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Number of prime divisors, counted with multiplicity, of the sum of two consecutive primes.
4

%I #17 Jan 22 2020 06:13:30

%S 1,3,3,3,4,3,4,3,3,4,3,3,4,4,4,5,5,7,3,6,4,5,3,3,4,4,4,6,3,6,3,3,4,7,

%T 5,4,7,4,4,6,6,4,8,4,5,3,3,5,5,4,4,7,4,3,5,4,6,3,4,4,8,6,3,6,5,7,3,5,

%U 5,5,4,4,4,5,3,3,3,4,6,5,6,4,8,4,5,3,3,5,5,4,3,4,3,5,3,4,3,5,5,7,6,7,3,5,4

%N Number of prime divisors, counted with multiplicity, of the sum of two consecutive primes.

%C Clearly sum of two consecutive primes prime(x) and prime(x+1) has more than 2 prime divisors for all x > 1.

%H Amiram Eldar, <a href="/A098037/b098037.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A001222(A001043(n)). - _Michel Marcus_, Feb 15 2014

%e Prime(2) + prime(3) = 2*2*2, 3 factors, the second term in the sequence.

%t PrimeOmega[Total[#]]&/@Partition[Prime[Range[110]],2,1] (* _Harvey P. Dale_, Jun 14 2011 *)

%o (PARI) b(n) = for(x=1,n,y1=(prime(x)+prime(x+1));print1(bigomega(y1)","))

%Y Cf. A071215.

%K easy,nonn

%O 1,2

%A _Cino Hilliard_, Sep 10 2004

%E Definition corrected by _Andrew S. Plewe_, Apr 08 2007