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A329564
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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.
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2
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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
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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EXAMPLE
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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.
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PROG
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(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<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|| setsearch(X, [n, k])|| [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. See the wiki page for a function S() which returns a vector: a(0..n-1) = S(5, 5; 0).
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CROSSREFS
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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.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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