A063470 -id:A063470 - 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

A112954 Number of numbers m such that phi(m) = n*tau(m), with phi=A000010 and tau=A000005.

Original entry on oeis.org

7, 9, 10, 9, 7, 17, 4, 17, 14, 15, 7, 19, 2, 16, 20, 21, 0, 29, 0, 29, 9, 13, 7, 32, 7, 11, 23, 21, 7, 39, 0, 19, 17, 4, 11, 44, 2, 0, 11, 41, 7, 24, 2, 19, 30, 11, 0, 55, 4, 23, 7, 21, 7, 46, 9, 27, 4, 11, 0, 61, 0, 0, 27, 29, 9, 30, 2, 10, 19, 31, 0, 57, 2, 9, 27, 4, 4, 30, 2, 50, 29, 9

Offset: 1

Reinhard Zumkeller, Oct 07 2005

nonn

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

Cf. A020491, A020488, A062516, A063469, A063470, A112955, A175667.

More terms 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

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

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

A308664 Numbers k such that tau(k) and phi(k) are the legs of a Pythagorean triple.

Original entry on oeis.org

20, 36, 60, 100, 300

Offset: 1

Antonio Roldán, Jul 14 2019

The sequence is finite since for all large enough n, we have tau(n) < n^(1/4) and phi(n) > n^(3/4) while, if x < y are the legs of a Pythagorean triangle, we always have y < x^2/2. - Giovanni Resta, Jul 27 2019

From Resta's inequality it can be deduced that phi(n) <= 2304. Then it's easy to see that the sequence is full. - Max Alekseyev, May 30 2024

			60 is in this sequence because tau(60) = 12 and phi(60) = 16, legs of the Pythagorean triple {12, 16, 20} (12^2 + 16^2 = 20^2).

Cf. A000005, A000010, A020488, A062516, A063469, A063470, A112954.

Select[Range[300], IntegerQ@Sqrt[DivisorSigma[0, #]^2 + EulerPhi[#]^2] &] (* Amiram Eldar, Jul 26 2019 *)

for(i = 1,  2000, a = eulerphi(i); b = numdiv(i); if(issquare(a^2 + b^2), print1(i,", ")))

"full" keyword added by Max Alekseyev, May 30 2024

A378089 Irregular triangle read by rows in which row n lists the numbers k such that phi(k)/tau(k) = n.

Original entry on oeis.org

1, 3, 8, 10, 18, 24, 30, 5, 9, 15, 28, 40, 72, 84, 90, 120, 7, 21, 26, 56, 70, 78, 108, 126, 168, 210, 34, 45, 52, 102, 140, 156, 252, 360, 420, 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

Offset: 1

Mohammed Yaseen, Nov 16 2024

			Triangle begins:
  n=1: 1, 3, 8, 10, 18, 24, 30;
  n=2: 5, 9, 15, 28, 40, 72, 84, 90, 120;
  n=3: 7, 21, 26, 56, 70, 78, 108, 126, 168, 210;
  n=4: 34, 45, 52, 102, 140, 156, 252, 360, 420;
  n=5: 11, 33, 88, 110, 198, 264, 330;
  n=6: 13, 35, 39, 63, 76, 104, 105, 130, 228, 234, 280, 312, 390, 504, 540, 630, 840;
  n=7: 58, 98, 174, 294;
  ...

Cf. A000005 (tau), A000010 (phi).
Cf. A020488 (row 1), A062516 (row 2), A063469 (row 3), A063470 (row 4).
Cf. A112954 (row lengths), A175667 (1st column), A112955 (right column), A020491 (ordered terms).

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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A112954 Number of numbers m such that phi(m) = n*tau(m), with phi=A000010 and tau=A000005.

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A063469 Numbers n such that tau(n)*3 = 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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A289585 Quotients as they appear as k increases when tau(k) divides phi(k).

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