Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


$A(q)=\sum \limits_{k=0}^\infty q^{2^k}$

Easily We can see that $$A(q)=q+A(q^2)\tag 1$$

Let's assume we redefine $A(q)$ as below

$A(q)=-\sum \limits_{k=1}^\infty c_k \ln{(1-q^k)}$

I would like to find $c_k$

Please see my attempt to find it.

$A(q^2)=-\sum \limits_{k=1}^\infty c_k \ln{(1-q^{2k})}=-\sum \limits_{k=1}^\infty c_k \ln{(1-q^{k})}-\sum \limits_{k=1}^\infty c_k \ln{(1+q^{k})}$

$A(q^2)=A(q)-\sum \limits_{k=1}^\infty c_k \ln{(1+q^{k})}$

If we use equation 1 then

$\sum \limits_{k=1}^\infty c_k \ln{(1+q^{k})}=q \tag2$

If we get derivative both sides

$\sum \limits_{k=1}^\infty k.c_k\cfrac{q^k}{1+q^{k}}=q \tag3$

$\sum \limits_{k=1}^\infty k.c_k q^k(1-q^{k}+q^{2k}-q^{3k}+.....)=q $

$\sum \limits_{k=1}^\infty k.c_k q^k-\sum \limits_{k=1}^\infty k.c_k q^{2k}+\sum \limits_{k=1}^\infty k.c_k q^{3k}-...=q $

If we find few terms of $c_k$


$2c_2-c_1=0$ ----> $c_2=\frac{1}{2}$

$3c_3+c_1=0$ ----> $c_3=-\frac{1}{3}$

$4c_4-2c_2-c_1=0$ ----> $c_4=\frac{1}{2}$

$5c_5+c_1=0$ ----> $c_5=-\frac{1}{5}$

$6c_6-3c_3+2c_2-c_1=0$ ----> $c_6=-\frac{1}{6}$

$7c_7+c_1=0$ ----> $c_7=-\frac{1}{7}$

$8c_8-4c_4-2c_2-c1=0$ ----> $c_8=\frac{1}{2}$

$9c_9+3c_3+c1=0$ ----> $c_9=0$

$10c_{10}-5c_5+2c_2-c_1=0$ ----> $c_{10}=-\frac{1}{10}$

$11c_{11}+c_1=0$ ----> $c_{11}=-\frac{1}{11}$

$12c_{12}-6c_6+4c_4-3c_3-2c_2-c_1=0$ ----> $c_{12}=-\frac{1}{6}$

I got an interesting result via using $c_k$

$\int_{0}^{1} \frac{A(q)}{q} dq=2 \tag3$

$\int_{0}^{1} \frac{-\sum \limits_{k=1}^\infty c_k \ln{(1-q^{k})}}{q} dq= -\sum \limits_{k=1}^\infty c_k \int_{0}^{1} \frac{ \ln{(1-q^{k})}}{q} dq\tag4$


$ \sum \limits_{k=1}^\infty \frac{c_k}{k} \int_{0}^{\infty} \frac{ u e^{-u}}{1-e^{-u}} du=2\tag5$

$ \sum \limits_{k=1}^\infty \frac{c_k}{k} \int_{0}^{\infty} u (e^{-u}+e^{-2u}+e^{-3u}+e^{-4u}+...) du=2\tag6$

$ (1+\frac{1}{2^2}+\frac{1}{3^2}+...)\sum \limits_{k=1}^\infty \cfrac{c_k}{k} =2\tag7$

$ \cfrac{\pi^2}{6}\sum \limits_{k=1}^\infty \cfrac{c_k}{k} =2\tag8$

$\sum \limits_{k=1}^\infty \cfrac{c_k}{k} =\cfrac{12}{\pi^2}\tag9$


I summarized my results and statements. I am trying to prove 3th,4th,5th statements

  1. if $k>2$ and it is prime number then $c_{k}=-\frac{1}{k}$
  2. if $k$ is $2^{m_0}p^{m_1}$ where $p>2$ and prime number ;$m_1>1$ and $m_0$ is non-negative integer then $c_{k}=0$ (example is $k=9,18,25,27,36,49,50,63,98,99,100$, need prove it)

  3. (for now without proof) It seems $c_{k}=\frac{1}{2}$ if $k=2^m$ where m is positive integer. I am very near to proof the statement. I will edit if I prove it.

  4. (for now without proof) my statement is if $ k=2^{m_0}.p_1^{m_1}.p_2^{m_2}.p_3^{m_3}...p_n^{m_n}$ where $p_1 , p_2 ,..,p_n$ are primes bigger than 2 and $m_0,m_1,m_2,m_3...m_n>0$ then $c_{k}=(-1)^n\frac{1}{2.p_1.p_2.p_3...p_n}$ (example is $k=6,10,12,24$, I need to prove it)
  5. (for now without proof) my statement is if $ k=p_1^{m_1}.p_2^{m_2}.p_3^{m_3}...p_n^{m_n}$ where $p_1 , p_2 ,..,p_n$ are primes bigger than 2 and $m1,m2,m3...m_n>0$ and integer then $c_{k}=(-1)^n\frac{1}{p_1.p_2.p_3...p_n}$ (example is $k=15,21,105$. I need to prove it)

Can my 2nd ,3rd, 4th and 5th statements be true? EDIT: Thanks to @Barry Cipra for the link and table for first 100 terms. I extended statement 2

Please help to prove or disprove them.

If I find the general rule of $c_k$, $e^{A(q)}$ can be expressed as product terms of $(1-q^k)^{c(k)}$.

Thanks for advises and helps.

share|cite|improve this question
I don't know if this will be of any help, but writing your $c_k$s in the form $a(k)/k$ and plugging the sequence 1,-1,-1,2,-1,-1,-1,4,0,-1,-1,-2 into the OEIS leads to which seems to satisfy your statement 3. – Barry Cipra May 2 '13 at 21:32
@Barry Cipra Thanks a lot for the link. It is very helpful. – Mathlover May 6 '13 at 12:39
@Mathlover, glad to help. But it looks to me that whynot has answered your questions. – Barry Cipra May 6 '13 at 14:17

1 Answer 1

In your equation $A(q)=−\sum_{k\ge 1} c_k \ln(1−q^k)$, expand the logarithm. You get $$ A(q)= \sum_{m=1} \frac{q^m}{m}\sum_{k\vert m} kc_k. $$ By Mobius inversion, you get $$ c_k = \frac1k \sum_{d\vert k} \mu(k/d) \delta(d) $$ where $\delta(m)=m$ if $m$ is a power of 2, and 0 otherwise. The expression for $c_k$ can be written as $$ c_k =\frac1k\sum_{u=0}^{v_2(k)}\mu(k/2^u) 2^u, $$ where $v_2(k)$ is the $2$-adic valuation of $k$.

It is easy to compute this sum (use the multiplicativity of $\mu$) and if I am not mistaken, $c_k= \frac{\mu(k)}k$ if $k$ is odd and $c_k=\frac{\mu(\ell)}{2\ell}$ if $k=2^{v_2(k)}\ell$ is even. Or something like that.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.