Search: a329413 -id:a329413
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A329453
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There are exactly two primes in {a(n+i) + a(n+j), 0 <= i < j <= 4} for any n: lexicographically earliest such sequence of distinct nonnegative integers.
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+10
18
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0, 1, 2, 8, 14, 4, 11, 6, 12, 10, 15, 5, 3, 7, 22, 9, 13, 17, 16, 18, 32, 21, 24, 20, 25, 19, 27, 23, 28, 26, 30, 29, 35, 34, 31, 36, 41, 33, 37, 40, 39, 45, 38, 42, 47, 43, 46, 44, 50, 54, 51, 49, 61, 53, 56, 57, 55, 59, 58, 68, 60, 48, 69, 62, 67, 64, 52, 63, 65, 66, 71, 70, 75, 73, 76, 72, 80, 78, 77, 81, 74
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OFFSET
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0,3
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COMMENTS
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That is, there are exactly two primes among the 10 pairwise sums of any five consecutive terms.
Conjectured to be a permutation of the nonnegative numbers. (Therefore the offset is chosen to have a(0) = 0. The restriction to positive indices is then a permutation of the positive integers, but not the smallest one which is A329413, conjectured.)
a(10^6) = 1000009 and all numbers below 999993 have appeared at that point.
Concerning the existence of the sequence, if the sequence is to be computed in a greedy manner, this means the following: for given n, we assume given P(n) := {a(n-1), a(n-2), a(n-3), a(n-4)} and thus S(n) := #{ primes x + y with x, y in P(n), x < y} which may equal 0, 1 or 2. We have to find a(n) such that we have exactly 2 - S(n) primes in a(n) + P(n). It is easy to prove that this is possible when 2 - S(n) is 0 or 1. When S(n) = 0, we must find two primes which are at a distance of |x - y| for some x, y in P(n). This is much weaker than to require the existence of twin primes or cousin primes etc., generally believed to hold and in part proved under hypotheses weaker than RH: First, we have the choice of several distances. Second, we don't require that there be no other primes in between. But more than that, it is not at all needed that the sequence be computable in a greedy manner! I.e., in the extremely improbable event that for some n with S(n) = 0 there might be no a(n) such that a(n) + P(n) contains 2 primes, we have infinitely many other choices for a(n-1) or even a(n-2), etc.! Given this additional freedom, the existence of a(n) for any n is beyond any doubt (and maybe not even difficult to prove, by contradiction).
Concerning the conjecture that all numbers will eventually occur: if a number m never appears, this means that m + P(n) never has the required number of 2 - S(n) primes. That is, for all n such that S(n) = 2, the set m + P(n) has at least one prime, and whenever S(n) = 1, the set m + P(n) has never exactly 1 prime. Given that each of the sets P(n) contains 4 numbers which were chosen essentially independently of m, it appears extremely improbable that all these infinitely many constraints could hold simultaneously for any m. Computational results so far also give only strong evidence in favor of this conjecture.
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LINKS
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EXAMPLE
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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 not generate any further prime. Given 0 and 1, primes and (primes - 1) are excluded, and a(3) = 8 is the smallest possible choice.
Then there are still two primes among the pairwise sums using {0, 1, 2, 8}: again, the next term must not produce any additional prime as sum with these. We find that a(4) = 14 is the smallest possibility.
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PROG
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(PARI) A329453(n, show=0, o=0, N=2, M=4, 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 (A329413), N, M: get N primes using M+1 consecutive terms.
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CROSSREFS
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Cf. A329413 (analog for positive integers), A329452 (2 primes among a(n+i)+a(n+j), 0 <= i < j < 4).
Cf. A329333 (1 odd prime among a(n+i)+a(n+j), 0 <= i < j < 3), A329450 (no primes among a(n+i)+a(n+j), 0 <= i < j < 3).
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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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A329456
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For any n >= 0, exactly four sums a(n+i) + a(n+j) are prime, for 0 <= i < j <= 4: lexicographically earliest such sequence of distinct nonnegative integers.
