%I
%S 1,2,2,3,6,46,1306,7695,17383720,2183805400,60512359083,447808566362,
%T 181830203704703
%N a(1) = 1; for n > 1: a(n) = smallest number >1 such that product of any two or more successive terms + 1 is prime.
%C a(14) > 2.5*10^14, if it exists.  _Don Reble_, Nov 22 2015
%C So far, up through a(13), the sequence is nondecreasing. I don't know a good reason why it should stay that way. (But since candidates for each successive value get rarer, the least candidate will tend to increase.)  _Don Reble_, Nov 22 2015
%C I don't think it will be easy to prove that this sequence is nondecreasing. The analogous sequence with other starting values often leads to nonmonotonic sequences, e.g., (3, 2, 2, 3, 26, 876, 15136, ...), (4, 3, 6, 6, 5, 14, 3597, 1218704, ...), or (5, 2, 3, 2, 3, 1176, 40, 142863, ...).  _M. F. Hasler_, Nov 24 2015
%e a(4) is not 2 since 2*2*2 + 1 = 9 is composite, but 2*3 + 1 = 7, 2*2*3 + 1 = 13, 1*2*2*3 + 1 = 13 are all prime, hence a(4) = 3.
%o (PARI) A096100(n, show=0, a=[1])={for(n=1, n1, show&&print1(a[n]", "); for(k=2, 9e9, my(p=1); for(i=0,n1,isprime(1+k*p*=a[ni])next(2)); a=concat(a,k); break)); a[n]} \\ Use 2nd or 3rd optional arg to print intermediate terms or to use other starting value(s) of the sequence. Not efficient enough to go beyond a(8).  _M. F. Hasler_, Nov 24 2015
%K more,nonn
%O 1,2
%A _Amarnath Murthy_, Jun 24 2004
%E Edited, corrected and extended by _Klaus Brockhaus_, Jul 05 2004
%E a(10) from _Donovan Johnson_, Apr 22 2008
%E a(11)a(13) from _Don Reble_, Nov 22 2015
