A262406 -id:A262406 - 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

A272798 Carmichael numbers k such that Euler totient function of k (phi(k)) is a perfect square.

Original entry on oeis.org

1729, 63973, 75361, 172081, 278545, 340561, 658801, 997633, 1773289, 3224065, 5310721, 8719309, 8719921, 12945745, 13187665, 15888313, 17586361, 27402481, 29020321, 39353665, 40430401, 49333201, 67371265, 84417985, 120981601, 128697361, 129255841, 130032865, 151530401, 151813201, 158864833

Offset: 1

Altug Alkan, May 06 2016

nonn

Subsequence of A262406.
If n is a Carmichael number, then phi(n) = Product_{primes p dividing n} (p-1).
So the question is: What are the Carmichael numbers n such that Product_{primes p dividing n} (p-1) is a square?
The number of prime divisors of terms of this sequence are 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 4, 4, 5, 5, 5, 4, 4, 4, 4, 4, ...
1299963601 = 601*1201*1801 is the second term that has three prime divisors and it is a member of this sequence since 600*1200*1800 = 2^10*3^4*5^6 is a square.
This sequence is infinite. See links section for more details. - Altug Alkan, Jan 16 2017

			1729 is a term because A000010(1729) = 1729*(1-1/7)*(1-1/13)*(1-1/19) = 1296 = 36^2.

Amiram Eldar, Table of n, a(n) for n = 1..10000
William D. Banks, Carmichael Numbers with a Square Totient, Canad. Math. Bull. 52 (2009), 3-8.
Index entries for sequences related to Carmichael numbers.

Cf. A000010, A000290, A002997, A039770, A262406, A290793.

isA002997(n) = {my(f); bittest(n, 0) && !for(i=1, #f=factor(n)~, (f[2, i]==1 && n%(f[1, i]-1)==1)||return) && #f>1}
lista(nn) = for(n=1, nn, if(isA002997(n) && issquare(eulerphi(n)), print1(n, ", ")));

a(30) corrected by Amiram Eldar, Aug 11 2017

cp's OEIS Frontend

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

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