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Prove that every practical number is either a power of two or a power of two times a non-trivial polygonal number, where a number $q$ is practical if and only if every integer less than or equal to $q$ can be represented as a sum of distinct divisors of $q$, the polygonal numbers are denoted as usual by $P_s(n)=\dfrac{n^2(s-2)-n(s-4)}{2}$, with $s \geq 3$, and a non-trivial polygonal number is one with $n \geq 3$, which ensures that the result is not trivially true (since $P_s(2)=s$).

Together with the result of G. Melfi in On Two Conjectures About Practical Numbers, proving the Goldbach conjecture analogue for practical numbers (every even integer is a sum of two practical numbers), the conjecture implies that every integer can be expressed as $2^{a_0-1}(P_{s_0}(n_0)+2^{a_1-a_0}P_{s_1}(n_1))$, with $a_1 \geq a_0 \geq 0$, $s_0, s_1 \geq 3$, $n_0, n_1 \geq 1$ and $n_0, n_1 \ne 2$.

My motivation was a very straightforward attempt to explain the high frequency of practical numbers of the form $n^2-1$. Practical numbers $>1$ are even, so we can rewrite this as $(2n+1)^2-1=8T_n$, where $T_n$ is the nth triangular number. Replacing 8 with any power of two, I found that large portion of small practical numbers had such a representation, but the dropoff is steep as we consider larger values. Replacing The triangular numbers with polygonal numbers I was unable to find a counterexample for practical numbers < 30,000.

Jaycob Coleman
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  • 220 views, but not even a single comment - I'm surprised. Just out of my curiosity, are you planning on doing any searches beyond 30,000? – Ryan Sep 08 '13 at 02:53
  • Actually I just realized that the method I used to generate practical numbers was flawed and missed a small number of them. I'm not aware of a quick method to find all of them, but I'll work on this and try retesting at least up to $30,000$. – Jaycob Coleman Sep 08 '13 at 05:11
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    There is a quick method to enumerate them but it uses a bit of memory. From the Stewart/Sierpinski characterisation we can generate them using a priority queue; for efficiency, store tuples $(n,p,\sigma(n),p^{a+1})$ where $p$ is the largest prime factor of $n$ and $a$ is its multiplicity in $n$. Initially insert $(1,2,1,2)$. When you pop a tuple, push $(np,p,\sigma(n)(p^{a+2}-1)/(p^{a+1}-1),p^{a+2})$ and $(nq,q,\sigma(n)(q+1),q^2)$ for all prime $p<q\le 1+sigma(n)$. I've implemented this in C# and get up to 30000 in 13 seconds. By not pushing tuples with $n>30000$ it takes 0.04 seconds. – Peter Taylor Sep 12 '13 at 18:35
  • @PeterTaylor Thank you, I'm not a fantastic programmer. Would you perhaps be interested in replacing the current oeis table of practical numbers with your own? – Jaycob Coleman Sep 12 '13 at 23:18
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    I might consider putting up a list somewhere else, but I don't think that OEIS tends to go in for excessively long lists. In the meantime, here's some code (I've optimised it and ported it to Java, which has a priority queue in the core library). – Peter Taylor Sep 13 '13 at 07:35
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    @Peter, I hope you will reconsider and submit a link to OEIS. They are willing to provide links to more extensive lists - there are links to lists of 100,000 primes, and 1200 highly composite numbers. I for one would like more practical numbers to use for pattern analysis. – half-integer fan Nov 19 '13 at 03:25
  • @half-integerfan Patterns like these? I've also found no non-practical records of the arithmetic derivative of the positive integers (which differ from the link only for $a(1)=1$) or highly abundant numbers $>10$. – Jaycob Coleman Nov 23 '13 at 13:11
  • @Jaycob, for the moment I was looking at lines of practical numbers for $x^2 \pm y$ or $P(x) \pm y$ (P being the pronic numbers), in a parallel to the patterns seen for primes in the Sacks number spiral. BTW, it was answering your other question that started me on the practical numbers and I haven't been able to give them up yet. – half-integer fan Nov 23 '13 at 14:24
  • @half-integerfan Yes, they're very interesting and I think there's much more to them than what is known. I've looked at the Ulam spiral for small practical numbers in the past, and they certainly seemed to show the same sort of patterns, but I figured I'd keep it under my hat until I (hopefully) have a better understanding of why such patterns might occur in general. – Jaycob Coleman Nov 23 '13 at 14:54
  • @half-integerfan, I've added lists of 1000 and 1000000 practical numbers to my personal site at http://cheddarmonk.org/maths/practical_numbers/practical-numbers-1000.txt and obvious substitution. I will refresh my memory on the current comment submission process for OEIS at a later date. – Peter Taylor Nov 25 '13 at 00:07
  • @PeterTaylor I will do it if you wish. I want more experience updating OEIS. Let me know. p.s. love the domain name, and those laser swords. – half-integer fan Nov 26 '13 at 02:35

