A072538

[franktaw at netscape.net]

I think this sequence is a total loss.
 
%I A072538
%S A072538 2,3,5,11,13,17,19,23,29
%N A072538 a(1) = 2, a(2) = 3 and a(n) = the smallest prime more than the previous term which is a linear combination of all previous terms. ( with nonzero coefficients whose modulus is smaller than its counterpart (the corresponding prime).
%C A072538 For the fourth term, 7 can not be so obtained as the coefficient of 2 could be either 1 or -1 hence the combination of 3 and 5 are required to give 5 or 9 which is impossible. Conjecture: 7 is the only prime which is not a member of this sequence.
%e A072538 17 is a member as 17 = 13 -11 + 3*3 + 3*2, the coefficients being 1,-1,3, and 3.=  for previous members (primes) 13, 11, 3 and 2. 7 is not a member as 7 = 10*2 - 3 - 2*5 but 10 > 2
%Y A072538 Cf. A072536, A072537, A062391.
%K A072538 hard,more,nonn
%O A072538 1,1
%A A072538 Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Aug 03 2002
 
First of all, the claim that 7 is not in the sequence is incorrect: 7 = 3*5 - 2*3 - 2.  Skipping over the incorrect example, once we get to 7, every number up to several times p is obtainable from linear combinations with the specified constraints; as there is always another prime less than 2p, the sequence will contain all primes.
 
Does anybody see another way to interpret this that makes sense?
 
Franklin T. Adams-Watters
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