OFFSET
1,2
COMMENTS
If instead of base 2 we take base 10, then we have the so-called Harshad or Niven numbers (i.e., positive integers divisible by the sum of their digits; A005349). - Emeric Deutsch, Apr 11 2007
A199238(a(n)) = 0. - Reinhard Zumkeller, Nov 04 2011
LINKS
Indranil Ghosh, Table of n, a(n) for n = 1..20000 (first 1000 terms from T. D. Noe)
Paul Dalenberg and Tom Edgar, Consecutive factorial base Niven numbers, Fibonacci Quarterly, Vol. 56, No. 2 (2018), pp. 163-166.
Jean-Marie De Koninck, Nicolas Doyon and Imre Kátai, On the counting function for the Niven numbers, Acta Arithmetica, Vol. 106, No. 3 (2003), pp. 265-275.
FORMULA
{k : A000120(k) | k}. - R. J. Mathar, Mar 03 2008
a(n) seems to be asymptotic to c*n*log(n) where 0.7 < c < 0.8. - Benoit Cloitre, Jan 22 2003
Heuristically, c should be 1/(2*log(2)), since a random d-bit number should have probability approximately 2/d of being in the sequence. - Robert Israel, Aug 22 2014
{a(n)} = {k : A199238(k) = 0}. - M. F. Hasler, Oct 09 2012
De Koninck et al. (2003) proved that the number of base-b Niven numbers not exceeding x, N_b(x), is asymptotically equal to ((2*log(b)/(b-1)^2) * Sum_{j=1..b-1} gcd(j, b-1) + o(1)) * x/log(x). For b=2, N_2(n) ~ (2*log(2) + o(1)) * x/log(x). Therefore, the constant c mentioned above is indeed 1/(2*log(2)). - Amiram Eldar, Aug 16 2020
EXAMPLE
20 is in the sequence because 20 is written 10100 in binary and 1 + 1 = 2, which divides 20.
21 is in the sequence because 21 is written 10101 in binary and 1 + 1 + 1 = 3, which divides 21.
22 is not in the sequence because 22 is written 10110 in binary 1 + 1 + 1 = 3, which does not divide 22.
MAPLE
a:=proc(n) local n2: n2:=convert(n, base, 2): if n mod add(n2[i], i=1..nops(n2)) = 0 then n else fi end: seq(a(n), n=1..300); # Emeric Deutsch, Apr 11 2007
MATHEMATICA
binHarshadQ[n_] := Divisible[n, Count[IntegerDigits[n, 2], 1]]; Select[Range[228], binHarshadQ] (* Jean-François Alcover, Dec 01 2011 *)
Select[Range[300], Divisible[#, DigitCount[#, 2, 1]]&] (* Harvey P. Dale, Mar 20 2016 *)
PROG
(PARI) for(n=1, 1000, b=binary(n); l=length(b); if(n%sum(i=1, l, component(b, i))==0, print1(n, ", ")))
(PARI) is_A049445(n)={n%norml2(binary(n))==0} \\ M. F. Hasler, Oct 09 2012
(PARI) isok(n) = ! (n % hammingweight(n)); \\ Michel Marcus, Feb 10 2016
(Haskell)
a049445 n = a049445_list !! (n-1)
a049445_list = map (+ 1) $ elemIndices 0 a199238_list
-- Reinhard Zumkeller, Nov 04 2011
(Python)
A049445 = [n for n in range(1, 10**5) if not n % sum([int(d) for d in bin(n)[2:]])] # Chai Wah Wu, Aug 22 2014
CROSSREFS
KEYWORD
nonn,easy,nice,base
AUTHOR
EXTENSIONS
More terms from Michael Somos
Edited by N. J. A. Sloane, Oct 07 2005 and May 16 2008
STATUS
approved