1 Answers1

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Suppose $x$ is a practical number which isn't a power of $2$. We want to verify that $$\exists j\ge0, s\ge 3, n\ge 3 : x = 2^j \frac{n^2(s-2) - n(s-4)}{2}$$

It's immediately apparent that $n$ divides $x$. Given a candidate value of $n$, we have $$s-2 = \frac{2^{1-j}xn^{-1} - 2}{n-1}$$ so for $s\ge 3$ we require $2^{1-j}xn^{-1} - 2 \ge n-1$, and for $s \in \mathbb N$ we require $2^{1-j}xn^{-1} \in \mathbb N$ and $2^{1-j}xn^{-1} \equiv 2 \pmod{n-1}$. Given the second condition, the first simplifies to $2^{1-j}xn^{-1} > 2$.

Empirically, for all possible values of $x \le 120000000$ this is true with $n$ equal either to $4$ or to the smallest odd prime factor of $x$. Let's consider these two cases.

Case: $n = 4$

We require $x > 2^{2+j}$, $2^{-1-j}x \in \mathbb N$ and $x \equiv 2^{j} \pmod{3}$.

The modular expression clearly fails if $x \equiv 0 \pmod 3$; otherwise it holds either for all odd $j$ or for all even $j$, so in particular it holds for one of $j=0$ or $j=1$. In either case, the fact that $n$ divides $x$ implies that $2^{-1-j}x \in \mathbb N$, and $x$ is large enough if it's greater than $8$.

Therefore this case covers all non-power-of-$2$ practical numbers except those smaller than 8 (i.e. $6$), those which are multiples of $3$, and those which aren't multiples of $4$. It turns out that all practical numbers are multiples of $4$ or $6$, so this simplifies to: the case $n=4$ covers all practical numbers except multiples of $3$, and we can reduce our consideration of the case $n = p_2$ to $p_2 = 3$.

Case $n = p_2 = 3$

We require $2^{1-j}x > 6$, $2^{1-j}3^{-1}x \in \mathbb N$ and $2^{1-j}x \equiv 0 \pmod 2$. If we set $j = 0$ then $x > 3$ is satisfied, $2x/3 \in \mathbb N$, and $2x \equiv 0 \pmod 2$, so this case is easy.

QED.

Peter Taylor
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    Or to put it in simpler terms: every practical number greater than $1$ is even and hence equivalent to $0$, $2$, or $4\pmod 6$. But polygonal numbers with $n=3$ enumerate all multiples of $3$ greater than $3$, so in particular all practical numbers which are equivalent to $0\pmod 6$; and polygonal numbers with $n=4$ enumerate the equivalence class of $4\pmod 6$ starting at $10$. That just leaves the practical numbers equivalent to $2\pmod 6$, and since they're multiples of $2$ half their value is equivalent to $4\pmod 6$. – Peter Taylor Sep 13 '13 at 11:50