A329415 Lexicographically earliest sequence of distinct positive numbers such that among the pairwise sums of any seven consecutive terms there are exactly two prime sums.

1, 2, 3, 7, 13, 19, 23, 5, 20, 9, 15, 25, 29, 10, 35, 21, 27, 11, 17, 28, 14, 37, 38, 4, 8, 6, 12, 18, 31, 24, 26, 32, 16, 22, 34, 33, 30, 40, 36, 42, 44, 45, 52, 46, 48, 50, 54, 39, 41, 69, 67, 43, 47, 56, 49, 51, 55, 59, 61, 53, 60, 62, 58, 65, 57, 63, 64, 71, 70, 77, 83, 72, 73, 76, 82, 88, 68, 66, 74

Offset: 1

Author

Eric Angelini and Jean-Marc Falcoz, Nov 14 2019

Keywords

• nonn

Comments

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

Examples

			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 septet {1,2,a(3),a(4),a(5),a(6),a(7)}.
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 septet {1,2,3,a(4),a(5),a(6),a(7)}.
a(4) = 7 as a(4) = 4, 5 or 6 would lead to a contradiction: indeed, the septets {1,2,3,4,a(5),a(6),a(7)}, {1,2,3,5,a(5),a(6),a(7)} and {1,2,3,6,a(5),a(6),a(7)} will produce more than the two required prime sums. With a(4) = 7 we have no contradiction as the septet {1,2,3,7,a(5),a(6),a(7)} has two prime sums so far: 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 septet); 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 septet {1,2,3,7,13,19,a(7)} shows two prime sums so far: 1 + 2 = 3 and 2 + 3 = 5.
a(7) = 23 as 23 is the smallest available integer not leading to a contradiction; indeed, the septet {1,2,3,7,13,19,23} shows now exactly two prime sums, which are 1 + 2 = 3 and 2 + 3 = 5.
a(8) = 5 as 5 is the smallest available integer not leading to a contradiction and producing exactly two prime sums with the septet {2,3,7,13,19,23,5}, which are 2 + 3 = 5 and 2 + 5 = 7.
And so on.
		

Links

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

Crossrefs

Cf. A329333 (3 consecutive terms, exactly 1 prime sum). See also A329450, A329452 onwards.

Programs

PARI

A329415(n, show=0, o=1, N=2, M=6, 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(#pM. F. Hasler, Nov 15 2019