A253073 Lexicographically earliest sequence of distinct numbers such that neither a(n) nor a(n-1)+a(n) is prime.

0, 1, 8, 4, 6, 9, 12, 10, 14, 16, 18, 15, 20, 22, 24, 21, 25, 26, 28, 27, 30, 32, 33, 35, 34, 36, 38, 39, 42, 40, 44, 46, 45, 48, 50, 49, 51, 54, 52, 56, 55, 57, 58, 60, 62, 63, 65, 64, 66, 68, 70, 72, 69, 74, 76, 77, 75, 78, 80, 81, 84, 82, 86, 85, 87, 88

Offset: 1

Author

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

Keywords

• nonn

Comments

Conjecture: this is a permutation of the nonprimes. [Proof outline given below by Semeon Artamonov and Pat Devlin.]

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. To make this cleaner, 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).

Links

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

Crossrefs

Cf. A254337, A002808, A253074.

Cf. A018252, A010051.

Programs

Haskell

a253073 n = a253073_list !! (n-1)
a253073_list = 0 : f 0 a018252_list 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