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A329413
Lexicographically earliest sequence of distinct positive numbers such that among the pairwise sums of any five consecutive terms there are exactly two prime sums.
3
1, 2, 3, 7, 13, 5, 8, 9, 17, 16, 4, 6, 11, 12, 10, 14, 15, 21, 18, 19, 20, 30, 22, 24, 29, 26, 23, 25, 36, 32, 33, 27, 28, 37, 31, 39, 35, 34, 38, 42, 44, 41, 40, 43, 45, 46, 47, 50, 52, 49, 65, 53, 51, 54, 55, 57, 48, 60, 56, 59, 61, 71, 70, 67, 58, 64, 62, 63, 68, 66, 73, 72, 69, 76, 75, 74, 78, 80
OFFSET
1,2
COMMENTS
Conjectured to be a permutation of the positive integers: a(10^6) = 10^6 + 9 and all numbers below 10^6 - 7 are used at that point. - M. F. Hasler, Nov 15 2019
LINKS
EXAMPLE
a(1) = 1 is the smallest possible choice for the first term.
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 5-tuple {1,2,a(3),a(4),a(5)}.
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 5-tuple {1,2,3,a(4),a(5)}.
a(4) = 7 as a(4) = 4, 5 or 6 would lead to a contradiction: indeed, the quintuplets {1,2,3,4,a(5)}, {1,2,3,5,a(5)} and {1,2,3,6,a(5)} will produce more than the two required prime sums. With a(4) = 7 we have no contradiction as the 5-tuple {1,2,3,7,a(5)} 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 5-tuple); in combination with any other term before it, a(5) = 13 will produce only composite sums.
a(6) = 5 as 5 is the smallest available integer not leading to a contradiction: indeed, the 5-tuple {2,3,7,13,5} shows exactly the two prime sums we are looking for: 2 + 3 = 5 and 2 + 5 = 7.
And so on.
PROG
(PARI) A329413(n, show=0, o=1, 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=0: start with a(0) = 0 (A329453), N, M: produce N primes using M+1 consecutive terms. - M. F. Hasler, Nov 15 2019
CROSSREFS
Cf. A329333 (3 consecutive terms, exactly 1 prime sum). See also A329450, A329452 onwards.
Sequence in context: A361910 A068134 A249051 * A225093 A278007 A081256
KEYWORD
nonn
AUTHOR
STATUS
approved