A112954 -id:A112954 - cp's OEIS frontend

A020488 Numbers n such that tau(n) (or sigma_0(n)) = phi(n).

1, 3, 8, 10, 18, 24, 30

Offset: 1

Numbers satisfying A000005(n) = A000010(n).
This sequence is complete because tau(n) < n^(2/3) for all n except a few small numbers, whereas phi(n) > n/(exp(gamma) * log(log(n)) + 3/(log(log(n))) for n > 2. log(log(n)) grows slowly, so phi(n) > tau(n) for all n greater than some relatively small constant. - Jud McCranie, Jun 17 2005
Subset of A112587. - Reinhard Zumkeller, Sep 14 2005
A. P. Minin proved in 1894 that these are the only terms. - Amiram Eldar, May 14 2017

			10 has four divisors: 1, 2, 5, 10, so tau(10) = 4. And four numbers less than 10 are coprime to 10: 1, 3, 7, 9, so phi(10) = 4. Since tau(10) = phi(10), 10 is in the sequence.
phi(12) = 4 also, but 12 has more than four divisors: 1, 2, 3, 4, 6, 12. So 12 is not in the sequence.

L. E. Dickson, History of the Theory of Numbers, Vol. 1, (1919), Chapter X, p. 313.
Jean-Marie De Koninck, Those Fascinating Numbers, translated by the author. Providence, Rhode Island (2009) American Mathematical Society, p. 3.
G. Pólya and G. Szegő, Problems and Theorems in Analysis II, Springer, 1976, Part VIII, Problem 45.

A. P. Minin, On integers N such that the number of divisors of N equals the number of integers less than N and prime to it, Math. Soc. Moscow, Vol. 17, (1894), pp. 537-544 (some front matter is in English and German, article is in Russian)

Cf. A064374, A064375, A064376, A064377, A000005, A000010.

Cf. A112954, A062516, A063469, A063470.

Filtered([1..1000],n->Tau(n)=Phi(n)); # Muniru A Asiru, Dec 20 2018

[n: n in [1..1000] | EulerPhi(n) eq NumberOfDivisors(n)]; // Marius A. Burtea, Dec 20 2018

select(k->tau(k)=phi(k),[$1..1000]); # Peter Luschny, Aug 26 2011

k = 1; s = Select[Range[100000], Equal[Sign[DivisorSigma[k - 1, #] - EulerPhi[#]^k ], 0 ] &]
Select[Range[1000], DivisorSigma[0, #] == EulerPhi[#] &] (* Alonso del Arte, Jan 15 2019 *)

isok(n) = numdiv(n) == eulerphi(n); \\ Michel Marcus, May 14 2017

A062516 Numbers k such that 2*tau(k) = phi(k).

Original entry on oeis.org

5, 9, 15, 28, 40, 72, 84, 90, 120

Offset: 1

Jason Earls, Jul 13 2001

Sequence is finite, since for large k and suitable constants and epsilon: phi(k) - 2*tau(k) > c1*k^(2/3) - 4*c2*k^(1/2) > 0 if k > c3, so phi(k) - 2*tau(k) > 0, QED. Moreover, phi(k) = m*tau(k) has at most finitely many solutions for any constant m or even for slowly increasing functions like m(k) = k^(epsilon). - Labos Elemer, Jul 20 2001

Cf. A112954, A020488, A063469, A063470.

Select[Range[150],2*DivisorSigma[0,#]==EulerPhi[#]&] (* Harvey P. Dale, Jun 28 2022 *)

for(n=1,1000000, if(numdiv(n)*2==eulerphi(n),print(n),))

"full" keyword from Max Alekseyev, Mar 01 2010

A063469 Numbers n such that tau(n)*3 = phi(n).

Original entry on oeis.org

7, 21, 26, 56, 70, 78, 108, 126, 168, 210

Offset: 1

Jason Earls, Jul 26 2001

"phi(n)=k*Tau[n] has at most finitely many solutions for any constant k or even for slowly increasing functions like k(n)=n^(epsilon)." - Labos Elemer, Jul 20 2001

Cf. A062516.

Cf. A112954, A020488, A063470.

Select[Range[300],3DivisorSigma[0,#]==EulerPhi[#]&] (* Harvey P. Dale, Sep 15 2016 *)

for(n=1,10^7, if(3*numdiv(n)==eulerphi(n),print(n)))

"full" keyword from Max Alekseyev, Mar 01 2010

A063470 Numbers n such that tau(n)*4 = phi(n).

