A352942 Let p = prime(n); a(n) = number of primes q with same number of binary digits as p that can be obtained from p by changing one binary digit.
1, 1, 1, 1, 0, 0, 1, 2, 2, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 3, 1, 1, 1, 1, 2, 3, 2, 0, 1, 1, 0, 2, 1, 2, 3, 1, 1, 4, 1, 0, 1, 1, 0, 1, 3, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 2, 2, 0, 3, 2, 1, 1, 2, 2, 1, 1, 0, 3, 0, 0, 2, 2, 0, 2, 2, 2, 3, 2, 2, 0, 2, 0, 1, 2, 0, 1
Offset: 1
Examples
prime(1) = 2, in binary 10, has one neighbor 11 in P(2, 2), so a(1) = 1. prime(14) = 43, in binary 101011, has neighbors 101001 (41), 101111 (47), 111011 (59), so a(14) = 3.
Links
- Paolo Xausa, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Maple
a:= n-> (p-> nops(select(isprime, {seq(Bits[Xor] (p, 2^i), i=0..ilog2(p)-1)})))(ithprime(n)): seq(a(n), n=1..100); # Alois P. Heinz, May 11 2022 -
Mathematica
A352942[n_] := Count[BitXor[#, 2^Range[0, BitLength[#] - 2]], _?PrimeQ] & [Prime[n]]; Array[A352942, 100] (* Paolo Xausa, Apr 23 2025 *)
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Python
from sympy import isprime, sieve def neighs(s): digs = "01" ham1 = (s[:i]+d+s[i+1:] for i in range(len(s)) for d in digs if d!=s[i]) yield from (h for h in ham1 if h[0] != '0') def a(n): return sum(1 for s in neighs(bin(sieve[n])[2:]) if isprime(int(s, 2))) print([a(n) for n in range(1, 88)])
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