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Bitwise OR of all divisors of n.
5

%I #15 Jan 29 2014 07:55:36

%S 1,3,3,7,5,7,7,15,11,15,11,15,13,15,15,31,17,31,19,31,23,31,23,31,29,

%T 31,27,31,29,31,31,63,43,51,39,63,37,55,47,63,41,63,43,63,47,63,47,63,

%U 55,63,51,63,53,63,63,63,59,63,59,63,61,63,63,127,77,127

%N Bitwise OR of all divisors of n.

%C a(n) <= A003817(n); a(A000040(n)) = A000040(n).

%H Reinhard Zumkeller, <a href="/A218388/b218388.txt">Table of n, a(n) for n = 1..8191</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/OR.html">OR</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Binary_and#OR">Bitwise operation OR</a>

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%e n=20: divisors(20) = {1, 2, 4, 5, 10, 20}, 00001 OR 00010 OR 00100 OR 00101 OR 01010 OR 10100 = 11111 -> a(20) = 31;

%e n=21: divisors(21) = {1, 3, 7, 21}, 00001 OR 00011 OR 00111 OR 10101 = 10111 -> a(21) = 23;

%e n=22: divisors(22) = {1, 2, 11, 22}, 00001 OR 00010 OR 01011 OR 10110 = 11111 -> a(22) = 31;

%e n=23: divisors(23) = {1, 23}, 00001 OR 10111 = 10111 -> a(23) = 23;

%e n=24: divisors(24) = {1, 2, 3, 4, 6, 8, 12, 24}, 00001 OR 00010 OR 00011 OR 00100 OR 00110 OR 01000 OR 01100 OR 11000 = 11111 -> a(24) = 31;

%e n=25: divisors(25) = {1, 5, 25}, 00001 OR 00101 OR 11001 = 11101 -> a(25) = 29.

%t Table[BitOr@@Divisors[n],{n,70}] (* _Harvey P. Dale_, Feb 27 2013 *)

%o (Haskell)

%o import Data.Bits ((.|.))

%o a218388 = foldl1 (.|.) . a027750_row :: Integer -> Integer

%Y Cf. A027750, A000225 (subsequence), A123345, A218403.

%K nonn,base,look

%O 1,2

%A _Reinhard Zumkeller_, Oct 27 2012