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%I #61 Feb 08 2024 07:11:13
%S 1,9,225,242,516,729,3872,13932,14406,17672,18225,20124,21780,29262,
%T 29616,45996,65025,76832,92778,95916,106092,106308,114630,114930,
%U 121872,125652,140130,140625,145794,149124,160986,179562,185100,234876,248652,252978,256860
%N Numbers k such that sigma(phi(k)) = phi(sigma(k)).
%C The largest term of this sequence that I found is 3^9550. Also, if (1/2)*(3^(k+1)-1) is prime (k+1 is a term of A028491) then 3^k is in the sequence, namely sigma(phi(3^k) = phi(sigma(3^k)) (the proof is easy). - _Farideh Firoozbakht_, Feb 09 2005
%D R. K. Guy, Unsolved Problems in Number Theory, 2nd edition, Springer Verlag, 1994, section B42, p. 99.
%H Donovan Johnson, <a href="/A033632/b033632.txt">Table of n, a(n) for n = 1..10000</a> (first 200 terms from T. D. Noe)
%H S. W. Golomb, <a href="/A006872/a006872_1.pdf">Equality among number-theoretic functions</a>, Unpublished manuscript. (Annotated scanned copy)
%H Walter Nissen, <a href="http://upforthecount.com/math/sigmaphi.html">sigma(phi(n)) = phi(sigma(n))</a>, Up for the Count !
%H Walter Nissen, <a href="http://upforthecount.com/math/sigphi1.html">sigma(phi(n)) = phi(sigma(n)): From "5" to "5 figures"</a>, Up for the Count !, Nov. 14, 2000
%H Walter Nissen, <a href="http://upforthecount.com/math/sigphi2.html">Addendum to : sigma(phi()): From "5" to "5 figures"</a>, Up for the Count !, June 8, 2008
%H Walter Nissen, <a href="http://upforthecount.com/math/sigphi3.html">Elaboration of : sigma(phi()): From "5" to "5 figures"</a>, Up for the Count !, Oct. 15, 2010
%F A062401(a(n)) = A062402(a(n)). - _Reinhard Zumkeller_, Jan 04 2013
%t Select[ Range[ 10^6 ], DivisorSigma[ 1, EulerPhi[ # ] ] == EulerPhi[ DivisorSigma[ 1, # ] ] & ]
%o (Haskell)
%o a033632 n = a033632_list !! (n-1)
%o a033632_list = filter (\x -> a062401 x == a062402 x) [1..]
%o -- _Reinhard Zumkeller_, Jan 04 2013
%o (PARI) is(n)=sigma(eulerphi(n))==eulerphi(sigma(n)) \\ _Charles R Greathouse IV_, May 09 2013
%o (Python)
%o from sympy import divisor_sigma as sigma, totient as phi
%o def ok(n): return sigma(phi(n)) == phi(sigma(n))
%o def aupto(nn): return [m for m in range(1, nn+1) if ok(m)]
%o print(aupto(10**4)) # _Michael S. Branicky_, Jan 09 2021
%Y Cf. A000203, A000010, A028491, A078148.
%K nonn,nice
%O 1,2
%A _Jud McCranie_
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