A306908 -id:A306908 - cp's OEIS frontend

A039770 Numbers k such that phi(k) is a perfect square.

1, 2, 5, 8, 10, 12, 17, 32, 34, 37, 40, 48, 57, 60, 63, 74, 76, 85, 101, 108, 114, 125, 126, 128, 136, 160, 170, 185, 192, 197, 202, 204, 219, 240, 250, 257, 273, 285, 292, 296, 304, 315, 364, 370, 380, 394, 401, 432, 438, 444, 451, 456, 468, 489, 504, 505

Offset: 1

Olivier Gérard

A004171 is a subsequence because phi(2^(2k+1)) = (2^k)^2. - Enrique Pérez Herrero, Aug 25 2011
Subsequence of primes is A002496 since in this case phi(k^2+1) = k^2. - Bernard Schott, Mar 06 2023
Products of distinct terms of A002496 form a subsequence. - Chai Wah Wu, Aug 22 2025

			phi(34) = 16 = 4*4.

D. M. Burton, Elementary Number Theory, Allyn and Bacon Inc., Boston MA, 1976, p. 141.

T. D. Noe, Table of n, a(n) for n = 1..10000
W. D. Banks, J. B. Friedlander, C. Pomerance and I. E. Shparlinski, Multiplicative structure of values of the Euler function, in High Primes and Misdemeanours: Lectures in Honour of the Sixtieth Birthday of Hugh Cowie Williams (A. Van der Poorten, ed.), Fields Inst. Comm. 41 (2004), pp. 29-47.
P. Pollack and C. Pomerance, Square values of Euler's function, submitted for publication.
Bernard Schott, Subfamilies and subsequences

Cf. A000010, A007614. A062732 gives the squares. A306882 (squares not totient).
Cf. A068560, A114063, A078164.
Subsequences: A054755, A002496, A004171, A262406, A324745, A324746, A247129, A324747, A306908.

with(numtheory); isA039770 := proc (n) return issqr(phi(n)) end proc; seq(`if`(isA039770(n), n, NULL), n = 1 .. 505); # Nathaniel Johnston, Oct 09 2013

Select[ Range[ 600 ], IntegerQ[ Sqrt[ EulerPhi[ # ] ] ]& ]

for(n=1, 120, if (issquare(eulerphi(n)), print1(n, ", ")))

from math import isqrt
from sympy import totient as phi
def ok(n): return isqrt(p:=phi(n))**2 == p
print([k for k in range(1, 506) if ok(k)]) # Michael S. Branicky, Aug 17 2025

a(n) seems to be asymptotic to c*n^(3/2) with 1 < c < 1.3. - Benoit Cloitre, Sep 08 2002

Banks, Friedlander, Pomerance, and Shparlinski show that a(n) = O(n^1.421). - Charles R Greathouse IV, Aug 24 2009

A054755 Odd powers of primes of the form q = x^2 + 1 (A002496).

Original entry on oeis.org

2, 5, 8, 17, 32, 37, 101, 125, 128, 197, 257, 401, 512, 577, 677, 1297, 1601, 2048, 2917, 3125, 3137, 4357, 4913, 5477, 7057, 8101, 8192, 8837, 12101, 13457, 14401, 15377, 15877, 16901, 17957, 21317, 22501, 24337, 25601, 28901, 30977, 32401

Offset: 1

Labos Elemer, Apr 25 2000

nonn

A002496 is a subset; the odd power exponent is 1.
From Bernard Schott, Mar 16 2019: (Start)
The terms of this sequence are exactly the integers with only one prime factor and whose Euler's totient is square, so this sequence is a subsequence of A039770. The primitive terms of this sequence are the primes of the form q = x^2 + 1, which are exactly in A002496.
Additionally, the terms of this sequence also have a square cototient, so this sequence is a subsequence of A063752 and A054754.
If q prime = x^2 + 1, phi(q) = x^2, phi(q^(2k+1)) = (x*q^k)^2, and cototient(q) = 1^2, cototient(q^(2k+1)) = (q^k)^2. (End)

			a(20) = 3125 = 5^5, q = 5 = 4^2+1 and Phi(3125) = 2500 = 50^2, cototient(3125) = 3125 - Phi(3125) = 625 = 25^2.

David A. Corneth, Table of n, a(n) for n = 1..18864 (terms <= 10^11)
Bernard Schott, Subfamilies and subsequences

Cf. A000010, A051953, A039770, A063752, A054754, A334745 (with 2 distinct prime factors), A306908 (with 3 distinct prime factors).