Original entry on oeis.org

34, 45, 52, 102, 140, 156, 252, 360, 420

Offset: 1

Jason Earls, Jul 26 2001

Phi(n) = k*tau(n) has at most finitely many solutions for any constant k or even for slowly increasing functions like k(n) = n^(epsilon). - Labos Elemer, Jul 20 2001

For n > 2, tau(n) > 2 and phi(n) <= n-1 so the least solution a(1) to tau(n)*k = phi(n), must be a(1) >= 2*k+1, for the case k=4, a(1) >= 2*4+1 = 9. - Enrique Pérez Herrero, May 12 2012

Cf. A062516.

Cf. A112954, A020488, A063469.

for(n=1,10^6, if(numdiv(n)*4==eulerphi(n),print(n)))

a(1) = A175667(4)

a(A112954(4)) = A112955(4). - Enrique Pérez Herrero, May 12 2012

"full" keyword from Max Alekseyev, Mar 01 2010

A112955 Greatest number m such that phi(m) = n*tau(m), with phi=A000010 and tau=A000005; a(n)=0 if no such m exists.

Original entry on oeis.org

30, 120, 210, 420, 330, 840, 294, 1260, 1080, 1320, 690, 2520, 318, 1470, 2310, 3360, 0, 3780, 0, 4620, 1290, 2760, 1410, 5460, 3000, 1590, 7560, 5880, 1770, 9240, 0, 10080, 4830, 1236, 3234, 10920, 894, 0, 2370, 13860, 2490, 6090, 1038, 9660, 11880

Offset: 1

Reinhard Zumkeller, Oct 07 2005

nonn

Each term is a multiple of 6. [Max Alekseyev, Mar 01 2010]

			a(1) = A020488(A112954(1)) = A020488(7) = 30;
a(2) = A062516(A112954(2)) = A062516(9) = 120;
a(3) = A063469(A112954(3)) = A063469(10) = 210;
a(4) = A063470(A112954(4)) = A063470(9) = 420.

Donovan Johnson, Table of n, a(n) for n = 1..10000

Cf. A112954.

Cf. A175667. [Enrique Pérez Herrero, Oct 22 2010]

More terms from Max Alekseyev, Mar 01 2010

A175667 Smallest number m such that phi(m) = n*tau(m), with phi=A000010 and tau=A000005; a(n)=0 if no such m exists.

Original entry on oeis.org

1, 5, 7, 34, 11, 13, 58, 17, 19, 55, 23, 65, 106, 29, 31, 85, 0, 37, 0, 41, 43, 115, 47, 119, 125, 53, 133, 145, 59, 61, 0, 388, 67, 274, 71, 73, 298, 0, 79, 187, 83, 203, 346, 89, 209, 235, 0, 97, 394, 101, 103, 169, 107, 109, 253, 113, 458, 295, 0, 287, 0, 0, 127, 514, 131

Offset: 1

Enrique Pérez Herrero, Aug 05 2010

nonn

If p = 2*n+1 is a prime, and if n > 1 then a(n)=p.
From R. J. Mathar, Aug 07 2010: (Start)
First column in the array
1,3,8,10,18,24,30: A020488
5,9,15,28,40,72,84,90,120: A062516
7,21,26,56,70,78,108,126,168,210: A063469
34,45,52,102,140,156,252,360,420: A063470
11,33,88,110,198,264,330,
13,35,39,63,76,104,105,130,228,234,280,312,390,504,540,630,840,
58,98,174,294,
17,51,128,136,170,176,224,260,306,384,408,468,510,528,672,780,1260,
19,57,74,135,152,182,190,222,342,456,546,570,756,1080,
55,82,99,124,165,246,308,350,372,440,792,924,990,1050,1320,
23,69,184,230,414,552,690,
65,117,148,195,238,315,364,380,444,520,684,714,864,936,1092,1140,1170,1560,2520,
... (End)

Enrique Pérez Herrero, Table of n, a(n) for n = 1..5000

Cf. A112954, A112955

Cf. A005384, A023212, A043297.

Table[SelectFirst[Range[10^5], EulerPhi@ # == n DivisorSigma[0, #] &] /.
k_ /; MissingQ@ k -> 0, {n, 120}] (* Michael De Vlieger, Aug 09 2017, Version 10.2 *)

From Enrique Pérez Herrero, Jan 01 2012: (Start)
If n > 1 then a(n) >= 2*n+1 or a(n)=0.
If p and q = 2*p+1 are both prime, A005384, then a(p) = 2*p+1.
If p > 3 and q = 4*p+1 are both prime, A023212, then a(p) = 8*p + 2 = 2*q.
If p > 2 is prime and both 2*p+1 and 4*p+1 are composite, A043297, then a(n)=0.
(End)

More terms from R. J. Mathar, Aug 07 2010

Comment corrected by Enrique Pérez Herrero, Aug 12 2010

A114063 Numbers k such that phi(k) = tau(k)^4, where tau(k) = A000005(k).

