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A329454
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There are exactly three primes among a(n+i) + a(n+j), 0 <= i < j <= 3, for any n >= 0: lexicographically earliest such sequence of distinct nonnegative integers.
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15
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0, 1, 2, 4, 5, 3, 8, 6, 11, 7, 10, 12, 9, 19, 22, 14, 15, 16, 13, 18, 21, 40, 43, 20, 27, 46, 17, 26, 33, 24, 35, 38, 32, 23, 29, 30, 31, 28, 25, 34, 36, 39, 37, 64, 42, 41, 67, 47, 60, 49, 48, 52, 45, 55, 44, 58, 69, 51, 50, 62, 53, 77, 54, 56, 83, 57, 66, 74, 65, 61, 102, 70, 71, 79, 78, 59, 68, 63, 72, 95, 86, 81, 76, 73, 75, 82, 106
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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 three primes (counted with multiplicity) among the 6 pairwise sums of any four consecutive terms.
It remains unknown whether this is a permutation of the nonnegative numbers. There is some hope that this could be the case, and it seems coherent to choose offset 0 not only in view of this. The restriction to positive indices would then be a permutation of the positive integers, but not the smallest one with the given property.
Concerning the existence of the sequence: If the sequence is to be computed in a greedy manner, this means that for given n, we assume given P(n) := {a(n-1), a(n-2), a(n-3)} and thus S(n) := #{ primes x + y with x, y in P(n), x < y} which may equal 0, 1, 2 or 3. We have to find a(n) such that we have exactly 3 - S(n) primes in a(n) + P(n). It is easy to prove that this is always possible when 3 - S(n) is 0 or 1, and similar to the case of A329452 and A329453 when S(n) = 1. When S(n) = 0, however, we must find three primes in the same configuration as the elements of P(n). It is almost surprising that this can always be found up to n = 10^6. However, the sequence does not need to be computable in a greedy manner. That is, if for given P(n) no a(n) would exist such that a(n) + P(n) contains 3 - S(n) primes, this simply means that the considered value of a(n-1) was incorrect, and the next larger choice has to be made. Given this freedom, well-definedness of this sequence up to infinity is far more probable than, for example, the k-tuple conjecture.
Computational results are as follows:
a(10^5) = 99954 and all numbers below 99915 have appeared at that point.
a(10^6) = 1000053 and all numbers below 999845 have appeared at that point.
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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 generate exactly one further prime. It appears that a(3) = 4 is the smallest possible choice.
Then there are again two primes among the pairwise sums using {1, 2, 4}, and the next term must again produce one additional prime as sum with these. We find that a(4) = 5 is the smallest possibility.
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MATHEMATICA
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Nest[Block[{k = 3}, While[Nand[FreeQ[#, k], Count[Subsets[Append[Take[#, -3], k], {2}], _?(PrimeQ@ Total@ # &)] == 3], k++]; Append[#, k]] &, {0, 1, 2}, 84] (* Michael De Vlieger, Nov 15 2019 *)
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PROG
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(PARI) A329454(n, show=0, o=0, N=3, M=3, 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; N, M: get N primes using M+1 consecutive terms.
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CROSSREFS
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Cf. A329452 (2 primes among a(n+i)+a(n+j), 0 <= i < j < 4), A329453 (2 primes among a(n+i)+a(n+j), 0 <= i < j < 5).
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).
Cf. A329405 ff: variants defined for positive integers.
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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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