Let $p_k(m)^2:=$ the square of $m^{th}$ number containing k prime factors, including repetitions.

Empirically for smallish numbers and as a conjecture, it appears that for every m and sufficiently large k, the formula

$p_k(m)^2 = p_{2k}(n)$ returns the same value of n.


$p_1(1)^2 = 4 = p_2(1)$,

$p_2(1)^2 = 16 = p_4(1)$,...

$p_1(5)^2 = 121 = p_2(40)$

$p_2(5)^2 = 196 = p_4(24)$

$p_3(5)^2 = 729 = p_6(23)$

$p_4(5)^2 = 2916 = p_8(23)$, ...

For m = 1-7 the elements of the sequence are: 1, 3, 8, 12, 23, 26, 32,...

The case m = 1 is easy enough. After that I'm not so sure.

Thanks for any insights into this property.

Edit: I think this generalizes a bit. For example,

$p_i(a)p_j(b)...p_k(c) = p_{i+j+...+k}(d)$, in which i, j,..., k are increased by 1 stepwise so their difference remains constant, also returns a constant d.

It also seems that the order of {a,b,...,c} does not change the value of d. For example,

$p_1(2)p_2(3)p_4(5) = p_7(23)$ and using the conjecture,

$p_3(2)p_4(3)p_6(5) = p_{13}(23)$ and if order unimportant$^*$,

$p_3(5)p_4(2)p_6(3) = p_{13}(23)$ and again using the conjecture,

$p_4(5)p_5(2)p_7(3) = p_{16}(23)$, verified computationally.

And as an aside, again for "sufficiently" large subscripts, if we know that, e.g.,

$p_1(2)p_3(5)p_8(6) = p_{12}(41)$ we can augment subscripts pairwise to get

$p_3(2)p_5(5)p_8(6) = p_{16}(41)$.

$^*$ Changing the order of {a,b,...c} may change the size of subscripts needed for the property to obtain. The subscripts {1,2,4} and {2,3,5} do not work for the product in the last line, but {3,4,6} and higher appear to.


The set of all numbers $n$ with $k$ prime factors (counted with multiplicity) is denumerable, hence for any such number $n$ there is an $i$ for which $n$ is the $i$th such number (here $k$ is fixed). Thus if

$$n=p_1^{r_1}\cdots p_l^{r_l}$$

has $r_1+\cdots+r_l=k$ factors, its square will have twice that, whence it is the $i$th number with $2k$ factors for some $i$, i.e. $p_{2k}(i)$.

More generally, there exists an isomorphism between the nonzero rationals and the direct sum

$$\mathbb{Q}^\times\cong\bigoplus_p\mathbb{Z} ~:~ \mapsto\big(v_2(n),v_3(n),v_5(n),\cdots),$$

where $v_p$ is the $p$-adic order, i.e. if $x\in\mathbb{Q}$ exists with $p^s$ in its generalized "prime factorization," the order will be $s$ (it takes any integer value). If we define $\ell(x)=v_2(x)+v_3(x)+v_5(x)+\cdots$ to be the generalized number of prime factors (counted with multiplicity, and also with reciprocals counted negatively; this is well-defined because only a finite number of primes can appear in the prime power decomposition of a rational number), then we can see that

$$\ell(xy)=\ell\left(\prod p^r \prod p^s\right)=\ell\left(\prod p^{r+s}\right)=\sum_{p}(r+s)=\left(\sum_p r\right)+\left(\sum_ps\right)=\ell(x)+\ell(y).$$

This means that $\ell$ is a group homomorphism from $\mathbb{Q}^\times$ to $(\mathbb{Z},+)$. It follows that if we define the preimages $\Gamma_k=\{x\in\mathbb{Q}:\ell(x)=k\}$ then we have a natural "graded" structure, represented by


(in the sense that $AB={ab:a\in A, b\in B}$). This is a higher generalization of your observations!

  • $\begingroup$ @daniel: Your more general question has the same answer, basically amounting to keeping track of the number of prime factors in the numbers involved. I have added some heavier number-theoretic description of a generalization of your observation to the rational numbers. And thanks for the bd 'grats! $\endgroup$ – anon Mar 16 '12 at 16:25
  • $\begingroup$ Generalization much appreciated, and also the link to 'direct sum' which I will visit. $\endgroup$ – daniel Mar 16 '12 at 16:38
  • 1
    $\begingroup$ @daniel: The article might be dense, so let me cut to the chase: $\bigoplus_p\mathbb{Z}$ is formed by all sequences $(a_2,a_3,a_5,\cdots)$ of integers (indexed by the primes) such that only a finite number of the $a_p$'s are nonzero. You can add these sequences together, and they form an abelian group under addition. $\endgroup$ – anon Mar 16 '12 at 16:56

For every $k$, call $a_k$ the integer such that $3^k = p_k(a_k)$. In the numbers with $k$ prime factors less than $3^k$, we have at least all the numbers $3^i\times 2^{k-i}$ where $0 \le i < k$, so that $a_k \ge k+1$.

Now, forall $k$ and $a$, if $a < a_{k+1}$, then $p_{k+1}(a) = 2 \times n$ for some $n$, because the smallest number with $k+1$ prime factors not including $2$ is $3^{k+1}$ and I supposed $a < a_{k+1}$. Then $n$ has to be the $a$-th smallest number with $k$ factors : $p_{k+1}(a) = 2 \times p_k(a)$. In particular, forall $a$, $a < a+1 \le a_a$, so $p_{a+k}(a) = 2^k \times p_a(a)$ forall $k$.

Now we can prove that forall $a,b$ there exists $c$ such that for $k_1$ and $k_2$ large enough, $p_{k_1}(a) \times p_{k_2}(b) = p_{k_1+k_2}(c)$ :

Let $l \ge a+b$ be an integer such that $2^{l-a-b}p_a(a)p_b(b) < 3^l$, and let $c$ be the integer such that $2^{l-a-b}p_a(a)p_b(b) = p_l(c)$. The definition of $l$ says that $c < a_l$, so we have that forall $k \ge l$, $p_{k}(c) = 2^{k-a-b}p_a(a)p_b(b)$

Then, forall $k_1,k_2$ such that $k_1 \ge a, k_2 \ge b, k_1+k_2 \ge l$, we have $p_{k_1}(a)p_{k_2}(b) = 2^{k_1+k_2-a-b}p_a(a)p_b(b) = p_{k_1+k_2}(c)$

For example, for $a=2$ and $b=3$, we get $p_2(2)=6, p_3(3) = 18$, we can pick $l=5$ and we get $108 = p_5(5)$, so $c=5$ : for $i,j$ large enough, we always have $p_i(2)p_j(3) = p_{i+j}(5)$. Similarly for $a=5$ and $b=5$ we get $c=23$, so that for all $i,j,k$ large enough, $p_i(2)p_j(3)p_k(5) = p_{i+j+k}(23)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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