Original entry on oeis.org

1, 17, 514, 8738, 32301, 33003, 36351, 41504, 42292, 43852, 51860, 62226, 549117, 561051, 571311, 599067, 617967, 629811, 634005, 657495, 673184, 674505, 683168, 701024, 705568, 718964, 722684, 732628, 745484, 759772, 774368

Offset: 1

Giovanni Resta, Feb 13 2006

For all large enough k, we have tau(k) < k^(1/5) and phi(k) > k^(4/5). Hence, tau(k)^4 < k^(4/5) < phi(k), implying that this sequence is finite. - Max Alekseyev, Mar 10 2016

Sequence is composed of 94030 terms. - Max Alekseyev, Jun 01 2024

			phi(33003) = 20736. tau(33003) = 12, 20736 = 12^4.
a(2) = A107655(4) = 17.

Max Alekseyev, Table of n, a(n) for n = 1..94030 (first 660 terms from Enrique Pérez Herrero; terms 661..7000 from Giovanni Resta)

Subsequence of A078164.

Cf. A020488, A062516, A063469, A063470, A107655, A112954, A112955, A175667, A068560, A068559.

Select[Range[10^6], EulerPhi[#] == DivisorSigma[0, #]^4 &] (* Paolo Xausa, May 31 2024 *)

isok(n) = eulerphi(n) == numdiv(n)^4; \\ Michel Marcus, Jan 22 2014

A043297 Primes p such that B(4*p) has denominator 30 where B(2n) are the Bernoulli numbers.

Original entry on oeis.org

2, 17, 19, 31, 47, 59, 61, 71, 101, 103, 107, 109, 137, 149, 151, 157, 167, 181, 197, 211, 223, 227, 229, 241, 257, 263, 269, 271, 283, 311, 313, 317, 331, 337, 347, 349, 353, 367, 379, 383, 389, 397, 401, 421, 439, 449, 457, 461, 463, 467, 479, 503, 521

Offset: 1

Benoit Cloitre, Mar 24 2002

Complement of A087634, primes p such that phi(k) = 4p has a solution, where phi is Euler's totient function.
The sequences a(n), A005384 and A023212 form a partition of the set of primes > 3: Using von Staudt-Clausen formula the divisors of 4p increased by 1 are {2,3,5,p+1,2p+1,4p+1}, p+1 is clearly an even number, and if 2p+1 and 4p+1 are not prime, then denom(B(4p))=30. - Enrique Pérez Herrero, Aug 15 2011
Also 2 with the primes p such that both 2*p+1 and 4*p+1 are composite: A210684. For these numbers k > 2 the equation: phi(n)=k*tau(n), where phi is A000010 and tau is A000005, has no solutions: A112954(a(n))=0. - Enrique Pérez Herrero, May 12 2012

E. Pérez Herrero, Table of n, a(n) for n=1..30000
Wikipedia, Von Staudt-Clausen theorem

Cf. A051225, A053176, A087634.
Cf. A005384, A023212.
Cf. A210684.

Select[Prime[Range[100]], Denominator[BernoulliB[4# ]]==30&] (* T. D. Noe, Feb 19 2004 *)
Select[Prime[Range[100]],!PrimeQ[4#+1]&&!PrimeQ[2#+1]||(#==2)&] (* Enrique Pérez Herrero, Aug 16 2011 *)

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

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.

cp's OEIS Frontend

A020488 Numbers n such that tau(n) (or sigma_0(n)) = phi(n).

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A062516 Numbers k such that 2*tau(k) = phi(k).

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A063469 Numbers n such that tau(n)*3 = phi(n).

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A063470 Numbers n such that tau(n)*4 = phi(n).

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A112955 Greatest number m such that phi(m) = n*tau(m), with phi=A000010 and tau=A000005; a(n)=0 if no such m exists.

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A175667 Smallest number m such that phi(m) = n*tau(m), with phi=A000010 and tau=A000005; a(n)=0 if no such m exists.

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A114063 Numbers k such that phi(k) = tau(k)^4, where tau(k) = A000005(k).

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A043297 Primes p such that B(4*p) has denominator 30 where B(2n) are the Bernoulli numbers.

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