cp's OEIS Frontend

Numbers n such that A055266(n) = n.

The sequence is well-defined (the terms must alternate in parity, and by Dirichlet's theorem a(n+1) always exists). - N. J. A. Sloane, Mar 07 2017

Does every positive integer eventually occur? - Dmitry Kamenetsky, May 27 2009. Reply from Robert G. Wilson v, May 27 2009: The answer is almost certainly yes, on probabilistic grounds.

It appears that this is the limit of the rows of A051237. That those rows do approach a limit seems certain, and given that that limit exists, that this sequence is the limit seems even more likely, but no proof is known for either conjecture. - Robert G. Wilson v, Mar 11 2011, edited by Franklin T. Adams-Watters, Mar 17 2011

The sequence is also a particular case of "among the pairwise sums of any M consecutive terms, N are prime", with M = 2, N = 1. For other M, N see A055266 & A253074 (M = 2, N = 0), A329333, A329405 - A329416, A329449 - A329456, A329563 - A329581, and the OEIS Wiki page. - M. F. Hasler, Feb 11 2020

That is, there are exactly four primes (counted with multiplicity) among the 6 pairwise sums of any four consecutive terms. This is the theoretical maximum: there can't be a sequence with more than 4 prime sums in any 4 consecutive terms, see the wiki page for details.

This map is defined with offset 0 as to have a permutation of the nonnegative integers in case each of these eventually appears, which is so far only conjectured, see below. The restriction to positive indices would then be a permutation of the positive integers, and as it happens, also the smallest one with the given property. (This is in contrast to most other cases where that one is not the restriction of the other one: see crossrefs).

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)} and thus 0 <= N(n) := #{ primes x + y with x, y in P(n), x < y} <= 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, ..., A329456, we see that P(n) is an "admissible constellation" in the sense that a(n-4) + P(n) already gave the number of primes required now. So a weaker variant of the k-tuple conjecture would ensure 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 a(n) would exist such that a(n) + P(n) contains 4 - N(n) primes, this simply 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.

About existence of this (infinite) sequence: If it is computed in greedy manner, this means that for given n we are given P(n) := {a(n-1), a(n-2)} and have to find a(n) such that we have exactly 1 or 2 primes in a(n) + P(n) depending on whether a(n-1) + a(n-2) is prime or not. It is easy to prove that this is always possible in the first case (1 prime required). In the second case, we must find two larger primes at given distance |a(n-1) - a(n-2)|, necessarily even, since a(n-3) + P(n) contains two primes. To have this infinitely many times, the twin prime conjecture or a variant thereof must hold. However, the sequence need not be computable in greedy manner! That is, if ever for given P(n) (with composite sum) no a(n) would exist such that a(n) + P(n) contains 2 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, there is no doubt about well-definedness of this sequence up to infinity. - M. F. Hasler, Nov 14 2019, edited Nov 16 2019

Could be extended to a(0) = 0 to yield a sequence of nonnegative integers with the same property, including lexicographic minimality, which is a permutation of the nonnegative integers iff this sequence is a permutation of the positive integers.

This is the first known example where the restriction of S(N,M;0) to [1..oo) gives S(N,M;1), where S(N,M;o) is the lexicographically smallest sequence with a(o)=o, N primes among pairwise sums of M consecutive terms, and no duplicate terms: For example, S(0,3;1) = A329405 is not A329450\{0}, S(2,4;1) = A329412 is not A329452\{0}, etc. The second such example is S(4,4;o) = A329449. - M. F. Hasler, Nov 16 2019

Differs from A055265 from a(30) = 33 on. See the wiki page for further considerations and variants. - M. F. Hasler, Nov 24 2019

Conjecture: this is a permutation of the nonnegative numbers. [Proof outline below due to Patrick Devlin and Semeon Artamonov]

Let x be a number that's missing.

Then eventually every term must be of the form PRIME - x. (Otherwise, x would appear as that next term.)

In particular, this means there are only finitely many multiples of x that appear in the sequence. Let Y be a multiple of x larger than all multiples of x appearing in the sequence.

Let q be a prime not dividing Y. Then since none of the terms Y, 2Y, 3Y, ..., 2qY appear, it must be that, eventually, every term in the sequence is of the form PRIME - Y and also of the form PRIME - 2Y and also of the form PRIME - 3Y, ... and also of the form PRIME - 2qY.

That means we have a prime p and a number Y such that p, p+Y, p+2Y, p + 3Y, p+4Y, ..., p+2qY are all prime. But take this sequence mod q. Since q does not divide Y, the terms 0, Y, ..., 2qY cover every residue class mod q twice. Therefore, p + kY covers each residue class mod q twice. Consequently, there are two terms congruent to 0 mod q. One can be q, but the other must be a multiple of it (contradicting its primality).

Essentially the same as A055266. - R. J. Mathar, Feb 13 2015

Simplified version of the proof: Assume x isn't in the sequence, then eventually all terms must be of the form PRIME - x, else x would appear next. In particular, no multiple of x can appear from there on. Assume k*x is the largest multiple of x in the sequence. Take a prime p not dividing x. Then m*x can't appear in the sequence for k+1 <= m <= k+p, and all terms are eventually of the form PRIME - m*x for all m in {k+1, ..., k+p}. Take one such term N > p, i.e., N + (k+1)*x, ..., N + (k+p)*x are all prime. Consider this sequence mod p. Since gcd(x,p)=1, the p terms cover each residue class mod p, so one is a multiple of p, in contradiction with their primality. - M. F. Hasler, Nov 25 2019

From M. F. Hasler, Nov 09 2013: (Start)

At each step, choose the smallest number not occurring earlier and such that a(n+1)+a(n) are palindromes, for all n.

Conjectured to be a permutation of the nonnegative integers.

See A062932 where injectivity is replaced by monotonicity; the sequences differ from a(16)=15 on.

This is an "arithmetic" analog to sequences A228407 and A228410, where instead of the sum, the union of the digits of subsequent terms is considered. (End)

That is, there are 6 primes, counted with multiplicity, among the 15 pairwise sums of any 6 consecutive terms.

Is this a permutation of the nonnegative integers?

If so, then the restriction to [1..oo) is a permutation of the positive integers, but not the lexicographically earliest one with this property, which starts (1, 2, 3, 4, 5, 7, 6, 8, 9, 10, 11, 13, 18, 19, 16, 12, 24, ...).

Inspired by an idea of Eric Angelini on the Sequence Fans list on Dec 28 2011.

Comments from N. J. A. Sloane, Aug 16 2021: (Start)

It is conjectured that this is a permutation of the positive integers. Is there a proof? The terms are distinct, by definition, and the sequence is clearly infinite. But does every number appear?

In the first 100000 terms, the only differences a(i)-a(i-1) that occur are -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11 (see A346610).

Also a(n) is surprisingly close to n - see A346611. (End)

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 positive integers.

This sequence is quite different from the restriction of the "nonnegative" variant A329564 to positive indices: it seems that the two 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.