OFFSET

1,2

COMMENTS

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

REFERENCES

R. K. Guy, Unsolved Problems in Number Theory, 2nd edition, Springer Verlag, 1994, section B42, p. 99.

LINKS

Donovan Johnson, Table of n, a(n) for n = 1..10000 (first 200 terms from T. D. Noe)

S. W. Golomb, Equality among number-theoretic functions, Unpublished manuscript. (Annotated scanned copy)

Walter Nissen, sigma(phi(n)) = phi(sigma(n)), Up for the Count !

Walter Nissen, sigma(phi(n)) = phi(sigma(n)): From "5" to "5 figures", Up for the Count !, Nov. 14, 2000

Walter Nissen, Addendum to : sigma(phi()): From "5" to "5 figures", Up for the Count !, June 8, 2008

Walter Nissen, Elaboration of : sigma(phi()): From "5" to "5 figures", Up for the Count !, Oct. 15, 2010

FORMULA

MATHEMATICA

Select[ Range[ 10^6 ], DivisorSigma[ 1, EulerPhi[ # ] ] == EulerPhi[ DivisorSigma[ 1, # ] ] & ]

PROG

(Haskell)

a033632 n = a033632_list !! (n-1)

a033632_list = filter (\x -> a062401 x == a062402 x) [1..]

-- Reinhard Zumkeller, Jan 04 2013

(PARI) is(n)=sigma(eulerphi(n))==eulerphi(sigma(n)) \\ Charles R Greathouse IV, May 09 2013

(Python)

from sympy import divisor_sigma as sigma, totient as phi

def ok(n): return sigma(phi(n)) == phi(sigma(n))

def aupto(nn): return [m for m in range(1, nn+1) if ok(m)]

print(aupto(10**4)) # Michael S. Branicky, Jan 09 2021

CROSSREFS

KEYWORD

nonn,nice

AUTHOR

STATUS

approved