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A329453
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There are exactly two primes in {a(n+i) + a(n+j), 0 <= i < j <= 4} for any n: lexicographically earliest such sequence of distinct nonnegative integers.
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18
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0, 1, 2, 8, 14, 4, 11, 6, 12, 10, 15, 5, 3, 7, 22, 9, 13, 17, 16, 18, 32, 21, 24, 20, 25, 19, 27, 23, 28, 26, 30, 29, 35, 34, 31, 36, 41, 33, 37, 40, 39, 45, 38, 42, 47, 43, 46, 44, 50, 54, 51, 49, 61, 53, 56, 57, 55, 59, 58, 68, 60, 48, 69, 62, 67, 64, 52, 63, 65, 66, 71, 70, 75, 73, 76, 72, 80, 78, 77, 81, 74
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OFFSET
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0,3
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COMMENTS
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That is, there are exactly two primes among the 10 pairwise sums of any five consecutive terms.
Conjectured to be a permutation of the nonnegative numbers. (Therefore the offset is chosen to have a(0) = 0. The restriction to positive indices is then a permutation of the positive integers, but not the smallest one which is A329413, conjectured.)
a(10^6) = 1000009 and all numbers below 999993 have appeared at that point.
Concerning the existence of the sequence, if the sequence is to be computed in a greedy manner, this means the following: for given n, we assume given P(n) := {a(n-1), a(n-2), a(n-3), a(n-4)} and thus S(n) := #{ primes x + y with x, y in P(n), x < y} which may equal 0, 1 or 2. We have to find a(n) such that we have exactly 2 - S(n) primes in a(n) + P(n). It is easy to prove that this is possible when 2 - S(n) is 0 or 1. When S(n) = 0, we must find two primes which are at a distance of |x - y| for some x, y in P(n). This is much weaker than to require the existence of twin primes or cousin primes etc., generally believed to hold and in part proved under hypotheses weaker than RH: First, we have the choice of several distances. Second, we don't require that there be no other primes in between. But more than that, it is not at all needed that the sequence be computable in a greedy manner! I.e., in the extremely improbable event that for some n with S(n) = 0 there might be no a(n) such that a(n) + P(n) contains 2 primes, we have infinitely many other choices for a(n-1) or even a(n-2), etc.! Given this additional freedom, the existence of a(n) for any n is beyond any doubt (and maybe not even difficult to prove, by contradiction).
Concerning the conjecture that all numbers will eventually occur: if a number m never appears, this means that m + P(n) never has the required number of 2 - S(n) primes. That is, for all n such that S(n) = 2, the set m + P(n) has at least one prime, and whenever S(n) = 1, the set m + P(n) has never exactly 1 prime. Given that each of the sets P(n) contains 4 numbers which were chosen essentially independently of m, it appears extremely improbable that all these infinitely many constraints could hold simultaneously for any m. Computational results so far also give only strong evidence in favor of this conjecture.
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LINKS
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EXAMPLE
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We start with a(0) = 0, a(1) = 1, a(2) = 2, the smallest possibilities which do not lead to a contradiction.
Now there are already 2 primes, 0 + 2 and 1 + 2, among the pairwise sums, so the next term must not generate any further prime. Given 0 and 1, primes and (primes - 1) are excluded, and a(3) = 8 is the smallest possible choice.
Then there are still two primes among the pairwise sums using {0, 1, 2, 8}: again, the next term must not produce any additional prime as sum with these. We find that a(4) = 14 is the smallest possibility.
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PROG
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(PARI) A329453(n, show=0, o=0, N=2, M=4, p=[], U, u=o)={for(n=o, n-1, show&& print1(o", "); U+=1<<(o-u); U>>=-u+u+=valuation(U+1, 2); p=concat(if(#p>=M, p[^1], p), o); my(c=N-sum(i=2, #p, sum(j=1, i-1, isprime(p[i]+p[j])))); if(#p<M && sum(i=1, #p, isprime(p[i]+u))<=c, o=u)|| for(k=u, oo, bittest(U, k-u) || sum(i=1, #p, isprime(p[i]+k))!=c || [o=k, break])); o} \\ Optional args: show=1: print a(o..n-1); o=1: start with a(1)=1 (A329413), N, M: get N primes using M+1 consecutive terms.
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CROSSREFS
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Cf. A329413 (analog for positive integers), A329452 (2 primes among a(n+i)+a(n+j), 0 <= i < j < 4).
Cf. A329333 (1 odd prime among a(n+i)+a(n+j), 0 <= i < j < 3), A329450 (no primes among a(n+i)+a(n+j), 0 <= i < j < 3).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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