OFFSET
0,3
COMMENTS
That is, there are exactly three primes (counted with multiplicity) among the 6 pairwise sums of any four consecutive terms.
It remains unknown whether this is a permutation of the nonnegative numbers. There is some hope that this could be the case, and it seems coherent to choose offset 0 not only in view of this. The restriction to positive indices would then be a permutation of the positive integers, but not the smallest one with the given property.
Concerning the existence of the sequence: If the sequence is to be computed in a greedy manner, this means that for given n, we assume given P(n) := {a(n-1), a(n-2), a(n-3)} and thus S(n) := #{ primes x + y with x, y in P(n), x < y} which may equal 0, 1, 2 or 3. We have to find a(n) such that we have exactly 3 - S(n) primes in a(n) + P(n). It is easy to prove that this is always possible when 3 - S(n) is 0 or 1, and similar to the case of A329452 and A329453 when S(n) = 1. When S(n) = 0, however, we must find three primes in the same configuration as the elements of P(n). It is almost surprising that this can always be found up to n = 10^6. However, the sequence does not need to be computable in a greedy manner. That is, if for given P(n) no a(n) would exist such that a(n) + P(n) contains 3 - S(n) primes, this simply means that the considered value of a(n-1) was incorrect, and the next larger choice has to be made. Given this freedom, well-definedness of this sequence up to infinity is far more probable than, for example, the k-tuple conjecture.
Computational results are as follows:
a(10^5) = 99954 and all numbers below 99915 have appeared at that point.
a(10^6) = 1000053 and all numbers below 999845 have appeared at that point.
LINKS
Eric Angelini, Prime sums from neighbouring terms, personal blog "Cinquante signes" (and post to the SeqFan list), Nov. 11, 2019.
Eric Angelini, Prime sums from neighbouring terms [Cached copy of html file, with permission]
Eric Angelini, Prime sums from neighbouring terms [Cached copy of pdf file, with permission]
EXAMPLE
We start with a(0) = 0, a(1) = 1, a(2) = 2, the smallest possibilities which do not lead to a contradiction.
Now there are already 2 primes, 0 + 2 and 1 + 2, among the pairwise sums, so the next term must generate exactly one further prime. It appears that a(3) = 4 is the smallest possible choice.
Then there are again two primes among the pairwise sums using {1, 2, 4}, and the next term must again produce one additional prime as sum with these. We find that a(4) = 5 is the smallest possibility.
MATHEMATICA
Nest[Block[{k = 3}, While[Nand[FreeQ[#, k], Count[Subsets[Append[Take[#, -3], k], {2}], _?(PrimeQ@ Total@ # &)] == 3], k++]; Append[#, k]] &, {0, 1, 2}, 84] (* Michael De Vlieger, Nov 15 2019 *)
PROG
(PARI) A329454(n, show=0, o=0, N=3, M=3, 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=1: start with a(1)=1; N, M: get N primes using M+1 consecutive terms.
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
M. F. Hasler, based on an idea from Eric Angelini, Nov 15 2019
STATUS
approved