|
%I #20 Feb 02 2018 02:39:45
%S 1,2,2,4,4,6,6,8,8,10,9,12,12,12,12,16,16,18,18,20,21,21,20,24,24,24,
%T 24,27,28,28,30,32,32,34,33,36,36,36,36,40,39,42,40,42,44,44,45,48,48,
%U 48,48,51,52,52,55,54,56,56,55,60,60,60,60,64,64,66,66,68,69,69,68,72,72
%N a(n) is the largest multiple of {the number of 1's in the binary representation of n} that is <= n.
%C a(n) = n - A199238(n). - _Reinhard Zumkeller_, Nov 04 2011
%H Reinhard Zumkeller, <a href="/A161764/b161764.txt">Table of n, a(n) for n = 1..10000</a>
%e 11 (decimal) in binary is 1011. There are three 1's. Because 9 is the largest multiple of 3 that is <= 11, then a(11) = 9.
%p a := proc (n) local n2, n1, j: n2 := convert(n, base, 2): n1 := add(n2[i], i = 1 .. nops(n2)): for j while j*n1 <= n do j*n1 end do end proc: seq(a(n), n = 1 .. 80); # _Emeric Deutsch_, Jun 26 2009
%o (PARI) a(n)=local(B=binary(n),w=B*vector(#B,x,1)~);n-n%w \\ _Hagen von Eitzen_, Jun 22 2009
%o (Haskell)
%o a161764 n = n - a199238 n -- _Reinhard Zumkeller_, Nov 04 2011
%Y Cf. A000120, A161765.
%K base,nonn
%O 1,2
%A _Leroy Quet_, Jun 18 2009
%E Extended by _Hagen von Eitzen_ and _Emeric Deutsch_, Jun 26 2009
|
