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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.

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up vote 1 down vote accepted

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!

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@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! – anon Mar 16 '12 at 16:25
Generalization much appreciated, and also the link to 'direct sum' which I will visit. – daniel Mar 16 '12 at 16:38
@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. – 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, $p_i(2)p_j(3)p_k(5) = p_{i+j+k}(23)$.

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