A329564 For all n >= 0, exactly five sums are prime among a(n+i) + a(n+j), 0 <= i < j < 5; lexicographically earliest such sequence of distinct nonnegative numbers.

0, 1, 2, 3, 6, 5, 8, 11, 7, 12, 29, 18, 19, 4, 13, 9, 22, 10, 21, 14, 57, 16, 15, 17, 26, 27, 20, 23, 33, 34, 38, 45, 25, 28, 51, 46, 31, 43, 58, 30, 24, 37, 49, 35, 36, 102, 47, 42, 55, 32, 41, 48, 65, 39, 62, 44, 40, 63, 69, 50, 68, 59, 80, 71, 54, 77, 60, 53, 56, 74, 75

Offset: 0

Author

M. F. Hasler, Feb 09 2020

Keywords

• nonn

Comments

That is, there are 5 primes, counted with multiplicity, among the 10 pairwise sums of any 5 consecutive terms.

Conjectured to be a permutation of the nonnegative integers.

If so, then the restriction to [1..oo) is a permutation of the positive integers, but not the smallest such, which is given in A329563. It seems that the two sequences have no common terms beyond a(6) = 8, except for the accidental a(22) = 15 and maybe some later coincidences of this type. There also appears to be no other simple relation between the terms of these sequences, in contrast to, e.g., A055265 vs. A128280. - M. F. Hasler, Feb 12 2020

Examples

			For n = 0, we consider pairwise sums among the first 5 terms a(0..4), among which we must have 5 primes. To get a(4), consider first a(0..3) = (0, 1, 2, 3) and the pairwise sums (a(i) + a(j), 0 <= i < j <= 3) = (1; 2, 3; 3, 4, 5) among which there are 4 primes, counted with multiplicity (i.e., the prime 3 is there two times). So the additional term a(4) must give exactly one more prime sum with all of a(0..3). We find that 4 or 5 would give two more primes, but a(4) = 6 gives exactly one more, 1 + 6 = 7.
Now, for n = 1 we forget the initial 0 and consider the pairwise sums of the remaining terms {1, 2, 3, 6}. There are 3 prime sums, so the next term must give two more. The term 4 would give two more (1+4 and 3+4) primes, but thereafter we would have {2, 3, 6, 4} with only 2 prime sums and impossibility to add one term to get three more prime sums: 2+x, 6+x and 4+x can't be all prime for x > 1.
Therefore 4 isn't the next term, and we try a(5) = 5 which indeed gives the required number of primes, and also allows us to continue.
		

Links

• M. F. Hasler, Prime sums from neighboring terms, OEIS wiki, Nov. 23, 2019

Crossrefs

Cf. A329425 (6 primes using 5 consecutive terms).

Cf. A055266 & A253074 (0 primes using 2 terms), A329405 & A329450 (0 primes using 3 terms), A055265 & A128280 (1 prime using 2 terms), A329333, A329406 - A329410 (1 prime using 3, ..., 10 terms), A329411 - A329416 and A329452, A329453 (2 primes using 3, ..., 10 terms), A329454 & A329455 (3 primes using 4 resp. 5 terms), A329449 & A329456 (4 primes using 4 resp. 5 terms), A329568 & A329569 (9 primes using 6 terms), A329572 & A329573 (12 primes using 7 terms), A329563 - A329581: other variants.

Programs

PARI

{A329564(n,show=1,o=0,N=5,M=4,X=[[4,4]],p=[],u,U)=for(n=o,n-1, 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