A329405 -id:A329405 - cp's OEIS frontend

A055265 a(n) is the smallest positive integer not already in the sequence such that a(n)+a(n-1) is prime, starting with a(1)=1.

1, 2, 3, 4, 7, 6, 5, 8, 9, 10, 13, 16, 15, 14, 17, 12, 11, 18, 19, 22, 21, 20, 23, 24, 29, 30, 31, 28, 25, 34, 27, 26, 33, 38, 35, 32, 39, 40, 43, 36, 37, 42, 41, 48, 49, 52, 45, 44, 53, 50, 47, 54, 55, 46, 51, 56, 57, 70, 61, 66, 65, 62, 69, 58, 73, 64, 63, 68, 59, 72, 67, 60

Offset: 1

Henry Bottomley, May 09 2000

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

			a(5) = 7 because 1, 2, 3 and 4 have already been used and neither 4 + 5 = 9 nor 4 + 6 = 10 are prime while 4 + 7 = 11 is prime.

Zak Seidov, Table of n, a(n) for n = 1..10000 (First 1000 terms from T. D. Noe)
N. J. A. Sloane, Table of n, a(n) for n = 1..100000 (computed using Orlovsky's Mma program)
M. F. Hasler, Prime sums from neighboring terms, OEIS Wiki, Nov. 23, 2019
Index entries for sequences that are permutations of the natural numbers

Inverse permutation: A117922; fixed points: A117925; A117923=a(a(n)). - Reinhard Zumkeller, Apr 03 2006
Cf. A036440, A051237, A051239, A055266, A088643. A010051.
Cf. A086527 (the primes a(n)+a(n-1)).
Cf. A070942 (n's such that a(1..n) is a permutation of (1..n)). - Zak Seidov, Oct 19 2011
See also A076990, A243625.
See A282695 for deviation from identity sequence.
A073659 is a version where the partial sums must be primes.

import Data.List (delete)
a055265 n = a055265_list !! (n-1)
a055265_list = 1 : f 1 [2..] where
   f x vs = g vs where
     g (w:ws) = if a010051 (x + w) == 1
                   then w : f w (delete w vs) else g ws
-- Reinhard Zumkeller, Feb 14 2013

A055265 := proc(n)
    local a,i,known ;
    option remember;
    if n =1 then
        1;
    else
        for a from 1 do
            known := false;
            for i from 1 to n-1 do
                if procname(i) = a then
                    known := true;
                    break;
                end if;
            end do:
            if not known and isprime(procname(n-1)+a) then
                return a;
            end if;
        end do:
    end if;
end proc:
seq(A055265(n),n=1..100) ; # R. J. Mathar, Feb 25 2017

f[s_List] := Block[{k = 1, a = s[[ -1]]}, While[ MemberQ[s, k] || ! PrimeQ[a + k], k++ ]; Append[s, k]]; Nest[f, {1}, 71] (* Robert G. Wilson v, May 27 2009 *)
q=2000; a={1}; z=Range[2,2*q]; While[Length[z]>q-1, k=1; While[!PrimeQ[z[[k]]+Last[a]], k++]; AppendTo[a,z[[k]]]; z=Delete[z,k]]; Print[a] (*200 times faster*) (* Vladimir Joseph Stephan Orlovsky, May 03 2011 *)

v=[1];n=1;while(n<50,if(isprime(v[#v]+n)&&!vecsearch(vecsort(v),n), v=concat(v,n);n=0);n++);v \\ Derek Orr, Jun 01 2015

U=-a=1; vector(100,k, k=valuation(1+U+=1<M. F. Hasler, Feb 11 2020

a(2n-1) = A128280(2n-1) - 1, a(2n) = A128280(2n) + 1, for all n >= 1. - M. F. Hasler, Feb 11 2020

Corrected by Hans Havermann, Sep 24 2002

A329450 Lexicographically earliest sequence of distinct nonnegative integers such that neither a(n) + a(n+1) nor a(n) + a(n+2) is prime for any n.

