

A055466


Numbers n such that d(n)^2 divides phi(n) + sigma(n).


1



1, 2, 4, 15, 39, 49, 55, 78, 81, 87, 95, 99, 110, 111, 119, 121, 125, 143, 159, 183, 184, 215, 247, 287, 295, 303, 319, 327, 335, 350, 357, 391, 407, 415, 423, 430, 447, 455, 471, 507, 511, 519, 527, 535, 543, 551, 559, 583, 591, 620, 623, 654, 655, 671, 679
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OFFSET

1,2


COMMENTS

Makowski proved that phi(n)+Sigma[n] = n*d[n] iff n is a prime (see in Sivaramakrishnan, Chapter I, page 8, Theorem 3).
Contains p^2 if p is a prime == 2 or 7 (mod 9), and p*q if p and q are distinct primes with p*q == 7 (mod 8).  Robert Israel, Jan 18 2018


REFERENCES

Sivaramakrishnan, R. (1989), Classical Theory of Arithmetical Functions, Marcel Dekker, Inc., New YorkBasel.


LINKS



FORMULA

Integer solutions of Phi[x]+Sigma[x] = kd[x]^2 or A000203(n)+A000010(n) = k*A000005(n)^2, where k is integer.


EXAMPLE

true for 2 (the only prime) and some composites. n = 78: 8 divisors, Sigma = 168, Phi = 24, 168+24 = 192 = 8*8*3


MAPLE

filter:= proc(n) uses numtheory;
phi(n)+sigma(n) mod tau(n)^2 = 0
end proc:


MATHEMATICA

okQ[n_] := Divisible[EulerPhi[n] + DivisorSigma[1, n], DivisorSigma[0, n]^2];


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



