

A263811


Numbers n such that n = tau(n) * phi(n1) + 1.


1




OFFSET

1,1


COMMENTS

The first 5 known Fermat primes from A019434 are in sequence.
The next term, if it exists, must be greater than 2*10^7.
A prime p is in the sequence iff p is a Fermat prime (A019434)  see proof in A171271.
Observation: the known composite terms are squares of primes.  Omar E. Pol, Nov 04 2015
Rearranging the definition gives (n1)/phi(n1) = tau(n), which means n1 is in A007694. Since n1 is thus 3smooth, there are two possibilities:
1) n1 is a power of 2 and tau(n) = 2, i.e. n is a Fermat prime,
2) n1 is a 3smooth number divisible by 6 and tau(n) = 3, i.e. n is a Pierpont number and the square of a prime.
In the second case, n1 factors as (p1)(p+1) for some p, and both parts are 3smooth if and only if p is in {2,3,5,7,17} (2 and 3 are excluded since in those cases n1 is not divisible by 6). Therefore, this sequence is complete if and only if there are no more Fermat primes. (End)


LINKS



EXAMPLE

17 is in this sequence because 17 = tau(17)*phi(16)+1 = 2*8+1.


MATHEMATICA

Select[Range[10^5], # == DivisorSigma[0, #] EulerPhi[#  1] + 1 &] (* Michael De Vlieger, Nov 05 2015 *)


PROG

(Magma) [n: n in [2..1000000]  n eq NumberOfDivisors(n) * EulerPhi(n1) + 1]
(PARI) for(n=1, 1e5, if( n1 == numdiv(n)*eulerphi(n1) , print1(n, ", "))) \\ Altug Alkan, Nov 05 2015


CROSSREFS

Cf. A263810 (numbers n such that n = tau(n) * phi(n2) + 1).


KEYWORD

nonn,hard,more


AUTHOR



STATUS

approved



