OFFSET
0,1
FORMULA
From Chai Wah Wu, Jan 03 2019: (Start)
a(n) >= 2^n+1, if it exists. a(n) is odd for n > 2, if it exists.
It is clear these are true for n <= 2. Suppose n > 2. If a(n) is even, then complementing any bits that is not LSB or MSB will result in an even nonprime number. If a(n) is odd, then complementing the LSB will result in an even nonprime number. So either case shows that a(n) has n+1 or more binary bits. It also shows that a(n) must be odd.
Conjecture: a(n) mod 10 == 5 for n > 7. (End)
EXAMPLE
a(1) = 4 because 4 in base 2 is 100 and 000 is 0, 110 is 6 and 101 is 5: hence only one prime.
a(2) = 6 because 6 in base 2 is 110 and 010 is 2, 100 is 4 and 111 is 7: hence two primes.
MAPLE
a:= proc(n) local k; for k from 2^n+1 while isprime(k) or n<>add(
`if`(isprime(Bits[Xor](k, 2^j)), 1, 0), j=0..ilog2(k)) do od; k
end:
seq(a(n), n=0..12); # Alois P. Heinz, Jan 03 2019
PROG
(Python)
from sympy import isprime
def A322743(n):
i = 4 if n <= 1 else 2**n+1
j = 1 if n <= 2 else 2
while True:
if not isprime(i):
c = 0
for m in range(len(bin(i))-2):
if isprime(i^(2**m)):
c += 1
if c > n:
break
if c == n:
return i
i += j # Chai Wah Wu, Jan 03 2019
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Paolo P. Lava, Dec 24 2018
EXTENSIONS
Definition clarified by Chai Wah Wu, Jan 03 2019
a(19) added and a(0) corrected by Rémy Sigrist, Jan 03 2019
a(20)-a(23) from Giovanni Resta, Jan 03 2019
STATUS
approved