A294994 Begin with 2; thereafter a(n) is the least prime not already in the sequence such that the Hamming distance between it and the preceding prime is at most 2.

2, 3, 5, 7, 11, 13, 29, 17, 19, 23, 31, 47, 37, 41, 43, 59, 61, 53, 101, 71, 67, 73, 79, 103, 97, 107, 109, 127, 191, 151, 131, 137, 139, 163, 167, 173, 157, 149, 181, 179, 211, 83, 89, 113, 241, 193, 197, 199, 223, 239, 227, 229, 233, 251, 379, 283, 271, 263, 257, 269, 277, 281, 313, 307, 311, 293

Offset: 1

Author

Robert G. Wilson v, Nov 12 2017

Keywords

• nonn
• base

Comments

The Hamming distance between two primes p and q is the Hamming distance between their binary expansions. - N. J. A. Sloane, May 27 2018

Conjecture: this sequence is a permutation of the primes.

By definition, the absolute difference of a(n) and a(n + 1) is in A048645. - David A. Corneth, Jan 12 2018

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..10000
• Rémy Sigrist, Scatterplot of (x, y) such that the prime numbers prime(x) and prime(y) have Hamming distance <= 2 for x = 1..1000 and y = 1..1000

Crossrefs

Cf. A000040, A004676, A048645, A163252.

See also A303593, A303594, A303595 (when n-th prime appears).

Programs

Mathematica

f[s_List] := Block[{p = s[[-1]], q = 3}, While[MemberQ[s, q] || Plus @@ IntegerDigits [BitXor[p, q], 2] > 2, q = NextPrime@q]; Append[s, q]]; s = {2}; Nest[f, s, 65]

PARI

s = 0; v = 2; for (n=1, 66, print1 (v ", "); s += 2^v; forprime (p=2, oo, if (!bittest(s, p) && hammingweight(bitxor(p, v))<=2, v = p; break))) \\ Rémy Sigrist, Jan 08 2018