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A329452
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There are exactly two primes in {a(n+i) + a(n+j), 0 <= i < j <= 3} for any n: lexicographically earliest such sequence of distinct nonnegative integers.
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29
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0, 1, 2, 8, 4, 5, 6, 3, 7, 11, 10, 9, 12, 13, 28, 15, 17, 16, 20, 14, 21, 22, 19, 23, 25, 24, 29, 30, 26, 18, 35, 31, 32, 27, 34, 36, 33, 38, 37, 40, 63, 39, 41, 44, 42, 45, 47, 50, 51, 43, 52, 49, 46, 48, 53, 54, 57, 55, 56, 58, 69, 62, 59, 65, 66, 61, 60, 67, 64, 68, 70, 81, 72, 76, 73, 75, 71
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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 6 pairwise sums of any four consecutive terms.
Conjectured to be a permutation of the nonnegative numbers.
a(100) = 97, a(1000) = 1001, a(10^4) = 9997, a(10^5) = 10^5, a(10^6) = 999984 and all numbers below 999963 have appeared at that point.
See the wiki page for considerations about existence and surjectivity of the sequence and variants thereof.
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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.
Now there is only one prime, 1 + 2 = 3, among the pairwise sums using {1, 2, 8}; the next term must produce exactly one additional prime as sum with these. We see that 3 is not possible (2 + 3 = 5 and 8 + 3 = 11), but a(4) = 4 is possible.
Now using {2, 8, 4} we have no prime as a pairwise sum, so the next term must produce two primes among the sums with these terms. Again, 3 would give three primes, but 5 yields exactly two primes, 2 + 5 = 7 and 8 + 5 = 13.
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PROG
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(PARI) A329452(n, show=0, o=0, 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(2<#p, p[^1], p), o); my(c=2-sum(i=2, #p, sum(j=1, i-1, isprime(p[i]+p[j])))); if(#p<3, o=u; next); for(k=u, oo, bittest(U, k-u) || sum(i=1, #p, isprime(p[i]+k))!=c || [o=k, break])); print([u]); o} \\ Optional args: show=1: print a(o..n-1); o=1: use indices & terms >= 1, i.e., compute A329412. See the wiki page for more general code returning a vector: S(n, 2, 4) = a(0..n-1).
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CROSSREFS
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Cf. A329412 (analog for positive integers), A329453 (2 primes in a(n+i)+a(n+j), i < j < 5).
Cf. A329333 (one odd prime among a(n+i)+a(n+j), 0 <= i < j < 3), A329450 (no prime in a(n+i)+a(n+j), i < j < 3).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Edited (deleted comments now found on the wiki) by M. F. Hasler, Nov 24 2019
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STATUS
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approved
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