login
A318507
a(n) = A032742(n) XOR A001065(n)-A032742(n), where XOR is bitwise-or (A003987) and A001065 = sum of proper divisors and A032742 = the largest proper divisor of n.
6
1, 1, 1, 3, 1, 0, 1, 7, 2, 6, 1, 12, 1, 4, 1, 15, 1, 5, 1, 6, 3, 8, 1, 20, 4, 14, 13, 0, 1, 20, 1, 31, 15, 18, 1, 55, 1, 16, 9, 10, 1, 52, 1, 4, 29, 20, 1, 44, 6, 11, 21, 14, 1, 60, 13, 56, 23, 30, 1, 80, 1, 28, 1, 63, 11, 12, 1, 58, 19, 4, 1, 115, 1, 38, 1, 60, 3, 20, 1, 106, 22, 42, 1, 72, 23, 40, 25, 28, 1, 78, 5, 48, 27, 44, 21, 92, 1
OFFSET
1,4
COMMENTS
Note that here zeros occur only on even perfect numbers (even terms of A000396), in contrast to A318457, which would be zero also for any hypothetical odd perfect number. - Antti Karttunen, Aug 29 2018
FORMULA
a(n) = A003987(A032742(n), A318505(n)).
For n > 1, a(n) = A001065(n) - 2*A318508(n).
PROG
(PARI)
A001065(n) = (sigma(n)-n);
A032742(n) = if(1==n, n, n/vecmin(factor(n)[, 1]));
A318505(n) = if(1==n, 0, A001065(n)-A032742(n));
A318507(n) = bitxor(A318505(n), A032742(n));
KEYWORD
nonn,base
AUTHOR
Antti Karttunen, Aug 28 2018
STATUS
approved