A015733 -id:A015733 - cp's OEIS frontend

A020491 Numbers k such that sigma_0(k) divides phi(k).

1, 3, 5, 7, 8, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 24, 26, 28, 29, 30, 31, 33, 34, 35, 37, 39, 40, 41, 43, 45, 47, 49, 51, 52, 53, 55, 56, 57, 58, 59, 61, 63, 65, 67, 69, 70, 71, 72, 73, 74, 76, 77, 78, 79, 82, 83, 84, 85, 87, 88, 89, 90, 91, 93, 95, 97, 98, 99, 101, 102, 103, 104

Offset: 1

David W. Wilson

nonn

In other words, numbers k such that d(k) divides phi(k).
From Enrique Pérez Herrero, Aug 11 2010: (Start)
sigma_0(k) divides phi(k) when:
k is an odd prime: A065091;
k is an odd squarefree number: A056911;
k = 2^m, where m <> 1 is a Mersenne number (A000225).
If d divides (p-1), with p prime, then p^(d-1) is in this sequence, as are p^(p-1), p^(p-2) and p^(-1+p^n).
(End)
phi(n) and d(n) are multiplicative functions, so if m and n are coprime and both of them are in this sequence then m*n is also in this sequence. - Enrique Pérez Herrero, Sep 05 2010
From Bernard Schott, Aug 14 2020: (Start)
The corresponding quotients are in A289585.
About the 3rd case of Enrique Pérez Herrero's comment: if k = 2^M_m, where M_m = 2^m - 1 is a Mersenne number >= 3 (A000225), then the corresponding quotient phi(k)/d(k) is the integer 2^(2^m-m-2) = A076688(m); hence, these numbers k, A058891 \ {2}, form a subsequence. (End)

Enrique Pérez Herrero, Table of n, a(n) for n = 1..5000
Psychedelic Geometry Blogspot, Fermat and Mersenne Numbers Conjecture-(2)

Cf. A000005, A000010.
Complement of A015733. [Enrique Pérez Herrero, Aug 11 2010]
Cf. A058891, A076688, A289585.

Select[ Range[ 105 ], IntegerQ[ EulerPhi[ # ]/DivisorSigma[ 0, # ] ]& ]

isok(k) = !(eulerphi(k) % numdiv(k)); \\ Michel Marcus, Aug 10 2020

A289585 Quotients as they appear as k increases when tau(k) divides phi(k).

Original entry on oeis.org

1, 1, 2, 3, 1, 2, 1, 5, 6, 2, 8, 1, 9, 3, 11, 1, 3, 2, 14, 1, 15, 5, 4, 6, 18, 6, 2, 20, 21, 4, 23, 14, 8, 4, 26, 10, 3, 9, 7, 29, 30, 6, 12, 33, 11, 3, 35, 2, 36, 9, 6, 15, 3, 39, 10, 41, 2, 16, 14, 5, 44, 2, 18, 15, 18, 48, 7, 10, 50, 4, 51, 6, 6, 13, 53, 3, 54, 5, 18, 56, 22, 12, 24, 2

Offset: 1

Bernard Schott, Jul 08 2017

nonn

Numbers k such that tau(k) divides phi(k) are in A020491.
Only for seven integers which are in A020488, we have a(n) = 1.
The integers such that a(n) = 2, 3, 4 are respectively in A062516, A063469, A063470.
When p is an odd prime then phi(p) = p-1, tau(p) = 2, so phi(p)/tau(p) = (p-1)/2 and A005097 is an infinite subsequence.
For k = A058891(m+1), that is 2^A000225(m), with m>=2, the corresponding quotient phi(k)/tau(k) is the integer A076688(m). - Bernard Schott, Aug 15 2020

			a(10) = 2 because A020491(10) = 15 and phi(15)/tau(15) = 8/4 = 2.

Robert G. Wilson v, Table of n, a(n) for n = 1..1000

Cf. A000010, A000005, A020491, A015733, A020488, A058891, A062516, A063469, A063470, A175667, A112955, A112954.
Cf. A000225, A058891, A076688.
Cf. A005097 (a subsequence).
Cf. A290634 (complement).

for n from 1 to 50 do q:=phi(n)/tau(n);
if q=floor(q) then print(n,q,phi(n),tau(n)) else fi; od:

f[n_] := Block[{d = EulerPhi[n]/DivisorSigma[0, n]}, If[ IntegerQ@d, d, Nothing]]; Array[f, 120] (* Robert G. Wilson v, Jul 09 2017 *)

lista(nn) = {for (n=1, nn, q = eulerphi(n)/numdiv(n); if (denominator(q)==1, print1(q, ", ")););} \\ Michel Marcus, Jul 10 2017

a(n) = A000010(A020491(n)) / A000005(A020491(n)). - David A. Corneth, Jul 09 2017

A286627 a(n) = exponent of the highest power of A000005(n) (number of divisors of n) dividing A000010(n) (totient function phi), a(1) = 1.

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, Jun 30 2017

nonn

a(1) = 1 by convention.

			A000005(5) = 2, A000010(5) = 4, 2^2 is the highest power of 2 which divides 4, thus a(5) = 2.
A000005(6) = 4, A000010(6) = 2, 4^0 = 1 is the highest power of 4 which divides 2, thus a(6) = 0.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000005, A000010, A286561, A286628, A289276.

Cf. A015733 (positions of zeros), A020491 (of nonzeros).

A286627(n) = valuation(eulerphi(n), numdiv(n));

a(n) = A286561(A000010(n), A000005(n)).