Original entry on oeis.org

0, 1, 8, 7, 2, 13, 12, 3, 6, 9, 15, 5, 10, 4, 11, 14, 16, 18, 17, 21, 19, 23, 25, 26, 20, 22, 24, 27, 28, 29, 34, 31, 32, 33, 30, 35, 39, 37, 38, 40, 36, 41, 44, 43, 42, 45, 46, 47, 48, 51, 54, 57, 58, 53, 52, 59, 56, 49, 50, 55, 60, 61, 62, 63, 66, 67, 68, 65, 64, 69, 71, 72, 70, 73, 74, 79, 76

Offset: 0

M. F. Hasler, based on an idea of Eric Angelini, Nov 13 2019

nonn

Equivalently: For any three consecutive terms, there is no prime among any of the pairwise sums. Or: For any n and 0 <= i < j <= 2, a(n+i) + a(n+j) is never prime.
For any n, a term a(n) which meets the requirements always exists: For any a(n-2), a(n-1), at least one in five consecutive values of k is such that one among {a(n-2) + k, a(n-1) + k} is divisible by 2 and the other one by 3.
Conjectured to be a permutation of the nonnegative integers. The restriction to positive indices is then a permutation of the positive integers with the same property, but not the lexicographically earliest given in A329405.
See the wiki page for additional considerations and other variants. - M. F. Hasler, Nov 24 2019

			After the smallest possible initial terms, a(0) = 0, a(1) = 1, the next term must be neither a prime nor a prime - 1. The smallest possibility is a(2) = 8.
The next term must not be a prime - 1 nor a prime - 8, which excludes 2, 4, 6 on one hand, and 3 and 5 on the other hand. The smallest possibility is a(3) = 7.

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

Cf. A329333 (always one odd prime among a(n+i)+a(n+j), 0 <= i < j <= 2).

Cf. A329405 (analog for positive integers).

Nest[Block[{k = 2}, While[Nand[FreeQ[#, k], ! PrimeQ[#[[-1]] + k], ! PrimeQ[#[[-2]] + k]], k++]; Append[#, k]] &, {0, 1}, 89] (* Michael De Vlieger, Nov 15 2019 *)

A329450(n,show=0,o=0,p=o,U=[])={for(n=o,n-1, show&&print1(p","); U=setunion(U,[p]); while(#U>1&&U[1]==U[2]-1, U=U[^1]); for(k=U[1]+1,oo, setsearch(U,k) || isprime(o+k) || isprime(p+k) || [o=p, p=k, break]));p} \\ Optional args: show=1: print a(o..n-1); o=1: start with a(1) = 1 (A329405). See the wiki page for more general code returning a vector: S(n,0,3) = A329450(0..n-1).

Edited by N. J. A. Sloane, Nov 14 2019

New definition corrected by M. F. Hasler, Nov 15 2019

A329449 For any n >= 0, exactly four sums a(n+i) + a(n+j) are prime, for 0 <= i < j <= 3: lexicographically earliest such sequence of distinct nonnegative integers.

Original entry on oeis.org

0, 1, 2, 3, 4, 9, 8, 15, 14, 5, 26, 17, 6, 11, 12, 7, 30, 29, 24, 13, 18, 19, 10, 43, 28, 31, 16, 25, 22, 21, 46, 37, 52, 27, 34, 45, 44, 39, 58, 69, 20, 51, 32, 41, 38, 35, 48, 23, 36, 53, 50, 47, 54, 59, 42, 55, 72, 65, 84, 67, 114, 79, 60, 49, 78, 71, 102, 61, 66, 91, 40, 73, 76, 33, 64, 63, 68

Offset: 0

M. F. Hasler, based on an idea from Eric Angelini, Nov 15 2019

nonn

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.

