A329412 Lexicographically earliest sequence of distinct positive numbers such that among the pairwise sums of any four consecutive terms there are exactly two prime sums.

1, 2, 3, 7, 5, 4, 8, 6, 9, 10, 11, 12, 13, 16, 15, 17, 20, 21, 18, 19, 23, 26, 24, 14, 27, 32, 22, 25, 28, 29, 30, 33, 31, 34, 36, 35, 38, 39, 40, 41, 42, 37, 43, 47, 46, 50, 53, 44, 49, 45, 48, 54, 55, 51, 56, 57, 59, 52, 61, 66, 58, 65, 62, 72, 60, 67, 63, 68, 69, 73, 64, 70, 75, 74, 76, 71, 105

Offset: 1

Author

Eric Angelini and Jean-Marc Falcoz, Nov 14 2019

Keywords

• nonn

Comments

Conjectured to be a permutation of the positive integers: a(10^5) = 10^5, a(10^6) = 999984 and all numbers below 99992 resp. 999963 have appeared by then. See A329452 for a more detailed discussion. - M. F. Hasler, Nov 15 2019

Examples

			a(1) = 1 is the smallest possible choice; there are no other restrictions so far.
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 (of the required two) with the quadruplet {1, 2, a(3), a(4)}.
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 quadruplet {1,2,5,a(4)}.
a(4) = 7 as a(4) = 4, 5 or 6 would lead to a contradiction: indeed, both the quadruplets {1, 2, 3, 4}, {1, 2, 3, 5} and {1, 2, 3, 6} will produce three prime sums (instead of two). With a(4) = 7 we have the quadruplet {1, 2, 3, 7} and the two prime sums we are looking for: 1 + 2 = 3 and 2 + 3 = 5.
a(5) = 5 as a(5) = 4 would again lead to a contradiction: indeed, the quadruplet {2, 3, 7, 4} will produce three prime sums (instead of two, they would be 2 + 3 = 5; 3 + 4 = 7 and 7 + 4 = 11). With a(5) = 5 the quadruplet {2, 3, 7, 4} shows exactly the two prime sums we are looking for: 2 + 3 = 5 and 3 + 4 = 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).

Cf. A329452 (variant using nonnegative integers).

Programs

PARI

A329412(n,show=0,o=1,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(2<#p,p[^1],p),o); my(c=2-sum(i=2,#p, sum(j=1,i-1,isprime(p[i]+p[j])))); if(#p<3, o=u; next); 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 (A329452). - M. F. Hasler, Nov 15 2019