2
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Continuing the investigation from this question on CGSE about pentagonal-pentagonal numbers:

Defining $p(n)$ as the $n$th pentagonal number (a positive integer of the form $n(3n−1)/2,\ n\geq 1$), and $pp(n)$ as the $n$th pentagonal-pentagonal number, where $pp(n)$ is the smallest pentagonal number which is also the sum of $n$ consecutive pentagonal numbers (if it exists), we have

$$pp(n)=\sum_{i=N}^{N+n-1}p(i)$$

Letting the definition of $pp(n)$ be the smallest solution for the above sum, gives, for example,

$$pp(4)=5+12+22+35=330$$

However, there is no solution for $pp(3)$.

The solutions get large quickly, as obviously $pp(n)\geq n(n-1)^2/2$, but are often much larger.

eg the smallest solution for $$pp(926)=68861900859697184438810992296964899584575857208833324126$$

Defining

\begin{equation} f(n)= \begin{cases} &1 \quad \text{ if }pp(n)\text{ has a solution}\\ &0 \quad \text{ otherwise}\\ \end{cases} \end{equation}

$\sum_{k=1}^{n}f(k)$ is then the counting function for the number of $pp(n)$ with a solution $\leq n$.

The following are the values I got for $f(n),\ 1\leq n\leq 1000$:

pp={1,1,0,1,1,1,1,0,1,0,1,1,0,0,1,0,0,1,1,1,1,0,0,0,1,1,1,0,1,1,0,0,0,0,1,1,1,0,1,1,1,1,1,0,0,1,1,0,1,1,0,0,0,0,1,1,0,0,0,1,1,1,0,1,0,0,1,1,1,1,1,1,0,1,1,1,0,0,1,0,1,0,0,1,1,1,0,0,1,1,1,1,0,0,0,0,1,0,0,1,0,1,0,0,0,1,1,0,1,1,1,1,0,1,0,0,0,0,1,0,1,0,0,1,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1,0,0,0,1,1,0,1,0,0,0,0,0,1,1,0,0,0,1,1,0,0,0,1,1,0,1,1,0,0,0,0,1,1,1,1,0,1,0,1,1,0,0,0,0,1,0,1,1,1,0,1,0,1,1,1,0,0,0,0,1,0,0,0,0,0,0,1,1,1,1,0,1,1,1,1,0,0,0,1,1,0,1,1,1,0,0,0,1,0,1,0,0,0,0,1,0,1,1,0,1,0,1,0,1,0,0,1,1,1,0,0,1,0,0,1,0,0,1,0,0,0,1,1,0,1,0,0,1,1,1,0,1,1,0,1,0,0,0,0,1,0,1,1,0,1,0,1,0,1,0,1,1,1,1,0,0,1,1,1,1,0,0,1,1,0,0,0,0,0,0,0,1,1,1,1,0,1,0,0,1,0,1,0,0,1,0,1,1,1,0,0,1,0,1,0,0,0,0,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,1,0,0,1,0,0,0,1,1,0,1,0,0,1,1,1,0,1,1,0,1,0,0,1,0,1,0,1,1,0,0,1,1,0,1,1,0,0,1,1,0,0,1,0,1,1,0,0,0,1,1,0,1,1,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,0,0,1,1,1,0,0,0,1,1,1,1,0,0,1,1,0,0,1,0,1,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,1,1,1,1,0,0,1,1,1,1,0,0,0,0,1,0,1,1,0,1,0,1,0,0,1,0,0,0,0,1,0,1,1,1,1,0,1,1,1,0,0,0,1,1,1,0,0,0,1,1,1,0,1,0,1,0,0,1,0,0,0,1,1,0,0,0,1,0,1,1,0,1,0,0,0,1,1,1,1,1,0,0,0,1,1,0,0,0,0,1,0,1,1,1,1,0,1,1,1,0,1,0,0,1,1,0,0,0,0,0,0,0,1,1,1,0,0,0,1,1,0,1,1,0,0,0,1,0,1,0,0,1,1,1,1,0,1,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,1,0,1,1,1,1,0,1,1,0,0,0,0,0,1,1,0,0,1,1,0,0,1,1,0,1,0,0,1,1,1,0,0,0,0,1,0,1,1,0,0,0,0,0,1,0,0,1,0,1,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,1,1,0,0,0,1,0,1,0,1,1,0,1,1,0,1,0,0,0,1,1,0,0,0,1,0,1,0,1,1,0,0,0,0,1,1,0,1,0,0,0,0,1,1,1,0,0,0,1,1,1,0,1,0,1,0,0,0,1,1,0,0,0,1,1,1,0,0,0,1,0,0,1,0,0,1,1,1,1,1,1,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,0,1,1,0,0,1,0,0,0,0,0,0,1,0,0,0,1,1,1,0,0,0,1,0,1,0,1,0,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,1,1,0,1,1,0,0,0,1,1,1,0,0,1,1,1,0,0,0,0,0,1,0,1,1,0,0,0,0,1,0,1,0,0,0,1,1,0,1,1,0,0,0,1,1,1,0,0,1,0,0,1,0,0,0,1,1,0,1,1,1,0,1,0,1,1,1,0,0,1,0,0,0,0,1,0,1,0,1,0,0,1,0,1,0,1,1,1,1,0,0,1,0,0,1,1,0,0,1,1,1,0,0,0,1,1,0,0,0,1,0,1,1,1,0,1,0,0,0,0,1,1,0,0,0,1,1,0,1,0,1,0,0,1,0,0,0,0,0,0,1,0,0,1,1,0,1,0,0,0,1,0,0,0,1,1,1,0,1,0,1,1,0,0,1,1,0,0,1,1};

Clearly, there are an infinite number of $pp(n)$ with solutions, since $$pp(3 n (n+1)+1)=3 (n+1) (3 n (n+1)+1) ((n+1) (3 n (n+1)+1)+1)$$

I am guessing from calculations that

$$\sum_{k=1}^{n}f(k)\sim \dfrac{n}{\log\log n +1/5}$$

Is this the case? If not, what is the asymptotic density of $f(n)$ in $\mathbb{N}$?

Can this be generalised to other polygonal numbers?

Improve this question
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  • 1
    $\begingroup$ Can you fix your definition of $pp(n)$? Your question will clearly make sense at some point, but currently I can make neither head nor tail of any of your definitions of $pp(n)$. $\endgroup$
    – eric
    Jan 5, 2016 at 18:57
  • $\begingroup$ Oh -- the definition (from the link in the question) is this: "$pp(n)$ is the smallest pentagonal number which is also the sum of $n$ consecutive pentagonal numbers, if it exists". Here a pentagonal number appears by definition to be positive, and so is a number of the form $(3n^2-n)/2$ for an integer $n>=1$. $\endgroup$
    – eric
    Jan 5, 2016 at 19:00
  • $\begingroup$ I would be slightly surprised if your computations of $f(n)$ are 100% accurate: I'm assuming you're just searching up to a certain range and then giving up. These are generalised Pell equations and solving them when $n$ is large will involve some subtleties I suspect. $\endgroup$
    – eric
    Jan 5, 2016 at 19:04
  • 1
    $\begingroup$ What does $\sum_{i \in \mathbb N}^{i + n}$ mean? Is it meant to be $\sum_{i = 1}^n$? $\endgroup$
    – LSpice
    Jan 5, 2016 at 22:15
  • 1
    $\begingroup$ The definition you have right now is the some of $n+1$ consecutive pentagonal numbers. Change the limits in your sum to $\sum_{i=N}^{N+n-1}$. $\endgroup$
    – Mark Fischler
    Jan 5, 2016 at 23:45

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