			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 we have 2 prime sums using {1, 2, 3}, so the next term must give two more prime when added to these. We find that a(4) = 4 is the smallest possible choice, with 1 + 4 = 5 and 3 + 4 = 7.
Then there are again 2 primes among the pairwise sums using {2, 3, 4}, so the next term must again produce two more prime sums. We find that a(5) = 9 is the smallest possibility, with 2 + 9 = 11 and 4 + 9 = 13.
a(10^4) = 9834 and all numbers up to 9834 occurred by then.
a(10^5) = 99840 and all numbers below 99777 occurred by then.
a(10^6) = 1000144 and all numbers below 999402 occurred by then.

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]
M. F. Hasler, Prime sums from neighboring terms, OEIS Wiki, Nov. 23, 2019.

Other sequences with N primes among pairwise sums of M consecutive terms, starting with a(o) = o, sorted by decreasing N and lowest possible M: A329581 (N=11, M=8, o=0), A329580 (N=10, M=8, o=0), A329569 (N=9, M=6, o=0), A329568 (N=9, M=6, o=1), A329425 (N=6, M=5, o=0), A329449 (N=4, M=4, o=0), A329411 (N=2, M=3, o=0 or 1), A128280 (N=1, M=2, o=0), A055265 (N=1, M=2, o=1), A055266 (N=0, M=2; o=1), A253074 (N=0, M=2; o=0).

For other variants see A329333 (N=1, M=3; o=0/1), A329405 (0,3;1) .. A329417 (3,4;1), A329449 (4,4;0) .. A329580 (10,8;0).

A329449(n, show=0, o=0, N=4, M=3, p=[], U, u=o)={for(n=o, n-1, if(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])))); for(k=u, oo, bittest(U, k-u) || min(c-#[0|p<-p, isprime(p+k)], #p>=M) || [o=k, break]));show&&print([u]); o} \\ Optional args: show=1: print a(o..n-1), show=-1: append a(o..n-1) to the global list L, in both cases print [least unused number] at the end; o=1: start with a(1)=1; N, M: get N primes using M+1 consecutive terms.

A128280 a(n) is the least number not occurring earlier such that a(n)+a(n-1) is prime, a(0) = 0.

Original entry on oeis.org

0, 2, 1, 4, 3, 8, 5, 6, 7, 10, 9, 14, 15, 16, 13, 18, 11, 12, 17, 20, 21, 22, 19, 24, 23, 30, 29, 32, 27, 26, 33, 28, 25, 34, 37, 36, 31, 40, 39, 44, 35, 38, 41, 42, 47, 50, 51, 46, 43, 54, 49, 48, 53, 56, 45, 52, 55, 58, 69, 62, 65, 66, 61, 70, 57, 74, 63, 64, 67, 60, 71, 68, 59

Offset: 0

Zak Seidov, May 03 2007

nonn

Original definition: start with a(1) = 2. See A055265 for start with a(1) = 1.
The sequence may well be a rearrangement of natural numbers. Interestingly, subsets of first n terms are permutations of 1..n for n = {2, 4, 8, 10, 18, 22, 24, 56, ...}. E.g., first 56 terms: {2, 1, 4, 3, 8, 5, 6, 7, 10, 9, 14, 15, 16, 13, 18, 11, 12, 17, 20, 21, 22, 19, 24, 23, 30, 29, 32, 27, 26, 33, 28, 25, 34, 37, 36, 31, 40, 39, 44, 35, 38, 41, 42, 47, 50, 51, 46, 43, 54, 49, 48, 53, 56, 45, 52, 55} are a permutation of 1..56.
Without altering the definition nor the existing values, one can as well start with a(0) = 0 and get (conjecturally) a permutation of the nonnegative integers. This sequence is in some sense the "arithmetic" analog of the "digital" variant A231433: Here we add subsequent terms, there the digits are concatenated. - M. F. Hasler, Nov 09 2013
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 A329333, A329405 ff, A329449 ff and the OEIS Wiki page. - M. F. Hasler, Nov 24 2019

Lars Blomberg, Table of n, a(n) for n = 0..9999
M. F. Hasler, Prime sums from neighboring terms, OEIS wiki, Nov. 23, 2019.

