Prove that every practical number is either a power of two or a power of two times a non-trivial polygonal number 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.
 A: 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.
