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Given k, the product of the first n primes and starting with the first prime at least 3 greater than k, a(n) is the number of consecutive primes where p-k is also prime.
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%I #9 Feb 03 2019 01:29:41

%S 2,6,10,19,25,29,36,42,43,41,63,65,45,69,86,98,90,77,94,132,118,112,

%T 132,120,131,133,150,128,203,220,163,175,161,176,168,171,206,233,236,

%U 250,201,230,293,270,297,283,256,253,272,318,266,277,296,308,349,353,312

%N Given k, the product of the first n primes and starting with the first prime at least 3 greater than k, a(n) is the number of consecutive primes where p-k is also prime.

%H <a href="http://h.xerol.org/f/1primecheck.txt">http://h.xerol.org/f/1primecheck.txt</a>Shows applicable sequences of primes for n = 2..9

%e For n = 2, k = 2*3 = 6. The first prime at least 3 greater than 6 is 11.

%e 11-6 = 5, 13-6 = 7, 17-6 = 11, 19-6 = 13, 23-6 = 17, 29-6 = 23, all of which are primes. 31-6 = 25 is not prime, so a(2) = 6 because there are 6 prime terms.

%p with(numtheory): a:=proc(n) local k,p,c,ct,j: k:=product(ithprime(i),i=1..n); p:=nextprime(k+2); c:=pi(p): ct:=0: for j from c while isprime(ithprime(j)-k)=true do ct:=ct+1: od: ct; end: seq(a(n),n=1..9); # _Emeric Deutsch_, Apr 13 2007

%K nonn

%O 1,1

%A Ellis M. Eisen (xerol(AT)xerol.org), Apr 02 2007

%E Corrected by _Emeric Deutsch_, Apr 13 2007

%E More terms from _R. J. Mathar_, Sep 23 2007