Subsequences: A002496 (primitive primes: m^2+1), A004171 (2^(2k+1)), A013710 (5^(2k+1)), A013722 (17^(2k+1)), A262786 (37^(2k+1)).

Select[Range[10^5], And[PrimeNu@ # == 1, IntegerQ@ Sqrt@ EulerPhi@ #] &] (* Michael De Vlieger, Mar 31 2019 *)

isok(m) = (omega(m)==1) && issquare(eulerphi(m)); \\ Michel Marcus, Mar 16 2019

upto(n) = {my(res = List([2]), q); forstep(i = 2, sqrtint(n), 2, if(isprime(i^2 + 1), listput(res, i^2 + 1) ) ); q = #res; forstep(i = 3, logint(n, 2), 2, for(j = 1, q, c = res[j]^i; if(c <= n, listput(res, c) , next(2) ) ) ); listsort(res); res } \\ David A. Corneth, Mar 17 2019

A000010(a(n)) = (q^(2k))*(q-1) and A051953(a(n)) = q^(2k), where q = 1 + x^2 and is prime.

A324746 Numbers k with exactly two distinct prime factors and such that phi(k) is square, when k = p^(2s+1) * q^(2t+1) with p < q primes, s,t >= 0.

Original entry on oeis.org

10, 34, 40, 57, 74, 85, 136, 160, 185, 202, 219, 250, 296, 394, 451, 489, 505, 513, 514, 544, 629, 640, 679, 802, 808, 985, 1000, 1057, 1154, 1184, 1285, 1354, 1387, 1417, 1576, 1717, 1971, 2005, 2047, 2056, 2125, 2176, 2509, 2560, 2594, 2649, 2761, 2885, 3097

Offset: 1

Bernard Schott, Mar 12 2019

nonn

An integer belongs to this sequence iff (p-1)*(q-1) = m^2.
This is the first subsequence of A324745, the second one is A324747.
Some values of (k,p,q,m): (10,2,5,2), (34,2,17,4), (40,2,5,4), (57,3,19,4), (74,2,37,6), (85,5,17,8).
The primitive terms of this sequence are the products p * q, with p < q which satisfy (p-1)*(q-1) = m^2; the first few are 10, 34, 57, 74, 85, 185. These primitives form exactly the sequence A247129. Then the integers (p*q) * p^2 and (p*q) * q^2 are new terms of the general sequence.
The number of semiprimes p*q whose totient is a square equal to (2*n)^2 can be found in A306722.

			629 = 17 * 37 and phi(629) = 16 * 36 = 9^2.
808 = 2^3 * 101 and phi(808) = (2^1 * 101^0 * 10)^2 = 20^2.

Cf. A039770, A062732, A221285, A054755, A324745, A324747, A306908.

Cf. A306722, A247129 (subsequence of primitives).

N:= 10^4:
Res:= {}:
p:= 1:
do
  p:= nextprime(p);
  if p^2 >= N then break fi;
  F:= ifactors(p-1)[2];
  dm:= mul(t[1]^ceil(t[2]/2),t=F);
  for j from (p-1)/dm+1 do
    q:= (j*dm)^2/(p-1) + 1;
    if q > N then break fi;
    if isprime(q) then Res:= Res union {seq(seq(
      p^(2*s+1)*q^(2*t+1),t=0..floor((log[q](N/p^(2*s+1))-1)/2)),
      s=0..floor((log[p](N/q)-1)/2))} fi
  od
od:
sort(convert(Res,list)); # Robert Israel, Mar 22 2019

Select[Range[6, 3100], And[PrimeNu@ # == 2, IntegerQ@ Sqrt@ EulerPhi@ #, IntegerQ@ Sqrt[Times @@ (FactorInteger[#][[All, 1]] - 1 )]] &] (* Michael De Vlieger, Mar 24 2019 *)

isok(k) = {if (issquare(eulerphi(k)), my(expo = factor(k)[,2]); if ((#expo == 2)&& (expo[1]%2) == (expo[2]%2), return (1)););} \\ Michel Marcus, Mar 18 2019

phi(p*q) = (p-1)*(q-1) = m^2 for primitive terms.

phi(k) = (p^s * q^t * m)^2 with k as in the name of this sequence.

cp's OEIS Frontend

A039770 Numbers k such that phi(k) is a perfect square.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

A054755 Odd powers of primes of the form q = x^2 + 1 (A002496).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A324746 Numbers k with exactly two distinct prime factors and such that phi(k) is square, when k = p^(2s+1) * q^(2t+1) with p < q primes, s,t >= 0.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

Formula