login
A387305
Least k such that the Hamming weight (A000120) of n*k is prime.
1
3, 3, 1, 3, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 19, 3, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 3, 19, 1, 3, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 9, 3, 1, 1, 1, 1, 7, 1, 5, 3, 1, 1, 3, 3, 1, 19, 1, 1, 67, 3, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 1, 5, 3, 1, 1, 1, 1, 5, 1, 5, 3
OFFSET
1,1
COMMENTS
a(n) is always odd.
LINKS
FORMULA
a(n) = min{k >= 1 : A000120(n*k) is prime}.
a(n) = 1 if A000120(n) is prime (see A052294).
For all m >= 0, a(2^m) = 3.
EXAMPLE
a(1) = 3 because A000120(1) = 1 (not prime), A000120(2) = 1, and A000120(3) = 2 (prime).
a(11) = 1 because A000120(11) = 3 (prime).
a(15) = 19 since 15*19 = 285 and A000120(285) = 5 (prime); for 1 <= k < 19 the value A000120(15*k) is not prime.
MATHEMATICA
A387305[n_] := Module[{k = -1}, While[!PrimeQ[DigitSum[(k += 2)*n, 2]]]; k];
Array[A387305, 100] (* Paolo Xausa, Sep 02 2025 *)
PROG
(PARI) a(n) = {my(k=1); while(!isprime(hammingweight(n*k)), k++); k};
vector(100, n, a(n)) \\ first 100 terms
(Python)
import sympy as sp
def a(n, kmax=10**6):
for k in range(1, kmax + 1):
if sp.isprime((n*k).bit_count()):
return k
return None
def A(N):
return [a(n) for n in range(1, N + 1)]
CROSSREFS
Sequence in context: A213662 A213657 A215596 * A268676 A354762 A180938
KEYWORD
nonn,base
AUTHOR
Pablo Cadena-Urzúa, Aug 25 2025
STATUS
approved