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+10
13
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0, 1, 2, 3, 24, 4, 5, 7, 8, 6, 9, 10, 11, 13, 18, 12, 16, 19, 29, 25, 42, 14, 15, 17, 20, 21, 22, 23, 26, 38, 45, 27, 28, 33, 40, 32, 31, 39, 30, 41, 48, 49, 36, 35, 34, 37, 43, 66, 47, 50, 46, 51, 52, 53, 55, 54, 44, 56, 83, 63, 59, 68, 64, 67, 72, 85, 57, 70, 79, 78, 58, 60, 61, 121, 76, 71, 90, 73
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OFFSET
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0,3
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COMMENTS
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That is, there are exactly four primes (counted with multiplicity) among the 10 pairwise sums of any five consecutive terms. (It is possible to have 4 primes among the pairwise sums of any 4 consecutive elements, see A329449.)
This map is defined with offset 0 so as to have a permutation of the nonnegative integers in case each of these eventually appears, which is not yet proved (cf. below). The restriction to positive indices would then be a permutation of the positive integers with the same property, but not the lexicographically earliest such, which starts (1, 2, 3, 4, 23, 8, 5, 6, 10, 7, 9, 11, 12, ...).
Concerning the existence of the sequence with infinite length: If the sequence is to be computed in a greedy manner, this means that for given P(n) := {a(n-1), a(n-2), a(n-3), a(n-4)} and thus N(n) := #{ primes x + y with x, y in P(n), x < y} in {0, ..., 4}, we have to find a(n) such that we have exactly 4 - N(n) primes in a(n) + N(n). It is easy to prove that this is always possible when 4 - N(n) = 0 or 1. Otherwise, similar to A329452, ..., A329455, we see that P(n) is an "admissible constellation" in the sense that a(n-5) + P(n) already gave the number of primes required now. So a (weaker) variant of the k-tuple conjecture ensures we can find this a(n). But the sequence need not be computable in greedy manner! That is, if ever for given P(n) no convenient a(n) would exist, this just means that the considered value of a(n-1) (and possibly a(n-2)) was incorrect, and the next larger choice has to be made. Given this freedom, there is no doubt that this sequence is well defined up to infinity.
Concerning surjectivity: If a number m would never appear, this means that m + P(n) will never have the required number of 4 - N(n) primes for all n with a(n) > m, in spite of having found for each of these n at least two other solutions, a(n-4) + P(n) and a(n) + P(n) which both gave 4 - N(n) primes. This appears extremely unlikely and thus as strong evidence in favor of surjectivity.
See examples for further computational evidence.
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LINKS
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EXAMPLE
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We start with a(0) = 0, a(1) = 1, a(2) = 2, a(3) = 3, the smallest possibilities which do not lead to a contradiction. Indeed, the four sums 0 + 2, 0 + 3, 1 + 2 and 2 + 3 are prime.
Now the next term must not give an additional prime when added to any of {0, 1, 2, 3}. We find that a(4) = 24 is the smallest possible choice.
Then there are 2 primes (1+2, 2+3) among the pairwise sums using {1, 2, 3, 24}, so the next term must produce two more prime sums. We find that a(5) = 4 is correct, with 1+4 and 3+4.
a(10^5) = 99948.
a(10^6) = 999923 and all numbers below 999904 occurred by then.
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PROG
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(PARI) A329455(n, show=0, o=0, N=4, M=4, p=[], U, u=o)={for(n=o, n-1, show>0&& 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])); show&&print([u]); o} \\ Optional args: show=1: print a(o..n-1), show=-1: print only [least unused number] at the end; o=1: start with a(1)=1; N, M: get N primes using M+1 consecutive terms.
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CROSSREFS
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Other sequences with N primes among pairwise sums of M consecutive terms, starting with a(o) = o, sorted by decreasing N: A329581 (N=11, M=8, o=0), A329580 (N=10, M=8, o=0), A329579 (N=9, M=7, o=0), A329577 (N=7, M=7, o=0), A329566 (N=6, M=6, o=0), A329449 (N=4, M=4, o=0), this A329456 (N=4, M=5, o=0), A329454 (3, 4, 0), A329455 (3, 5, 0), A329411 (2, 3, o=1 and 0), A329452 (2, 4, 0), A329412 (2, 4, 1), A329453 (2, 5, 0), A329413 (2, 5, 1), A329333 (N=1, M=3, o=0 and 1), A329450 (0, 3, 0), A329405 (0, 3, 1).
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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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