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A329567
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For all n >= 0, exactly seven sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6: lexicographically earliest such sequence of distinct nonnegative numbers.
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1
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0, 1, 2, 3, 4, 8, 5, 11, 26, 15, 9, 32, 14, 17, 20, 21, 27, 10, 16, 19, 7, 12, 13, 24, 6, 23, 35, 25, 37, 18, 36, 22, 31, 61, 28, 30, 39, 40, 43, 33, 64, 38, 45, 34, 29, 63, 50, 44, 53, 42, 59, 47, 54, 48, 41, 90, 49, 55, 52, 108, 58, 46, 51, 121, 73, 78, 76, 100, 79, 81, 151, 60, 67, 112, 70, 69, 82, 62, 87, 57, 80, 111, 56, 71, 66, 68, 86, 83, 65
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OFFSET
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0,3
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COMMENTS
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That is, there are 7 primes, counted with multiplicity, among the 15 pairwise sums of any 6 consecutive terms.
Conjectured to be a permutation of the nonnegative integers. The restriction to [1,oo) is then a permutation of the positive integers with the same property, but not the smallest one which is A329576 = (1, 2, 3, 4, 5, 8, 11, ...).
For n > 5, a(n) is the smallest number not used earlier such that the set a(n) + {a(n-5}, ..., a(n-1)} has the same number of primes as a(n-6) + {a(n-5), ..., a(n-1)}. Such a number always exists, by definition of the sequence. (If it would not exist for a given n, this means the term a(n-1) (or earlier) "was wrong and must be corrected", so to say. Of course this only refers to an incorrect computation.)
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LINKS
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EXAMPLE
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Using the smallest possible 5 initial terms a(0..4) = (0, 1, 2, 3, 4), we have a total of 6 primes among the pairwise sums, namely 0+2, 0+3, 1+2, 1+4, 2+3 and 3+4. To satisfy the definition for n = 0, the next term a(5) must give exactly one more prime when added to these 5 initial terms. The smallest number with this property is 6, but this choice of a(5) would make it impossible to find a suitable a(7): Indeed, for n = 1 we must consider the pairwise sums of (1, 2, 3, 4, a(5), a(6)). If we had (1, 2, 3, 4, 6, a(6)), the first 5 terms would give 5 prime sums 1+2, 1+4, 1+6, 2+3 and 3+4. Then a(6) should give two more prime sums, which is easily possible, either with even a(6) such that 1+a(6) and 3+a(6) are prime, or odd a(6) such that two among {2, 4, 6} + a(6) are prime. Thereafter, for n = 2, we drop the 1 and include a(7) instead, which must produce the same number of prime sums when added to {2, 3, 4, 6, a(6)} as was the case for 1. For even a(6) this was 4 (1+2, 1+4, 1+6 and 1+a(6)), which is impossible to achieve with a(7) > 1, since 2+x, 4+x and 6+x can't be all prime for x > 1. For odd a(6) it is 3 (1+2, 1+4 and 1+6), which is also impossible as well for odd a(7) (same reason as before) as for even a(7) (since only 3 and a(6) are odd and can give a prime sum).
This shows that we can't take a(5) equal to 6, and must consider the next larger possibility, which is a(5) = 8 (with prime sum 3+8 = 11, while 7 would give more than one, 0+7 and 4+7). Now we find that the smallest possible a(6) = 5 yields a solution and all subsequent terms can also be chosen greedily.
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PROG
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(PARI) {A329567(n, show=1, o=0, N=7, M=5, X=[[4, 6]], p=[], u=o, U)=for(n=o, n-1, show>0&& print1(o", "); show<0&& listput(L, 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|| setsearch(X, [n, k])|| [o=k, break])); show&&print([u]); o} \\ optional args: show=1: print a(o..n-1), show=-1: append them on global list L, in both cases print [least unused number] at the end. See the wiki page for a function S() which returns a vector: a(0..n-1) = S(7, 6; 0).
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CROSSREFS
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Cf. related sequences with N prime sums using M consecutive terms, labeled (N,M): A329425 (6,5), A329566 (6,6), A329449 (4,4), A329456 (4,5), A329454 & A329416 (3,4), A329455 (3,5), A329411 (2,3), A329452 (2,4), A329453 (2,5), A329333 (1,3), A128280 & A055265 (1,2); A055266 & A253074 (0,2), A329405 & A329450 (0,3), A329406 - A329416: (1,4) ... (2,10).
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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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