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A329414
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Lexicographically earliest sequence of distinct positive numbers such that among the pairwise sums of any six consecutive terms there are exactly two prime sums.
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1
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1, 2, 3, 7, 13, 19, 5, 8, 9, 17, 16, 40, 4, 6, 11, 12, 10, 14, 22, 18, 15, 20, 24, 26, 25, 29, 28, 52, 30, 35, 21, 23, 33, 31, 32, 27, 39, 37, 38, 43, 36, 48, 44, 46, 34, 45, 42, 50, 41, 54, 49, 69, 51, 47, 57, 60, 53, 55, 59, 58, 61, 56, 66, 65, 63, 67, 62, 78, 68, 70, 64, 71, 72, 73, 75, 81, 82, 80
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OFFSET
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1,2
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COMMENTS
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Condition a(1) = 1 follows from minimality. Conjectured to be a permutation of the positive integers: a(10^6) = 999994 and all numbers up to there have appeared at that point. - M. F. Hasler, Nov 15 2019
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LINKS
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EXAMPLE
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a(1) = 1 by minimality.
a(2) = 2 as 2 is the smallest available integer not leading to a contradiction. Note that as 1 + 2 = 3 we already have one prime sum (on the required two) with the sextuplet {1,2,a(3),a(4),a(5),a(6)}.
a(3) = 3 as 3 is the smallest available integer not leading to a contradiction. Note that as 2 + 3 = 5 we now have the two prime sums required with the sextuplet {1,2,3,a(4),a(5),a(6)}.
a(4) = 7 as a(4) = 4, 5 or 6 would lead to a contradiction: indeed, the sextuplets {1,2,3,4,a(5),a(6)}, {1,2,3,5,a(5),a(6)} and {1,2,3,6,a(5),a(6)} will produce more than the two required prime sums. With a(4) = 7 we have no contradiction as the sextuplet {1,2,3,7,a(5),a(6)} has now exactly two prime sums: 1 + 2 = 3 and 2 + 3 = 5.
a(5) = 13 as a(5) = 4, 5, 6, 8, 9, 10, 11 or 12 would again lead to a contradiction (more than 2 prime sums with the sextuplet); in combination with any other term before it, a(5) = 13 will produce only composite sums.
a(6) = 19 as 19 is the smallest available integer not leading to a contradiction: indeed, the sextuplet {1,2,3,7,13,19} shows exactly the two prime sums we are looking for: 1 + 2 = 3 and 2 + 3 = 5.
a(7) = 5 as 5 is the smallest available integer not leading to a contradiction; indeed, the sextuplet {2,3,7,13,19,5} shows exactly two prime sums, which are 2 + 3 = 5 and 2 + 5 = 7.
And so on.
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PROG
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(PARI) A329414(n, show=0, o=1, N=2, M=5, 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])); print([u]); o} \\ Optional args: show=1: print terms a(o..n-1); o=0: start with a(0)=0; N, M: produce N primes using M+1 consecutive terms. - M. F. Hasler, Nov 15 2019
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CROSSREFS
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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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