A015734 Odd numbers k such that d(k) does not divide phi(k).

Original entry on oeis.org

25, 27, 75, 81, 121, 189, 225, 275, 289, 297, 343, 363, 405, 425, 513, 529, 567, 575, 605, 621, 725, 729, 825, 837, 841, 867, 1025, 1029, 1053, 1089, 1161, 1175, 1225, 1269, 1275, 1325, 1331, 1377, 1445, 1475, 1539, 1587, 1593, 1681, 1725

Offset: 1

Robert G. Wilson v

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000

Intersection of A005408 and A015733.

Cf. A000005 (d), A000010 (phi).

Select[Range[1, 2000, 2], !Divisible[EulerPhi[#], DivisorSigma[0, #]] &] (* Amiram Eldar, May 15 2024 *)

is(k) = if(k%2, my(f = factor(k), d = numdiv(f), p = eulerphi(f)); p % d, 0); \\ Amiram Eldar, May 15 2024

Offset corrected by Amiram Eldar, May 15 2024

A342665 Numbers k for which phi(k)+1 is a multiple of d(k), where phi is Euler totient function (A000010) and d(n) gives the number of divisors of n (A000005).

Original entry on oeis.org

1, 2, 4, 25, 81, 121, 289, 529, 841, 1681, 2209, 2809, 3481, 5041, 6889, 7921, 10201, 11449, 12100, 12769, 17161, 18769, 22201, 27889, 28561, 28900, 29929, 32041, 36481, 38809, 51529, 54289, 57121, 63001, 66049, 69169, 72361, 78961, 84100, 85849, 96721, 100489, 120409, 124609, 128881, 146689, 151321, 160801, 175561

Offset: 1

Antti Karttunen, Mar 30 2021

nonn

Numbers k such that A124331(k) = k. This is also a subsequence of the records of A124331 (both their values and their positions).

Terms other than 2 are a perfect square. Proof: phi(k) is even for k > 2. So phi(k)+1 is odd for k > 2. d(k) is odd only if k is a perfect square. So for any term k > 2 we need k to be a perfect square. Checking cases <= 2 leaves only 2 as the nonsquare in this sequence. - David A. Corneth, Mar 31 2021

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A000005, A000010, A020491.

Fixed points of A124331. After 1, a subsequence of A015733.

Select[Join[{1, 2}, Range[2, 420]^2], Divisible[EulerPhi[#] + 1, DivisorSigma[0, #]] &] (* Amiram Eldar, Mar 31 2021 *)

isA342665(n) = !((eulerphi(n)+1) % numdiv(n));

A290634 Positive integers which are never the quotient of phi(n)/tau(n).

Original entry on oeis.org

17, 19, 31, 38, 47, 59, 61, 62, 71, 85, 91, 101, 103, 107, 109, 118, 121, 133, 137, 149, 151, 157, 167, 181, 187, 197, 211, 217, 218, 223, 227, 229, 241, 247, 257, 259, 263, 266, 269, 271, 283, 289, 305, 311, 313, 314, 317, 327, 331, 334, 337, 347, 349, 353, 355, 361, 367

Offset: 1

Bernard Schott, Aug 08 2017

nonn

For phi(n)/tau(n) see A279287/A279288.
Numbers that do not appear in A175667.
The first nine terms of this sequence are exactly A119480(3) through A119480(11), and many other terms are common to these two sequences.

Cf. A000010, A000005, A020491, A015733, A020488, A062516, A175667, A112954, A112955, A119480, A289585, A279287/A279288.

A358061 a(n) = phi(n) mod tau(n).

Original entry on oeis.org

0, 1, 0, 2, 0, 2, 0, 0, 0, 0, 0, 4, 0, 2, 0, 3, 0, 0, 0, 2, 0, 2, 0, 0, 2, 0, 2, 0, 0, 0, 0, 4, 0, 0, 0, 3, 0, 2, 0, 0, 0, 4, 0, 2, 0, 2, 0, 6, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 4, 0, 2, 0, 4, 0, 4, 0, 2, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 2, 4, 0, 0, 0, 0, 2, 0, 0, 0, 0

Offset: 1

Ctibor O. Zizka, Oct 28 2022

nonn

a(n) > 0 for n in A015733, a(n) = 0 for n in A020491.

			For n = 4; a(4) = A000010(4) mod A000005(4) = 2 mod 3 = 2.

Cf. A000005 (tau), A000010 (phi), A015733, A020491.

a[n_] := Mod[EulerPhi[n], DivisorSigma[0, n]]; Array[a, 100] (* Amiram Eldar, Oct 28 2022 *)

from math import prod
from sympy import factorint
def A358061(n):
    f = factorint(n).items()
    d = prod(e+1 for p, e in f)
    return prod(pow(p,e-1,d)*((p-1)%d) for p, e in f) % d # Chai Wah Wu, Oct 29 2022

a(n) = A000010(n) mod A000005(n).

cp's OEIS Frontend

A020491 Numbers k such that sigma_0(k) divides phi(k).

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A289585 Quotients as they appear as k increases when tau(k) divides phi(k).

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A286627 a(n) = exponent of the highest power of A000005(n) (number of divisors of n) dividing A000010(n) (totient function phi), a(1) = 1.

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A015734 Odd numbers k such that d(k) does not divide phi(k).

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A342665 Numbers k for which phi(k)+1 is a multiple of d(k), where phi is Euler totient function (A000010) and d(n) gives the number of divisors of n (A000005).

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A290634 Positive integers which are never the quotient of phi(n)/tau(n).

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A358061 a(n) = phi(n) mod tau(n).

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