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.

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

Jean-Marc Falcoz, Table of n, a(n) for n = 1..10000

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

Adjacent sequences: A329410 A329411 A329412 * A329414 A329415 A329416

Keyword

nonn

Author

Eric Angelini and Jean-Marc Falcoz, Nov 14 2019

Status

approved