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A329565 For all n >= 0, exactly five sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6; lexicographically earliest such sequence of distinct nonnegative numbers.

%I A329565 #14 Feb 09 2020 22:48:27
%S A329565 0,1,2,3,6,24,4,5,8,9,10,11,7,13,12,17,16,14,15,19,22,18,21,20,26,23,
%T A329565 25,27,33,34,28,29,32,38,39,30,31,41,40,36,35,42,61,44,43,66,37,52,45,
%U A329565 47,46,51,50,57,48,49,53,55,56,59,54,58,72,95,62,65,67,63,84,64,60,68,89,71,69,73,80,78,70,79,87,76,75,74,88,77,81,82,189,85
%N A329565 For all n >= 0, exactly five sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6; lexicographically earliest such sequence of distinct nonnegative numbers.
%C A329565 That is, there are 5 primes, counted with multiplicity, among the 15 pairwise sums of any 6 consecutive terms.
%C A329565 Conjectured to be a permutation of the nonnegative integers.
%C A329565 If so, then the restriction to [1..oo) is a permutation of the positive integers.
%H A329565 M. F. Hasler, <a href="/wiki/User:M._F._Hasler/Prime_sums_from_neighboring_terms">Prime sums from neighboring terms</a>, OEIS wiki, Nov. 23, 2019
%e A329565 For n = 0, we consider pairwise sums of the first 6 terms a(0..5) = (0, 1, 2, 3, 6, 24): We have (a(i) + a(j), 0 <= i < j < 6) = (1; 2, 3; 3, 4, 5; 6, 7, 8, 9; 24, 25, 26, 27, 30) among which there are 5 primes, counted with repetition. If one tries to take a(4) equal to 4 or 5, this yields already 6 primes among the pairwise sums of the first 5 terms, so the smallest possible choice is a(4) = 6, and thereafter any a(5) less than 24 would again yield too many prime sums. So (0, 1, 2, 3, 6, 24) is indeed the start of the lexicographically earliest nonnegative sequence with the required properties.
%e A329565 Then one finds that a(6) = 4 is possible, giving also 6 prime sums for n = 1, so this is the correct continuation (modulo later confirmation that the sequence can be continued without contradiction given this choice).
%e A329565 Next one finds that a(7) = 5 is also possible, and so on.
%o A329565 (PARI) {A329565(n,show=0,o=0,N=5/*#primes*/,M=5,p=[],U,u=o)=for(n=o,n-1, if(show>0,print1(o", "), show<0,listput(L,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]));show&&print([u]);o} \\ optional args: show=1: print a(o..n-1), show=-1: append them on global list L, in both cases print [least unused number] at the end; o=1: start at a(1)=1; N, M: find N primes using M+1 terms. See the wiki page for a function S() which returns a vector: a(0..n-1) = S(n,5,6).
%Y A329565 Cf. A329425 (6 primes using 5 consecutive terms), A329566 (6 primes using 6 consecutive terms).
%Y A329565 Cf. A329449 (4 primes using 4 consecutive terms), A329456 (4 primes using 5 consecutive terms).
%Y A329565 Cf. A329454 (3 primes using 4 consecutive terms), A329455 (3 primes using 5 consecutive terms).
%Y A329565 Cf. A329411 (2 primes using 3 consecutive terms), A329452 (2 primes using 4 consecutive terms), A329453 (2 primes using 5 consecutive terms).
%Y A329565 Cf. A329333 (1 (odd) prime using 3 terms), A128280 & A055265 (1 prime using 2 terms); A055266 & A253074 (0 primes using 2 terms), A329405 & A329450 (0 primes using 3 terms), A329406 ff: other variants.
%K A329565 nonn
%O A329565 0,3
%A A329565 _M. F. Hasler_, Feb 09 2020