

A334045


Bitwise NOR of binary representation of n and n1.


3



0, 0, 0, 0, 2, 0, 0, 0, 6, 4, 4, 0, 2, 0, 0, 0, 14, 12, 12, 8, 10, 8, 8, 0, 6, 4, 4, 0, 2, 0, 0, 0, 30, 28, 28, 24, 26, 24, 24, 16, 22, 20, 20, 16, 18, 16, 16, 0, 14, 12, 12, 8, 10, 8, 8, 0, 6, 4, 4, 0, 2, 0, 0, 0, 62, 60, 60, 56, 58, 56, 56, 48, 54, 52, 52
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OFFSET

1,5


COMMENTS

All terms are even.
a(1) = 0, a(2) = 0, and a(2^n + 1) = 2^n  2 for n > 0. Are there any other cases where n  a(n) < 4?  Charles R Greathouse IV, Apr 13 2020
The answer to the above question is no. Write n as n = (2m+1)*k, i.e. k = A006519(n) is the highest power of 2 dividing n. If m = 0, a(n) = 0 and n  a(n) = n. If m > 0, then a(n) = 2v*k, where v is the 1's complement of m. Thus na(n) = (2(mv)+1)*k. Since m in binary has a leading 1, m  v >= 1 and thus n  a(n) >= 3 with n  a(n) = 3 when n > 2, k = 1 and m  v = 1, i.e. m is a power of 2 and n is of the form 2^r + 1.  Chai Wah Wu, Apr 13 2020


LINKS



EXAMPLE

a(11) = 11 NOR 10 = bin 1011 NOR 1010 = bin 100 = 4.


MAPLE

a:= n> Bits[Nor](n, n1):


PROG

(Python)
def norbitwise(n):
a = str(bin(n))[2:]
b = str(bin(n1))[2:]
if len(b) < len(a):
b = '0' + b
c = ''
for i in range(len(a)):
if a[i] == b[i] and a[i] == '0':
c += '1'
else:
c += '0'
return int(c, 2)
(Python)
m = n(n1)
return 2**(len(bin(m))2)1m # Chai Wah Wu, Apr 13 2020
(PARI) a(n) = my(x=bitor(n1, n)); bitneg(x, #binary(x)); \\ Michel Marcus, Apr 13 2020


CROSSREFS



KEYWORD

easy,base,nonn


AUTHOR



STATUS

approved



