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Is there a closed formula to the problem $f(1)=1, f(2n)=f(n), f(2n+1)=f(2n)+1$. So far i have found a solution for $n$, which is the number of power of $2$'s needed to add up to the number starting with the greatest power of $2$. Also $n=$ the number of $1$'s needed to represent $n$ in binary form. So for example $f(12)=2, 2^3 + 2^2 = 12$, $12$ in binary is represented as $1100$, two $1$'s so $n=2$. My problem is i know the process but can't find a simple formula to this problem.

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Are you sure that there is a simpler method than counting the number of $1$s in the binary expansion? –  robjohn Dec 3 '11 at 2:05
A related function is "fusc", a name coined by Dijkstra. See also here. –  Zev Chonoles Dec 3 '11 at 13:36

4 Answers 4

There is no known closed form solution for the number of ones in the binary representation of a number, also known as the Hamming weight of the number. You may want to check out this related question on efficient ways to compute f(n): Best algorithm to count the number of set bits in a 32-bit integer?

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It's obvious that f(n) = number of 1s in n's binary representation

# the following python code would do
f = lambda x:str(bin(x)).count('1')
f(5)  # 2
f(8)  # 1
f(15) # 4

And for efficiency, the following code in C was given in chapter 5.1 of Hacker's Delight

/* returns number of 1s in x's binary reperesantation */
static inline int cnt1(int x)
    x = (x & 0x55555555) + ((x >> 1) & 0x55555555);
    x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
    x = (x & 0x0f0f0f0f) + ((x >> 4) & 0x0f0f0f0f);
    x = (x & 0x00ff00ff) + ((x >> 8) & 0x00ff00ff);
    x = (x & 0x0000ffff) + ((x >> 16) & 0x0000ffff);
    return x;

EDIT: If you want a more mathematical representation, f[x_] := Sum[Floor[Mod[x, 2^(i + 1)]/2^i], {i, 0, Infinity}] in Mathematica looks like a ordinary sum formula. But for a closed form solution, sadly I don't think there exist one.

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@Sean I've edited my answer to add a mathematical looking formula. But a closed form solution may not exist :-( –  Pengyu CHEN Dec 3 '11 at 1:25

$$f\left(\sum\limits_{n=0}^{+\infty}2^ix_i\right)=\sum\limits_{n=0}^{+\infty}x_i,\qquad x_i\in\{0,1\}. $$

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there is not close formula, but the algorithm can be known review the works of Richard Hamming.

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