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<DIV>I think this sequence is a total loss.</DIV>
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<DIV>%I A072538<BR>%S A072538 2,3,5,11,13,17,19,23,29<BR>%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).<BR>%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.<BR>%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<BR>%Y A072538 Cf. A072536, A072537, A062391.<BR>%K A072538 hard,more,nonn<BR>%O A072538 1,1<BR>%A A072538 Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Aug 03 2002</DIV>
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<DIV>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.</DIV>
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<DIV>Does anybody see another way to interpret this that makes sense?</DIV>
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<DIV>Franklin T. Adams-Watters<BR></DIV></DIV></DIV>
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