%I #14 Dec 23 2016 21:57:53
%S 0,0,1,1,2,2,2,5,6,6,6,9,9,7,7,7,8,12,12,16,16,16,16,19,19,14,14,19,
%T 19,25,25,25,26,26,26,26,26,25,25,25,25,33,33,32,32,32,32,29,29,32,32,
%U 32,32,35,35,35,35,35,35,46,46,45
%N Number of 0's in the binary expansion of the least common multiple of the first n integers.
%H Indranil Ghosh, <a href="/A279515/b279515.txt">Table of n, a(n) for n = 0..10000</a>
%e For n = 10, the LCM of all the numbers from 1 to 10 is 2520 = 100111011000_2, which has a total of 6 0's, so a(10) = 6.
%t Map[DigitCount[#, 2, 0] &, {1}~Join~Table[LCM @@ Range@ n, {n, 61}]] (* _Michael De Vlieger_, Dec 16 2016 *)
%o (Python)
%o def gcd(a, b):
%o while b:
%o a, b = b, a % b
%o return a
%o def lcm(a, b):
%o return a * b // gcd(a, b)
%o def c(*ar):
%o return reduce(lcm, ar)
%o def a(n):
%o if n==0:
%o return 0
%o x=bin(c(*range(1, n+1)))[2:]
%o return x.count("0")
%o for i in range(0, 10001):
%o print str(i)+" "+str(a(i))
%o (PARI) a(n) = my(lcmn = lcm(vector(n, k, k))); #binary(lcmn) - hammingweight(lcmn); \\ _Michel Marcus_, Dec 23 2016
%Y Cf. A003418, A279506.
%K nonn,base
%O 0,5
%A _Indranil Ghosh_, Dec 13 2016