A253073 -id:A253073 - cp's OEIS frontend

A253074 Lexicographically earliest sequence of distinct numbers such that a(n-1)+a(n) is not prime.

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

Offset: 1

N. J. A. Sloane, Feb 01 2015, based on a suggestion from Patrick Devlin

nonn

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

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A055266, A254337, A253073, A010051.

a253074 n = a253074_list !! (n-1)
a253074_list = 0 : f 0 [1..] where
   f u vs = g vs where
     g (w:ws) | a010051' (u + w) == 1 = g ws
              | otherwise = w : f w (delete w vs)
-- Reinhard Zumkeller, Feb 02 2015

A253074_upto(n=99, a, u, U)={vector(n,n, for(k=u,oo, bittest(U,k-u)|| isprime(a+k)||[a=k,break]); (a>u && U+=1<<(a-u))|| U>>=-u+u+=valuation(U+2,2); a)+if(default(debug),print([u]))} \\ additional args allow to tweak computation. If debug > 0, print least unused number at the end. - M. F. Hasler, Nov 25 2019

A254337 Lexicographically earliest sequence of distinct numbers such that no sum of consecutive terms is prime.

Original entry on oeis.org

0, 1, 8, 6, 10, 14, 12, 4, 20, 16, 24, 18, 22, 28, 26, 34, 30, 32, 36, 40, 42, 46, 38, 44, 52, 48, 54, 50, 58, 56, 62, 64, 60, 66, 68, 72, 70, 74, 80, 76, 78, 86, 82, 84, 90, 92, 94, 88, 98, 96, 104, 100, 102, 108, 110, 112, 114, 106, 116, 122, 118, 120, 124, 126, 130, 132, 134, 128, 138, 136, 142, 140, 144, 146, 148, 150, 154, 152, 156, 158

Offset: 0

M. F. Hasler, Jan 28 2015

In other words, no sum a(i)+a(i+1)+a(i+2)+...+a(n) may be prime. In particular, the sequence may not contain any primes.
I conjecture that the sequence contains all even numbers > 2 and no odd number beyond 1. If so, we must simply ensure that the sum a(1)+...+a(n) is not prime, which is always possible for one of the three consecutive even numbers {2n, 2n+2, 2n+4}. As a consequence, it would follow that a(n) ~ 2n.
Is there even a proof that the smallest odd composite number, 9, does not appear?
The variant A254341 has the additional restriction of alternating parity, which avoids excluding the odd numbers.
The least odd composite number a'(n+1) that could occur as the next term after a(n) and such that sum(a(i),i=k...n)+a'(n+1) is composite for all k <= n is (for n = 0, 1, 2,...): 9, 9, 25, 21, 39, 25, 69, 65, 45, 119, 95, 77, 55, 27, 595, 561, 531, 865, 1519, 1479, 1437, 1391, 1353, 1309, 1257, 1209, 1155, 1105, 1047, 2317, 2255, 2191, 3565, 5719, 13067, 12995, 12925, 12851, 12771, 12695, 12617, 12531, 12449, 12365, 12275, ... The growth of this sequence shows how it is increasingly unlikely that an odd number could occur, since the next possible even term is only about 2n.

			To explain the beginning of the sequence, observe that starting with the smallest possible terms 0, 1 does not appear to lead to a contradiction (and in fact never does), so we start there.
The next composite would be 4 but 1+4=5 is prime, as is 1+6, but 1+8=9 is not, so we take a(2) = 8 to be the next term.
4 is impossible for a(3) since 1+8+4=13 is prime, but neither 1+8+6=15 nor 8+6 is prime, so a(3)=6.

Robert G. Wilson v and M. F. Hasler, Table of n, a(n) for n = 0..5000 (terms 0..999 from M. F. Hasler)

Cf. A254341, A153136, A254211, A002808, A253073, A253074, A054408, A084834.

Cf. A025044 (no pairwise sum is prime), A025043 (no pairwise difference is prime).

f[lst_List] := Block[{k = 1}, While[ PrimeQ@ k || MemberQ[lst, k] || Union@ PrimeQ@ Accumulate@ Reverse@ Join[lst, {k}] != {False}, k++]; Append[lst, k]]; Nest[f, {0}, 70] (* Robert G. Wilson v, Jan 31 2015 *)

a=[];u=0; for(i=1,99, a=concat(a,0); until( ! isprime(s) || ! a[i]++, while( isprime(a[i]) || bittest(u,a[i]), a[i]++); s=a[k=i]; while( k>1 && ! isprime( s+=a[k--]),)); u+=2^a[i]; print1(a[i]","))

It appears that a(n) ~ 2n.

cp's OEIS Frontend

A253074 Lexicographically earliest sequence of distinct numbers such that a(n-1)+a(n) is not prime.

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