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A039770
Numbers k such that phi(k) is a perfect square.
39
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
1,2
COMMENTS
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
REFERENCES
D. M. Burton, Elementary Number Theory, Allyn and Bacon Inc., Boston MA, 1976, p. 141.
LINKS
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.
FORMULA
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
EXAMPLE
phi(34) = 16 = 4*4.
MAPLE
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
MATHEMATICA
Select[ Range[ 600 ], IntegerQ[ Sqrt[ EulerPhi[ # ] ] ]& ]
PROG
(PARI) for(n=1, 120, if (issquare(eulerphi(n)), print1(n, ", ")))
CROSSREFS
Cf. A000010, A007614. A062732 gives the squares. A306882 (squares not totient).
Sequence in context: A166955 A286808 A121294 * A236019 A247426 A047618
KEYWORD
nonn,easy,nice
STATUS
approved