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A020488
Numbers n such that tau(n) (or sigma_0(n)) = phi(n).
32
1, 3, 8, 10, 18, 24, 30
OFFSET
1,2
COMMENTS
Numbers satisfying A000005(n) = A000010(n).
This sequence is complete because tau(n) < n^(2/3) for all n except a few small numbers, whereas phi(n) > n/(exp(gamma) * log(log(n)) + 3/(log(log(n))) for n > 2. log(log(n)) grows slowly, so phi(n) > tau(n) for all n greater than some relatively small constant. - Jud McCranie, Jun 17 2005
Subset of A112587. - Reinhard Zumkeller, Sep 14 2005
A. P. Minin proved in 1894 that these are the only terms. - Amiram Eldar, May 14 2017
REFERENCES
L. E. Dickson, History of the Theory of Numbers, Vol. 1, (1919), Chapter X, p. 313.
Jean-Marie De Koninck, Those Fascinating Numbers, translated by the author. Providence, Rhode Island (2009) American Mathematical Society, p. 3.
G. Pólya and G. Szegő, Problems and Theorems in Analysis II, Springer, 1976, Part VIII, Problem 45.
LINKS
A. P. Minin, On integers N such that the number of divisors of N equals the number of integers less than N and prime to it, Math. Soc. Moscow, Vol. 17, (1894), pp. 537-544 (some front matter is in English and German, article is in Russian)
EXAMPLE
10 has four divisors: 1, 2, 5, 10, so tau(10) = 4. And four numbers less than 10 are coprime to 10: 1, 3, 7, 9, so phi(10) = 4. Since tau(10) = phi(10), 10 is in the sequence.
phi(12) = 4 also, but 12 has more than four divisors: 1, 2, 3, 4, 6, 12. So 12 is not in the sequence.
MAPLE
select(k->tau(k)=phi(k), [$1..1000]); # Peter Luschny, Aug 26 2011
MATHEMATICA
k = 1; s = Select[Range[100000], Equal[Sign[DivisorSigma[k - 1, #] - EulerPhi[#]^k ], 0 ] &]
Select[Range[1000], DivisorSigma[0, #] == EulerPhi[#] &] (* Alonso del Arte, Jan 15 2019 *)
PROG
(PARI) isok(n) = numdiv(n) == eulerphi(n); \\ Michel Marcus, May 14 2017
(Magma) [n: n in [1..1000] | EulerPhi(n) eq NumberOfDivisors(n)]; // Marius A. Burtea, Dec 20 2018
(GAP) Filtered([1..1000], n->Tau(n)=Phi(n)); # Muniru A Asiru, Dec 20 2018
KEYWORD
nonn,fini,full
STATUS
approved