A090395(n): return n//gcd(n,divisor_count(n)) # Chai Wah Wu, Jun 20 2022

a(n) = n/g with g = A009191(n) = gcd(A000005(n), n). This explains the "rays" in the graph, e.g., g = 1 for odd squarefree n, g = 2 for even semiprimes n = 2p > 4 and n = 4p, p > 3. - M. F. Hasler, Apr 15 2022

More terms from Robert G. Wilson v, Feb 04 2004

A348172 a(n) is the number of positive k (can be greater than n) such that A000005(n)/n = A000005(k)/k.

Original entry on oeis.org

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

Offset: 1

Tejo Vrush, Oct 04 2021

The first 1 occurs at a(4). The corresponding k value is 4.
The first 2 occurs at a(1). The corresponding k values are 1, 2.
The first 3 occurs at a(9). The corresponding k values are 9, 18, 24.
The first 4 occurs at a(243). The corresponding k values are 243, 405, 486, 810.
Does every number appear in this sequence?
The first 5 is at a(3189375) with k-values {3189375, 5467500, 6378750, 6561000, 8748000} while the first 6 is at 3176523 with k-values {3176523, 4084101, 6353046, 6806835, 8168202, 13613670}. If 7 appears it does so past 10^8. - Charles R Greathouse IV, Oct 05 2021

			For n = 9, the k such that A000005(9)/9 = 1/3 = A000005(k)/k are 9, 18, and 24. Therefore, a(9) = 3.

David A. Corneth, Table of n, a(n) for n = 1..10000
Charles R Greathouse IV, PARI/GP script

Cf. A000005, A090387, A090395, A005382, A005383, A348184.

Array[Function[r, Count[Range[Ceiling[4/r^2]], _?(DivisorSigma[0, #]/# == r &)]][DivisorSigma[0, #]/#] &, 105] (* or *)
Block[{nn = 7, m, s}, m = 2^(2 nn); s = KeySort@ PositionIndex[Array[DivisorSigma[0, #]/# &, m]]; s = Reverse@ KeyDrop[s, TakeWhile[Keys@ s, 4/#^2 > m &]]; Length /@ Array[Lookup[s, DivisorSigma[0, #]/#] &, 2^nn]] (* Michael De Vlieger, Oct 04 2021 *)

a(n) = {my(q=numdiv(n)/n); sum(i=1, 4/q^2, numdiv(i)/i == q);} \\ Michel Marcus, Oct 04 2021

a(n) = my(q=numdiv(n)/n, s=denominator(q), res = 0); forstep(i=s, 4/q^2, s, if(numdiv(i) == q * i, res++)); res \\ David A. Corneth, Oct 07 2021

PARI
```
\\ See Greathouse link
```

A136641 a(n) is the smallest positive integer that is coprime to n and has n divisors.

Original entry on oeis.org

1, 3, 4, 15, 16, 175, 64, 105, 100, 567, 1024, 1925, 4096, 3645, 784, 945, 65536, 13475, 262144, 6237, 1600, 295245, 4194304, 25025, 1296, 2657205, 4900, 40095, 268435456, 3776773, 1073741824, 10395, 25600, 215233605, 5184, 175175, 68719476736, 1937102445, 102400

Offset: 1

Leroy Quet, Apr 14 2008

nonn

Is this the same as the least index m where A090387(m) = n? - Michel Marcus, Mar 25 2022

For p prime, a(p) = 2^(p-1) for p > 2, a(2*p) = 3^(p-1)*5 for p > 5, a(3*p) = 2^(p-1)*25 for p > 3, a(5*p) = 2^(p-1)*3^4 for p >5, ... . - Michael S. Branicky, Mar 26 2022

			The sequence of positive integers each with 9 divisors starts: 36, 100, 196, 225, 256, ... Now 36 is not coprime to 9. But 100, the next bigger value with 9 divisors, is. So a(9) = 100.

Michael S. Branicky, Table of n, a(n) for n = 1..65

Cf. A000005, A073904, A090387.

a(n) = my(k=1); while ((gcd(n,k) != 1) || (numdiv(k) != n), k++); k; \\ Michel Marcus, Mar 25 2022

from math import gcd
from sympy import divisor_count
def a(n):
    k = 1
    while gcd(n, k) != 1 or divisor_count(k) != n: k += 1
    return k
print([a(n) for n in range(1, 19)]) # Michael S. Branicky, Mar 25 2022

a(11)-a(36) from Sean A. Irvine, May 03 2010

a(37) and beyond from Michael S. Branicky, Mar 26 2022

A368523 Positive integers in decreasing order of tau(k)/k, where tau(k) = A000005(k).