Cf. A083236.

Cf. A055265 for the variant starting with a(1) = 1, and A329333, A329405, ..., A329425 and A329449, ..., A329581 for other variants. - M. F. Hasler, Nov 24 2019

PARI
```
{a=0;u=0; for(n=1, 99, u+=1<
				
```

a(2n-1) = A055265(2n-1) + 1, a(2n) = A055265(2n) - 1, for all n >= 1. - M. F. Hasler, Feb 11 2020

Initial a(0) = 0 prefixed by M. F. Hasler, Nov 09 2013

A329425 For all n >= 0, six among (a(n+i) + a(n+j), 0 <= i < j < 5) are prime: lexicographically first such sequence of distinct nonnegative integers.

Original entry on oeis.org

0, 1, 2, 3, 4, 9, 8, 10, 33, 14, 93, 20, 17, 23, 44, 6, 24, 35, 65, 5, 18, 32, 11, 12, 29, 30, 7, 31, 72, 16, 22, 25, 37, 15, 46, 64, 43, 28, 85, 19, 54, 13, 88, 34, 49, 39, 40, 27, 100, 57, 26, 52, 111, 21, 38, 45, 62, 41, 51, 56, 47, 116, 50, 81, 63, 68, 59, 170, 69, 71

Offset: 0

M. F. Hasler, following an idea from Eric Angelini, Nov 24 2019

nonn

The restriction to [1, oo) is the lexicographically first such sequence of positive integers. (This is rather exceptional, cf. A128280 vs A055265, A329405 vs A329450, ..., see the wiki page for more.)
Conjectured to be a permutation, i.e., all n >= 0 appear. The restriction to [1, oo) is then the lexicographically first such permutation of the positive integers.
Among pairwise sums of 5 consecutive terms, there cannot be more than 2 x 3 = 6 primes: see the wiki page for this and further considerations and variants.

Robert Israel, Table of n, a(n) for n = 0..10000
Éric Angelini, Prime sums from neighbouring terms, SeqFan list, and personal blog "Cinquante signes", 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]
M. F. Hasler, Prime sums from neighboring terms, OEIS wiki, Nov. 23, 2019.

Cf. A055265, A128280 (1 prime from 2 terms), A329333 (1 prime from 3 terms), A329405-A329416 (N primes from M terms >= 1), A329449, ..., A329581 (N primes from M terms >= 0).

R:= 0,1,2,3,4:
S:= {R}:
for i from 1 to 100 do
  for x from 5 do
    if member(x,S) then next fi;
    n1:= nops(select(isprime,[seq(seq(R[i+j]+R[i+k],j=1..k-1),k=1..4)]));
    if nops(select(isprime,[seq(R[i+j]+x,j=1..4)]))+n1 = 6 then
      R:= R, x; S:= S union {x}; break
    fi
od od:
R; # Robert Israel, Dec 29 2022

A329425_upto(N) = S(N,6,5,0) \\ see the wiki page for the function S().

A329411 Among the pairwise sums of any three consecutive terms there are exactly two prime sums: lexicographically earliest such sequence of distinct positive numbers.

Original entry on oeis.org

1, 2, 3, 4, 7, 6, 5, 8, 9, 10, 13, 16, 15, 14, 17, 12, 11, 18, 19, 22, 21, 20, 23, 24, 29, 30, 31, 28, 25, 33, 34, 26, 27, 32, 35, 36, 37, 42, 41, 38, 45, 44, 39, 40, 43, 46, 51, 50, 47, 53, 54, 48, 49, 52, 55, 57, 82, 56, 75, 62, 64, 87, 63, 76, 61, 66, 65, 71, 86, 60, 77, 67, 72, 59, 68, 69, 58, 70

