A072538

[Joshua Zucker]

Hi Franklin,
it says "modulus smaller than its counterpart" which I think means
that the coefficient of 2 is supposed to have an absolute value less
than 2 -- hence -1 or 0 or 1.  Oh, wait, it says nonzero, so it must
be -1 or 1.  Similarly the coefficient of 3 must be -2 or -1 or 1 or
2.

However that is contradicted by the example for 17, in which there are
zero 5s and too many 3s and 2s.

But 17 = 1*13 + -1*11 + 2*5 + 1*3 + 1*2 seems to work, assuming my
adding skills are OK, so I'd say that the example needs to be edited.

And 7 = 3*5 + -2 * 3 + -1 * 2, as you say, so 7 is in the sequence.
(the comment saying it's impossible to make 9 out of 3 and 5 is
wrong.)

So now that 7 is in the sequence, 17 has to be made by
17 = 1*13 - 1*11 + 1*7 + 2*5 - 2*3 + 1*2
I think?  Anyway it's still makeable.

I'd write the definition as something like
a(1) = 2, a(2) = 3, and a(n+1) = the smallest prime p of the form c1 *
a(1) + c2 * a(2) + ... + cn * a(n) with integers ci, |ci| < a(i), and
all ci nonzero.

Does that work?

But now it seems pretty clear (as you already have remarked) that
every prime will be in the sequence, so the whole thing just becomes a
comment on the sequence of primes:
every prime p is a linear combination of previous primes p(n) with
nonzero coefficients c(n) and |c(n)| < p(n).

--Joshua Zucker

On 5/17/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
>
>
>
> 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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