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A065413 Number of positive solutions to "numbers that are n times their number of binary 1's". 7
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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified February 28 08:28 EST 2020. Contains 332323 sequences. (Running on oeis4.)