

A259558


Numbers n such that prime(n)1 and prime(n+1)1 have the same number of distinct prime factors.


2



2, 4, 5, 8, 9, 12, 15, 16, 18, 19, 23, 24, 25, 28, 29, 31, 36, 38, 39, 40, 42, 44, 52, 56, 58, 59, 60, 63, 64, 71, 73, 74, 76, 80, 85, 88, 91, 92, 94, 96, 98, 99, 102, 103, 106, 107, 109, 110, 111, 112, 113, 117, 126, 129, 130, 131, 132, 133, 134, 136, 139, 141, 142, 143, 144, 151, 152, 159, 160, 161, 165, 168, 169, 173
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OFFSET

1,1


COMMENTS

Unlike A105403, this sequence appears to be infinite.
Dickson's conjecture would imply that there are infinitely many p such that p, p+6, 2*p+1 and 2*p+13 are prime and there are no primes between 2*p+1 and 2*p+13. Then n is in the sequence where 2*p+1=prime(n).  Robert Israel, Jun 30 2015


LINKS

Robert Israel, Table of n, a(n) for n = 1..10000


EXAMPLE

The prime factors of prime(5)1 are 2,5. The prime factors of prime(6)1 are 2,3,3 and they have the same number of distinct prime factors.


MAPLE

N:= 2000: # to use primes <= N
Primes:= select(isprime, [2, seq(2*i+1, i=1..floor((N1)/2))]):
npf:= map(t > nops(numtheory:factorset(Primes[t]1)), [$1..nops(Primes)]):
select(t > npf[t+1]=npf[t], [$1..nops(Primes)1]); # Robert Israel, Jun 30 2015


MATHEMATICA

Select[Range@ 173, PrimeNu[Prime[#]  1] == PrimeNu[Prime[# + 1]  1] &] (* Michael De Vlieger, Jul 01 2015 *)


PROG

(PARI) lista(nn) = {forprime(p=2, nn, if (omega(p1)==omega(nextprime(p+1)1), print1(primepi(p), ", ")); ); } \\ Michel Marcus, Jul 01 2015


CROSSREFS

Cf. A105403, A259559.
Sequence in context: A002541 A239953 A321324 * A189140 A189134 A189019
Adjacent sequences: A259555 A259556 A259557 * A259559 A259560 A259561


KEYWORD

nonn


AUTHOR

Pratik Koirala, Otis Tweneboah, Nathan Fox, Eugene Fiorini, Jun 30 2015


STATUS

approved