Offset: 1

Eric Angelini and Jean-Marc Falcoz, Nov 14 2019

nonn

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

			a(1) = 1 is the smallest possible choice; there's no restriction on the first term.
a(2) = 2 as 2 is the smallest available integer not leading to a contradiction. Note that as 1 + 2 = 3 we already have one prime sum (out of the required two) with the pair {1, 2}.
a(3) = 3 as 3 is the smallest available integer not leading to a contradiction. Since 2 + 3 = 5 we now have our two prime sums with the triplet {1, 2, 3}.
a(4) = 4 as 4 is the smallest available integer not leading to a contradiction. Since 3 + 4 = 7 we now have our two prime sums with the triplet {2, 3, 4}: they are 2 + 3 = 5 and 3 + 4 = 7.
a(5) = 7 because 5 or 6 would lead to a contradiction: indeed, both the triplets {3, 4, 5} and {3, 4, 6} will produce only one prime sum (instead of two). With a(5) = 7 we have the triplet {3, 4, 7} and the two prime sums we were looking for: 3 + 4 = 7 and 4 + 7 = 11.
And so on.

Jean-Marc Falcoz, Table of n, a(n) for n = 1..10000
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]
M. F. Hasler, Prime sums from neighboring terms, OEIS wiki, Nov. 23, 2019

Cf. A055265 (sum of two consecutive terms is always prime: differs from a(30) on).
Cf. A329412 .. A329416 (exactly 2 prime sums using 4, ..., 10 consecutive terms).
Cf. A329333, A329406 .. A329410 (exactly 1 prime sum using 3, 4, ..., 10 consecutive terms).
Cf. A055266 (no prime sum among 2 consecutive terms), A329405 (no prime among the pairwise sums of 3 consecutive terms).
See also "nonnegative" variants: A253074, A329450 (0 primes using 2 resp. 3 terms), A128280 (1 prime from 2 terms), A329452, A329453 (2 primes from 4 resp. 5 terms), A329454, A329455 (3 primes from 4 resp. 5 terms), A329449, A329456 (4 primes from 4 resp. 5 terms). See the Wiki page for more.

a[1]=1;a[2]=2;a[n_]:=a[n]=(k=1;While[Length@Select[Plus@@@Subsets[{a[n-1],a[n-2],++k},{2}],PrimeQ]!=2||MemberQ[Array[a,n-1],k]];k);Array[a,100] (* Giorgos Kalogeropoulos, May 09 2021 *)

A329411(n,show=0,o=1,N=2,M=2,p=[],U,u=o)={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])))); for(k=u,oo, bittest(U,k-u)|| min(c-#[0|p<-p, isprime(p+k)], #p>=M) ||[o=k,break]));show&&print([u]);o} \\ Optional args: show=1: print a(o..n-1), show=-1: append a(o..n-1) to the (global) list L, in both cases print [least unused number] at the end; o=0: start with a(o)=o; N, M: find N primes using M+1 consecutive terms. - M. F. Hasler, Nov 16 2019

A329454 There are exactly three primes among a(n+i) + a(n+j), 0 <= i < j <= 3, for any n >= 0: lexicographically earliest such sequence of distinct nonnegative integers.

Original entry on oeis.org

0, 1, 2, 4, 5, 3, 8, 6, 11, 7, 10, 12, 9, 19, 22, 14, 15, 16, 13, 18, 21, 40, 43, 20, 27, 46, 17, 26, 33, 24, 35, 38, 32, 23, 29, 30, 31, 28, 25, 34, 36, 39, 37, 64, 42, 41, 67, 47, 60, 49, 48, 52, 45, 55, 44, 58, 69, 51, 50, 62, 53, 77, 54, 56, 83, 57, 66, 74, 65, 61, 102, 70, 71, 79, 78, 59, 68, 63, 72, 95, 86, 81, 76, 73, 75, 82, 106

Offset: 0

M. F. Hasler, based on an idea from Eric Angelini, Nov 15 2019

nonn

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.

			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.

