OFFSET
1,1
COMMENTS
Makowski proved that phi(k) + sigma(k) = k*d(k) if and only if k is a prime (see in Sivaramakrishnan, Chapter I, page 8, Theorem 3). Generally, when phi(k) + sigma(k) = m*d(k) there are special cases in which phi(k) + sigma(k) is divisible by higher powers of the number of divisors d(k).
This sequence is infinite: it includes all the semiprimes p*q such that p == 1 (mod 128), and q == 127 (mod 128). - Amiram Eldar, Mar 25 2024
REFERENCES
R. Sivaramakrishnan, Classical Theory of Arithmetic Functions, Marcel Dekker, Inc., New York and Basel, 1989.
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
A. Makowski, Problem 339, Elemente der Mathematik, Vol. 13 (1958), p. 115; alternative link.
C. A. Nicol, Problem E 1674, The American Mathematical Monthly, Vol. 71, No. 3 (1964), p. 317; Another characterization of prime number, Solutions to Problem E 1674 by Martin J. Cohen and J. A. Fridy, ibid., Vol. 72, No. 2 (1965), pp. 186-187.
FORMULA
EXAMPLE
511 is a term since it has 4 divisors, phi(511) = 432, sigma(511) = 592, and 432 + 592 = 1024 = 4 * 4^4.
MATHEMATICA
Select[Range[10000], CompositeQ[#] && Divisible[EulerPhi[#] + DivisorSigma[1, #], DivisorSigma[0, #]^4] &] (* Amiram Eldar, Mar 25 2024 *)
PROG
(PARI) is(n)=my(f=factor(n)); (eulerphi(f)+sigma(f))%numdiv(f)^4==0 && !isprime(n) \\ Charles R Greathouse IV, Mar 01 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Jun 27 2000
STATUS
approved