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Number of prime factors with multiplicity of the difference between consecutive primes.
0

%I #8 Mar 16 2020 14:02:09

%S 0,1,1,2,1,2,1,2,2,1,2,2,1,2,2,2,1,2,2,1,2,2,2,3,2,1,2,1,2,2,2,2,1,2,

%T 1,2,2,2,2,2,1,2,1,2,1,3,3,2,1,2,2,1,2,2,2,2,1,2,2,1,2,2,2,1,2,2,2,2,

%U 1,2,2,3,2,2,2,2,3,2,3,2,1

%N Number of prime factors with multiplicity of the difference between consecutive primes.

%F a(n) = bigomega(p(n+1)-p(n)).

%e p(5)-p(4)=4, bigomega(4)=2.

%t PrimeOmega/@Differences[Prime[Range[100]]] (* _Harvey P. Dale_, Mar 16 2020 *)

%o (PARI) a(n) = bigomega(prime(n+1) - prime(n)); \\ _Michel Marcus_, Aug 26 2013

%Y Cf. A001222, A001223.

%K easy,nonn

%O 1,4

%A Paul Finley (pfinley(AT)touro.edu), Sep 11 2007