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]

Cf. A329452 (2 primes among a(n+i)+a(n+j), 0 <= i < j < 4), A329453 (2 primes among a(n+i)+a(n+j), 0 <= i < j < 5).
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).
Cf. A329405 ff: variants defined for positive integers.

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 *)

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

A329416 Among the pairwise sums of any ten consecutive terms there are exactly two prime sums: lexicographically earliest such sequence of distinct positive numbers.

Original entry on oeis.org

1, 2, 3, 7, 13, 19, 23, 25, 31, 32, 17, 8, 26, 37, 43, 49, 14, 38, 55, 61, 11, 20, 35, 67, 73, 79, 57, 9, 5, 15, 21, 42, 27, 12, 33, 30, 39, 45, 47, 18, 48, 6, 51, 24, 63, 69, 72, 75, 16, 36, 54, 60, 22, 66, 10, 4, 40, 29, 28, 34, 44, 41, 46, 50, 52, 58, 64, 53, 70, 71, 59, 62, 76, 56, 82, 88, 94, 65, 100

Offset: 1

Eric Angelini and Jean-Marc Falcoz, Nov 14 2019

nonn

Conjectured to be a permutation of the positive integers: a(10^6) = 10^6 + 2 and all numbers up to 10^6 - 7 have appeared at that point. - M. F. Hasler, Nov 15 2019

			a(1) = 1 is the smallest possible choice, there's no restriction on the first term.
a(2) = 2 as 2 is the smallest available integer not leading to a contradiction. Note that as 1 + 2 = 3 we already have one prime sum (on the required two) with the 10-set {1,2,a(3),a(4),a(5),a(6),a(7),a(8),a(9),a(10)}.
a(3) = 3 as 3 is the smallest available integer not leading to a contradiction. Note that as 2 + 3 = 5 we now have the two prime sums required with the 10-set {1,2,a(3),a(4),a(5),a(6),a(7),a(8),a(9),a(10)}.
a(4) = 7 as a(4) = 4, 5 or 6 would lead to a contradiction: indeed, the 10-sets {1,2,3,4,a(5),a(6),a(7),a(8),a(9),a(10)}, {1,2,3,5,a(5),a(6),a(7),a(8),a(9),a(10)} and {1,2,3,6,a(5),a(6),a(7),a(8),a(9),a(10)} will produce more than the two required prime sums. With a(4) = 7 we have no contradiction as the 10-set {1,2,3,7,a(5),a(6),a(7),a(8),a(9),a(10)} has now two prime sums so far: 1 + 2 = 3 and 2 + 3 = 5.
a(5) = 13 as a(5) = 4, 5, 6, 8, 9, 10, 11 or 12 would again lead to a contradiction (more than 2 prime sums with the 10-set); in combination with any other term before it, a(5) = 13 will produce only composite sums.
a(6) = 19 as 19 is the smallest available integer not leading to a contradiction: indeed, the 10-set {1,2,3,7,13,19,a(7),a(8),a(9),a(10)} shows two prime sums so far: 1 + 2 = 3 and 2 + 3 = 5.
a(7) = 23 as 23 is the smallest available integer not leading to a contradiction; indeed, the 10-set {1,2,3,7,13,19,23,a(8),a(9),a(10)} shows only two prime sums so far, which are 1 + 2 = 3 and 2 + 3 = 5.
a(8) = 25 as 25 is the smallest available integer not leading to a contradiction and producing two prime sums so far with the 10-set {1,2,3,7,13,19,23,25,a(9),a(10)}; etc.

Jean-Marc Falcoz, Table of n, a(n) for n = 1..10000

Cf. A329333 (3 consecutive terms, exactly 1 prime sum).
Cf. A329405 (no prime among the pairwise sums of 3 consecutive terms).
Cf. A329406 .. A329410 (exactly 1 prime sum using 4, ..., 10 consecutive terms).
Cf. A329411 .. A329415 (exactly 2 prime sums using 3, ..., 7 consecutive terms).
See also "nonnegative" variants: A329450 (0 primes using 3 terms), A329452 (2 primes using 4 terms), A329453 (2 primes using 5 terms), A329454 (3 primes using 4 terms), A329449 (4 primes using 4 terms), A329455 (3 primes using 5 terms), A329456 (4 primes using 5 terms).

