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A272887
Number of ways to write prime(n) as (4*x + 2)*y + 4*x + 1 where x and y are nonnegative integers.
1
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).
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
KEYWORD
nonn
AUTHOR
EXTENSIONS
More terms from Alois P. Heinz, May 17 2016
STATUS
approved