%I #49 Dec 21 2015 02:56:50
%S 3,13,43,33013,10243213,670302913,207263519443,
%T 119244359152460559009973,565918396036931688582304453,
%U 529942848372194145518502999686683108035251455827553
%N Primes obtained from recursively adding to a 0th term of 1 the product of the first k (current number of addends) primes not dividing the predecessor.
%C An unsubmitted supersequence with an initial term of value 1 and index 0 is implicitly referenced. Term k+1 in this supersequence results by adding to its k-th term the product of the smallest k primes not dividing it. Thus, each term in the sequence results by adding either the k-th primorial or the prior primorial times some prime larger than the k-th prime. To a point beyond the possibility of evaluating the primality status for the sum, only two early additions are not of primorials; and in both of these cases the next prime after the k-th is introduced.
%C This sequence was partly motivated by considering the search for primorial primes, this being one of the simpler ways of also ensuring non-divisibility by the first so many primes. Other than giving terms in the sum different numbers of prime factors beginning with 0, there is nothing special about starting with 1: A similar sequence may be produced with any integer starting value.
%C Many strings of 4 or more consecutive digits the same, including six successive 8s, five successive 5s, and two same-digit quadruples very near to each other, were found noteworthy by the author in the terms calculated.
%C The program below gives [A187750-values]:a(n):[Length] and a sentence for the exceptional adds. A187750 holds the supersequence indices for this sequence.
%e a(1) = 1 + 2# = 1 + 2 = 3. Because 3 divides this and 1 + 2 + 2*5 = 13 is prime, this is a(2). a(3) = a(2) + 5# = 13 + 30 = 43. The next 3 terms in sum are 7#, 7#*13 (11 divides 43 + 7# = 253) and 13#, with a(4) resulting from the last because the 4th and 5th terms in the unsubmitted supersequence are composite.
%o (PARI)
%o {
%o n=1;i=1;
%o while(1,
%o a=prod(j=1,i-1,prime(j));k=i;
%o while(n%prime(k)==0,k++;next());n+=a*prime(k);
%o if(ispseudoprime(n),
%o print1(i":"n"::"ceil(log(n)/log(10))"\n"));
%o if(k!=i,
%o print1("Term "i+1" of the sum is not a primorial.\n"));
%o i++;
%o next())
%o }
%Y Cf. A187750.
%K nonn
%O 1,1
%A _James G. Merickel_, Jan 17 2013