A272887 Number of ways to write prime(n) as (4*x + 2)*y + 4*x + 1 where x and y are nonnegative integers.

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, 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, 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

Offset

1,3

Comments

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.

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

Conjectures:

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, ...

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

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

Links

Table of n, a(n) for n=1..105.

Formula

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

Example

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.

Mathematica

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

Crossrefs

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.

Sequence in context: A137735 A365576 A290983 * A143966 A260617 A178695

Adjacent sequences: A272884 A272885 A272886 * A272888 A272889 A272890

Keyword

nonn

Author

Juri-Stepan Gerasimov, May 16 2016

Extensions

More terms from Alois P. Heinz, May 17 2016

Status

approved