|
%I #10 Jan 18 2014 13:30:21
%S 2,7,43,157,7,72775847,139,523,751,193,19183,22651,140165120353,1051,
%T 37057,433,7459,8263,19248899859613286187907,1564207235629,10453,877,
%U 1993,45183625504351,121453,89248200525047,1505879629
%N Consider N = numerator( 1/p! + 1/q! ) where p = prime(n), q = prime(n+1) for n = 1,2,3,.... Append N to sequence if it is a prime.
%H K. D. Bajpai, <a href="/A235714/b235714.txt">Table of n, a(n) for n = 1..1300</a>
%e 43 is in the sequence because ( 1/5! + 1/7! ) = (1/120 + 1/5040) = 43/5040: numerator(43/5040) = 43 which is prime.
%p KD := proc() local a,b,d,e; a:=ithprime(n)!; b:= ithprime(n+1)!; d:=(1/a) + (1/b); e:=numer(d); if isprime(e) then RETURN (e); fi;end: seq(KD(), n=1..100);
%t k={}; Do[p=Prime[n];q=Prime[n+1]; p2=Numerator[1/p!+1/q!]; If[PrimeQ[p2],AppendTo[k,p2]],{n,150}]; k
%Y Cf. A000040, A120833.
%K nonn
%O 1,1
%A _K. D. Bajpai_, Jan 15 2014
|
