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Number of ways to write prime(n) as (4*x + 2)*y + 4*x + 1 where x and y are nonnegative integers.
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%I #70 Jul 24 2016 21:20:37

%S 0,1,2,1,2,2,3,2,2,4,1,2,4,2,2,4,4,2,2,3,2,2,4,6,3,4,2,4,4,4,1,4,4,4,

%T 6,2,2,2,4,4,6,4,2,2,6,3,2,2,4,4,6,4,3,6,4,4,8,2,2,4,2,6,4,4,2,4,2,3,

%U 4,6,4,6,2,4,4,2,8,2,4,4,8,2,4,4,4,4,9,2,8,2,6,4,2,4,4,6,8,6,2,2,2,6,4,8,4

%N Number of ways to write prime(n) as (4*x + 2)*y + 4*x + 1 where x and y are nonnegative integers.

%C Number of distinct values of k such that k/p_n + k divides (k/p_n)^(k/p_n) + k, (k/p_n)^k + k/p_n and k^(k/p_n) + k/p_n where p_n = prime(n) is n-th prime.

%C a(1) = 0, a(n+1) = number of odd divisors of 1+prime(n+1).

%C Conjectures:

%C 1) a(Fermat prime(n)) >= n, i.e. a(A019434(1)=3) = 1, a(A019434(2)=5) = 2, a(A019434(3)=17) = 3, a(A019434(4)=257) = 4, a(A019434(5)=65537) = 12 > 5, ...

%C 2) a(2^(2^n)+1) > n;

%C 3) a(2^(2^n)+1) < a(2^(2^(n+1))+1).

%F a(n+1) = A001227(A000040(n+1) + 1).

%e a(3) = 2 because (4*0+2)*2+4*0+1 = 5 for (x=0, y=2) and (4*1+2)*0+4*1+1 = 5 for (x=1, y=0) where 5 is the 3rd prime.

%t {0}~Join~Table[Count[Divisors[Prime[n + 1] + 1], _?OddQ], {n, 120}] (* _Michael De Vlieger_, Jun 24 2016 *)

%Y Cf. A000215 {Fermat numbers), A001227, A000668 (Mersenne primes n such that a(n)=1), A019434 (Fermat primes), A069283, A192869 (primes n such that a(n) = 1 or 2), A206581 (primes n such that a(n)=2), A254748.

%K nonn

%O 1,3

%A _Juri-Stepan Gerasimov_, May 16 2016

%E More terms from _Alois P. Heinz_, May 17 2016