OFFSET
1,2
COMMENTS
If 2^k <= n < 2^(k+1), then also 2^k <= a(n) < 2^(k+1), since any proper divisor of n is < 2^k.
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..8191
PROG
(PARI) a(n)=local(ds, r); ds=divisors(n); for(k=1, #ds, r=bitxor(r, ds[k])); r
(Haskell)
import Data.Bits (xor)
a178910 = foldl1 xor . a027750_row :: Integer -> Integer
-- Reinhard Zumkeller, Nov 17 2012
CROSSREFS
KEYWORD
AUTHOR
Franklin T. Adams-Watters, Jun 22 2010
STATUS
approved