Original entry on oeis.org

1, 2, 4, 3, 6, 8, 12, 5, 10, 9, 18, 24, 16, 20, 7, 14, 15, 30, 36, 28, 48, 40, 60, 21, 42, 32, 11, 22, 72, 13, 26, 27, 54, 56, 84, 44, 45, 90, 120, 80, 96, 33, 66, 25, 50, 17, 34, 52, 35, 70, 108, 64, 19, 38, 144, 39, 78, 180, 63, 126, 168, 88, 132, 100, 112

Offset: 1

Keith J. Bauer, Dec 28 2023

nonn

Because tau(k)/k is bounded by 2/sqrt(k), this sequence is well-defined.
In the case of ties, terms are sorted from least to greatest.
Let c be the j-th distinct value of tau(a(n))/a(n). Terms of this sequence for which tau(a(n))/a(n) >= c are the proper divisors of A059992(j + 1) for 1 <= j <= 4. True for j = 0 if the 0th value of c is taken to be infinity. Pattern breaks for j > 4.

			tau(1)/1 = tau(2)/2 = 1
tau(4)/4 = 3/4
tau(3)/3 = tau(6)/6 = 2/3
tau(8)/8 = tau(12)/12 = 1/2
tau(5)/5 = tau(10)/10 = 2/5
tau(9)/9 = tau(18)/18 = tau(24)/24 = 1/3

Cf. A000005, A090387, A090395, A323383, A354768.

length = 100
result = {}
for n = 1, length do
  local div_count = 0
  local root_n = math.sqrt(n)
  for d = 1, root_n do
    if n % d == 0 then
      div_count = div_count + 2
    end
  end
  if (root_n == math.floor(root_n)) then
    div_count = div_count - 1
  end
  result[n] = {n, div_count / n}
end
function compare(a, b)
  if a[2] ~= b[2] then
    return a[2] > b[2]
  else
    return a[1] < b[1]
  end
end
table.sort(result, compare)
i = 1
bound = 2 / math.sqrt(length)
while result[i][2] >= bound do
  io.write(result[i][1] .. ',')
  i = i + 1
end

nmax = 100; s = Sort[Table[{k, DivisorSigma[0, k]/k}, {k, 1, nmax^2}], #1[[2]] >= #2[[2]] &]; Table[s[[j, 1]], {j, 1, nmax}] (* Vaclav Kotesovec, Jan 04 2024 *)

A355140 n/d(n) rounded to the nearest integer, where d(n) is the number of divisors of n (A000005).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 4, 2, 3, 3, 6, 2, 7, 4, 4, 3, 9, 3, 10, 3, 5, 6, 12, 3, 8, 7, 7, 5, 15, 4, 16, 5, 8, 9, 9, 4, 19, 10, 10, 5, 21, 5, 22, 7, 8, 12, 24, 5, 16, 8, 13, 9, 27, 7, 14, 7, 14, 15, 30, 5, 31, 16, 11, 9, 16, 8, 34, 11, 17, 9, 36, 6, 37, 19, 13, 13, 19

Offset: 1

Sameer Khan, Jun 20 2022

nonn

In the ambiguous case, fractions are rounded up.

			a(1) = round (1 / 1) = 1;
a(4) = round (4 / 3) = 1;
a(5) = round (5 / 2) = 3;

Cf. A000005, A078709 (floor), A334762 (ceiling), A090395 (numerators), A090387 (denominators).

Table[Floor[n/DivisorSigma[0,n]+1/2],{n,100}] (* Harvey P. Dale, Dec 22 2022 *)

from sympy import divisor_count
def A355140(n): return (2*n+(d:=divisor_count(n)))//(2*d) # Chai Wah Wu, Jun 20 2022

a(n) = round (n / A000005(n)).

cp's OEIS Frontend

A090395 Denominator of d(n)/n, where d(n) is the number of divisors of n (A000005).

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A348172 a(n) is the number of positive k (can be greater than n) such that A000005(n)/n = A000005(k)/k.

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A136641 a(n) is the smallest positive integer that is coprime to n and has n divisors.

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A368523 Positive integers in decreasing order of tau(k)/k, where tau(k) = A000005(k).

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Lua

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A355140 n/d(n) rounded to the nearest integer, where d(n) is the number of divisors of n (A000005).

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