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 A039770 Numbers n such that phi(n) is a perfect square. 26

%I

%S 1,2,5,8,10,12,17,32,34,37,40,48,57,60,63,74,76,85,101,108,114,125,

%T 126,128,136,160,170,185,192,197,202,204,219,240,250,257,273,285,292,

%U 296,304,315,364,370,380,394,401,432,438,444,451,456,468,489,504,505

%N Numbers n such that phi(n) is a perfect square.

%C A004171 is a subsequence because phi(2^(2n+1)) = (2^n)^2. - _Enrique Pérez Herrero_, Aug 25 2011

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

%H T. D. Noe, <a href="/A039770/b039770.txt">Table of n, a(n) for n = 1..10000</a>

%H W. D. Banks, J. B. Friedlander, C. Pomerance and I. E. Shparlinski, <a href="http://math.dartmouth.edu/~carlp/PDF/banksfinal2.pdf">Multiplicative structure of values of the Euler function</a>, 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.

%H P. Pollack and C. Pomerance, <a href="http://www.math.dartmouth.edu/~carlp/squaretotients8.pdf">Square values of Euler's function</a>, submitted for publication.

%F a(n) seems to be asymptotic to c*n^(3/2) with 1<c<1.3 - _Benoit Cloitre_, Sep 08 2002

%F Banks, Friedlander, Pomerance, and Shparlinski show that a(n) = O(n^1.421). - _Charles R Greathouse IV_, Aug 24 2009

%e phi(34)=16=4*4.

%p 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

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

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

%Y Cf. A000010, A007614. A062732 gives the squares.

%Y Cf. A068560, A004171, A114063, A262406.

%K nonn,easy,nice

%O 1,2

%A _Olivier Gérard_

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