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Let $f(n)=1+x^n+x^{2n}$

Let $p(x)$ be $1+x+x^2+x^5+x^7+...$ where the exponents are the pentagonal numbers.

Let $a(n)$ be the sequence of integers such that the coefficients of the series $f(a(1)) f(a(2)) f(a(3))...$ are congruent mod $2$ to the coefficients of $p(x)$

The first few values of $a(n)$ are: $1,5,6,7,9,11,13,17,18,19,23,25$.

Question 1: Is it true that $a(n+1)-a(n)$ always is $1$, $2$, or $4$?

The first few values of $a(n+1)-a(n)$ are : $4, 1, 1, 2, 2, 2, 4, 1, 1, 4, 2, 4, 1, 1, 4$

I've checked and it is so for the first $900$ elements of the sequence.

Question 2: Is the sequence $a(2)-a(1) , a(3)- a(2) , a(4)- a(3),...$ periodic?

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    $\begingroup$ You can actually describe exactly which numbers appear as a value of $a(n)$. For a positive integer $m$ relatively prime to $6$, and nonnegative integers $a,b$, the integer $2^a3^bm$ is a member of the $a(n)$ sequence if and only if $2^a$ appears in the binary expansion of $b+1$. This implies a positive answer to question 1 and a negative answer to question 2. (Apologies for leaving a comment instead of a detailed answer. If no one has expanded on this I can write a proof when I have access to a keyboard.) $\endgroup$
    – Gjergji Zaimi
    Commented Apr 27, 2018 at 22:05
  • $\begingroup$ You mean $f(a(1))\cdot f(a(2))\cdot f(a(3))...$? $\endgroup$
    – Fedor Petrov
    Commented Apr 28, 2018 at 12:57
  • $\begingroup$ @Fedor Petrov I've made a correction in response to your comment , but I'm not sure if it answers your question. $\endgroup$
    – David S. Newman
    Commented Apr 28, 2018 at 15:10
  • $\begingroup$ @GjergjiZaimi Will similar arguments work if f(n) is taken to be 1+x^n+x^2n+x^3n+...+x^kn . ? $\endgroup$
    – David S. Newman
    Commented Apr 29, 2018 at 0:56
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    $\begingroup$ @FedorPetrov I finally understood your question and corrected the problem to read as you have indicated. I would like to think that those extra commas were not put there by me, but by some other well meaning editor. $\endgroup$
    – David S. Newman
    Commented Apr 30, 2018 at 4:35

1 Answer 1

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Start by noticing that the generating function of pentagonal numbers when working in $\mathbb Z/2\mathbb Z$ is given by $$p(x)=1+\sum_{k=1}^{\infty}\left(x^{\frac{k(3k-1)}{2}}+x^{\frac{k(3k+1)}{2}}\right)=\prod_{n\geq 1}(1+x^n)=\prod_{n\geq 1}\frac{1}{(1+x^{2n-1})}$$ The first equation comes from Euler's pentagonal number theorem, and the second from Euler's bijection between partitions into distinct parts and partitions into odd parts.

Next we notice that we have a simple telescoping factorization $$\frac{1}{1+x^m}=\left(\frac{1+x^{3m}}{1+x^m}\right)\left(\frac{1+x^{9m}}{1+x^{3m}}\right)\cdots=\prod_{k\geq 0}f(3^km)$$ and applying the previous equation gives us $$p(x)=\prod_{n\geq 1, k\geq 0}f(3^k(2n-1))=\prod_{k\geq 0}\prod_{(m,6)=1}f(3^km)^{k+1}.$$ Our last ingredient is the fact that mod 2 we have $(x+y)^{\sum 2^{a_i}}=\prod(x^{2^{a_i}}+y^{2^{a_i}})$. Define the set $S$ as the integers which can be written in the form $2^r3^km$ with $(m,6)=1$ and $2^r$ appearing in the binary expansion of $k+1$. Then we have proved: $$p(x)=\prod_{n\in S}f(n).$$ From here notice that all integers relatively prime to $6$ are in $S$, which bounds the differences between consecutive members by $4$. Next notice that the only way for $3$ to appear as a consecutive difference is if $6k-2$ or $6k+2$ is in $S$ but they are both ruled out by our characterization. Therefore Question 1 has an affirmative answer. With a little more work you can also show that Question 2 has a negative answer.

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