A329416(n, show=0, o=1, N=2, M=9, 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(#pM. F. Hasler, Nov 15 2019

A329455 There are exactly three primes in {a(n+i) + a(n+j), 0 <= i < j <= 4} for any n >= 0: lexicographically earliest such sequence of distinct nonnegative integers.

Original entry on oeis.org

0, 1, 2, 4, 8, 6, 3, 10, 14, 11, 5, 9, 15, 26, 12, 17, 13, 7, 18, 16, 20, 21, 19, 23, 27, 40, 22, 31, 24, 25, 29, 28, 30, 32, 33, 39, 34, 36, 35, 38, 41, 46, 37, 43, 48, 42, 55, 47, 44, 45, 52, 49, 50, 53, 56, 58, 54, 57, 51, 73, 76, 61, 59, 63, 64, 68, 60, 69, 67, 62, 65, 66, 70, 71, 72, 79, 77, 74, 81, 86, 78, 89, 82, 85, 80, 99, 84, 83, 75, 92, 87, 88, 90, 91, 93, 94, 100

Offset: 0

M. F. Hasler, based on an idea from Eric Angelini, Nov 15 2019

nonn

That is, there are exactly three primes among the 10 pairwise sums of any five consecutive terms.
It is unknown whether this is a permutation of the nonnegative numbers. There is 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), a(n-4)} 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) = 0 or 1, and for S(n) = 0 or 1, this is similar to the case of A329452 or A329453. 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.

			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 3 primes among the pairwise sums using {0, 1, 2, 4}, and the next term must not produce an additional prime as sum with these. The terms 0 and 1 exclude primes and (primes - 1). We find that a(4) = 8 is the smallest possibility.
Then there are 2 primes (1+2 and 1+4) among the pairwise sums using {1, 2, 4, 8}, and the next term must produce exactly one additional prime as sum with these terms. We find that a(5) = 6 is the smallest possibility (since 5+2 and 5+8 would give 2 primes).

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]

Cf. A329454 (3 primes among a(n+i)+a(n+j), 0 <= i < j <= 3).
Cf. A329452 (2 primes among a(n+i)+a(n+j), 0 <= i < j <= 3), A329453 (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 <= 2), A329450 (0 primes among a(n+i)+a(n+j), 0 <= i < j <= 2).
Cf. A329405 ff: variants defined for positive integers.

A329455(n, show=0, o=0, N=3, 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

A329456 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.

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, based on an idea from Eric Angelini, Nov 15 2019

nonn

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.

			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.

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]

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).

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

cp's OEIS Frontend

A055265 a(n) is the smallest positive integer not already in the sequence such that a(n)+a(n-1) is prime, starting with a(1)=1.

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A329450 Lexicographically earliest sequence of distinct nonnegative integers such that neither a(n) + a(n+1) nor a(n) + a(n+2) is prime for any n.

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A329449 For any n >= 0, exactly four sums a(n+i) + a(n+j) are prime, for 0 <= i < j <= 3: lexicographically earliest such sequence of distinct nonnegative integers.

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A128280 a(n) is the least number not occurring earlier such that a(n)+a(n-1) is prime, a(0) = 0.

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A329425 For all n >= 0, six among (a(n+i) + a(n+j), 0 <= i < j < 5) are prime: lexicographically first such sequence of distinct nonnegative integers.

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A329411 Among the pairwise sums of any three consecutive terms there are exactly two prime sums: lexicographically earliest such sequence of distinct positive numbers.

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A329454 There are exactly three primes among a(n+i) + a(n+j), 0 <= i < j <= 3, for any n >= 0: lexicographically earliest such sequence of distinct nonnegative integers.

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