[go: up one dir, main page]

login
A344856
Bitwise XOR of prime(n) and n^2.
1
3, 7, 12, 23, 18, 41, 32, 83, 70, 121, 102, 181, 128, 239, 206, 309, 282, 377, 298, 471, 496, 427, 578, 537, 528, 705, 702, 891, 804, 1013, 958, 1155, 1224, 1039, 1116, 1415, 1476, 1287, 1366, 1773, 1570, 1617, 1926, 1873, 1836, 2179, 2162, 2527, 2434, 2337
OFFSET
1,1
COMMENTS
This is effectively the bitwise XOR of A000040 and A000290.
FORMULA
a(n) = prime(n) XOR n^2.
a(n) = A003987(A000040(n), A000290(n)).
EXAMPLE
For n=3, a(3) is prime(3) XOR 3^2 = 5 XOR 9 or b(0101) XOR b(1001) = (b)1100, which in base 10 is 12.
MAPLE
a:= n-> Bits[Xor](n^2, ithprime(n)):
seq(a(n), n=1..50); # Alois P. Heinz, May 30 2021
MATHEMATICA
a[n_] := BitXor[n^2, Prime[n]]; Array[a, 50] (* Amiram Eldar, Jun 05 2021 *)
PROG
(Python)
from sympy import primerange, prime
import numpy
def a_vector(n):
primes = list(primerange(0, prime(n)))
squares = [x ** 2 for x in range(1, n)]
return numpy.bitwise_xor(primes, squares)
(PARI) A344856(n) = bitxor(prime(n), n*n); \\ Antti Karttunen, Jun 05 2021
(Python)
from sympy import prime
def A344856(n): return prime(n) ^ n**2 # Chai Wah Wu, Jun 12 2021
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Chris von Csefalvay, May 30 2021
STATUS
approved