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A263811 Numbers k such that k = tau(k) * phi(k-1) + 1.

%I A263811 #27 Mar 16 2025 02:38:50
%S A263811 3,5,17,25,49,257,289,65537
%N A263811 Numbers k such that k = tau(k) * phi(k-1) + 1.
%C A263811 Numbers k such that k = A000005(k) * A000010(k-1) + 1.
%C A263811 The first 5 known Fermat primes from A019434 are in this sequence.
%C A263811 The next term, if it exists, must be greater than 2*10^7.
%C A263811 A prime p is in the sequence iff p is a Fermat prime (A019434) - see proof in A171271.
%C A263811 Observation: the known composite terms are squares of primes. - _Omar E. Pol_, Nov 04 2015
%C A263811 From _Charlie Neder_, Mar 02 2019: (Start)
%C A263811 Rearranging the definition gives (k-1)/phi(k-1) = tau(k), which means k-1 is in A007694. Since k-1 is thus 3-smooth, there are two possibilities:
%C A263811 1) k-1 is a power of 2 and tau(k) = 2, i.e., k is a Fermat prime,
%C A263811 2) k-1 is a 3-smooth number divisible by 6 and tau(k) = 3, i.e., k is a Pierpont number and the square of a prime.
%C A263811 In the second case, k-1 factors as (p-1)(p+1) for some p, and both parts are 3-smooth if and only if p is in {2,3,5,7,17} (2 and 3 are excluded since in those cases k-1 is not divisible by 6). Therefore, this sequence is complete if and only if there are no more Fermat primes. (End)
%e A263811 17 is in this sequence because 17 = tau(17)*phi(16) + 1 = 2*8 + 1.
%t A263811 Select[Range[10^5], # == DivisorSigma[0, #] EulerPhi[# - 1] + 1 &] (* _Michael De Vlieger_, Nov 05 2015 *)
%o A263811 (Magma) [n: n in [2..1000000] |  n eq NumberOfDivisors(n) * EulerPhi(n-1) + 1];
%o A263811 (PARI) for(n=1, 1e5, if( n-1 == numdiv(n)*eulerphi(n-1) , print1(n, ", "))) \\ _Altug Alkan_, Nov 05 2015
%Y A263811 Cf. A000005, A000010, A019434.
%Y A263811 Cf. A263810 (numbers k such that k = tau(k) * phi(k-2) + 1).
%K A263811 nonn,hard,more
%O A263811 1,1
%A A263811 _Jaroslav Krizek_, Nov 04 2015