
COMMENTS

If 2n+1 is 1 or a prime, set a(n) = 2n+1. If no prime is ever reached, set a(n) = 1.
Comment from Klaus Brockhaus, Aug 01 2007:
"The sequence, starting from the beginning but writing 0 when a number with more than 90 digits is reached, is:
1,3,5,7,3,11,13,53,17,19,73,23,5,313,29,31,113,0,37,197,41,43,0,47,7,173,53,0,193,
59,61,0,0,67,233,71,73,0,391393,79,9313991471335749211973,83,0,293,89,137,313,
0,97,0,101,103,0,107,109,373,113,0,391393,593,11,1993,0,127,433,131,197,0,137,139,
0,0,0,0,149,151,511793,0,157,1141931201,6113,163,0,167,13,0,173,0,593,179,181,
613,12575251553,0,0,191,193,0,197,199,673,..."
Contribution from Andrew Carter (acarter09(AT)newarka.edu), Dec 16 2008: (Start)
If this sequence were to do all terms, all even terms except 2 and 4 would be 1:
a simple proof is that an even number's smallest proper divisor is 2,
all even numbers except 2 and 4 have factors in front so the resulting number would be 10n+2 and n is an integer greater than 1,
and therefore an even number endng in 2 other than two  rinse and repeat (End)
