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

1,23

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]

Links

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

Example

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.

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

Crossrefs

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

Sequence in context: A229038 A229143 A330018 * A107131 A027200 A035654

Adjacent sequences: A065410 A065411 A065412 * A065414 A065415 A065416

Keyword

base,nonn

Author

Henry Bottomley, Nov 23 2001

Status

approved