A065413 Number of positive solutions to "numbers that are n times their number of binary 1's".

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 3, 1, 0, 0, 2, 1, 1, 1, 1, 1, 1, 1, 0, 1, 2, 0, 1, 1, 2, 1, 1, 1, 1, 3, 0, 1, 0, 0, 1, 0, 2, 2, 0, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 2, 0, 2, 1, 1, 2, 1, 0, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 1, 2, 3, 1, 0, 0, 1, 0, 0, 1, 0, 2, 1, 0, 0, 2

Offset: 1

Author

Henry Bottomley, Nov 23 2001

Keywords

• base
• nonn

Comments

Equivalently, this is the number of ways to write n as an arithmetic mean of distinct powers of 2. [Brian Kell, Mar 01 2009]

Examples

			a(23)=3 since 69, 92 and 115 are written in binary as 1000101, 1011100 and 1110011 and 69=23*3, 92=23*4 and 115=23*5.
		

Links

• Peter Kagey, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000120, A037478, A058898, A272797 (greedy inverse).

Programs

Maple

N:= 1000: # to get a(1) to a(N)
A:= Vector(N):
for x from 1 while x/(1+ilog2(x)) <= N do
  v:= x/convert(convert(x,base,2),`+`);
  if v::integer and v <= N then
    A[v]:= A[v]+1
  fi
od:
seq(A[i],i=1..N); # Robert Israel, May 06 2016
# alternative program
read("transforms") :
A065413 := proc(n)
    local bdgs,a,x;
    a := 0 ;
    for bdgs from 1 do
        x := n*bdgs ;
        # x must have bdgs bits set, so x =bdgs*n >= 2^bdgs-1.
        if n < (2^bdgs-1)/x then
            break;
        elif wt(x) = bdgs then
            a := a+1 ;
        end if;
    end do:
    a ;
end proc: # R. J. Mathar, May 11 2016