Lemur Numbers

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Lemur Numbers

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#1 h May 22, 2009, 5:48 PM • 2 Y

Y by Adventure10, Mango247

Let the $n^{\text{th}}$ Lemur set, $L_n$ , be the set composed of all positive integers that are equal to the sum of the squares of their first $n$ divisors. For example, $L_1 = \{1\}$ , $L_2 = \{\}$ , and $L_4 = \{130\}$ .

A) Find $L_3$ , $L_5$ , and $L_6$ .
B) Describe all $n$ for which $L_n$ is empty.
C) Describe all $n$ for which $L_n$ is infinite.
D) Provide a method for finding members of non-empty Lemur sets.

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#2 h Jun 11, 2009, 5:49 PM • 2 Y

Y by Adventure10, Mango247

$L_3$ is empty.
Proof: Suppose $6\mid n$ . Then we would need $n = 1 + 4 + 9 = 14$ which isnt' divisible by 6.
Now suppose $n$ is divisible by 4 and not 6. Then $n = 1 + 4 + 16 = 21$ doesn't work.

Suppose $n$ is even and not divisible by 4 or 6. Then $n = 1 + 4 + p^2$ for some odd prime p which divides $n$ .
Then $2p\mid 5 + p^2 \Rightarrow p\mid 5 \Rightarrow p = 5$ , but $1 + 4 + 25 = 30$ does not work.

Suppose the first three divisors are $1,p,p^2$ . Then we get $p\mid 1$ , contradiction.

So now, we have $n = 1 + q^2 + p^2$ for $p > q$ the smallest two prime divisors of an odd integer $n$ . Consider the equation $kpq = 1 + p^2 + q^2$ for some $k$ . Eventually, we will take $k$ not divisible by a prime less than $q$ , and $p,q$ primes, but for now let them all just be positive integers, so we can Vieta jump.

Fix $k$ and suppose $p,q$ satisfy the above equation with $p > q$ . Then the quadratic $x^2 - kqx + q^2 + 1$ has two roots, $p$ , and $p_0 = kq - p = \frac {q^2 + 1}p$ (which is an integer). Note that $\frac {q^2 + 1}p$ must be less than q. Thus if $(p,q)$ is a solution with $p > q$ , then $(q,p_0)$ is a solution with $q > p_0$ . Thus, we can repeat this process until we get down to $(1,2)$ . For the pair $(1,2)$ , we find $k = 3$ and thus all solutions have $k = 3$ .

Since now $3pq = p^2 + q^2 + 1$ , so $3\mid n$ thus $q = 3$ and $9p = p^2 + 10$ so $p\mid10$ , and checking the cases, we have a contradiction.

P.S: Lemur numbers is a stupid name.

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#3 h Jun 16, 2009, 10:50 PM • 2 Y

Y by Adventure10, Mango247

Actually, I just thought of a simpler proof that $L_3$ is empty, relying on the fact that $p^2 \equiv 1 \pmod{6} \text{ } \forall \text{ } p > 3 \in \mathbb{P}$ .

We can break the problem up into the following six cases:

Case One: smallest three positive divisors are $1, 2, 3$ $\longrightarrow$ $6 \nmid 14$ .
Case Two: smallest three positive divisors are $1, 2, 4$ $\longrightarrow$ $4 \nmid 21$ .
Case Three: smallest three positive divisors are $1, 3, 9$ $\longrightarrow$ $9 \nmid 91$ .
Case Four: smallest three positive divisors are $1, 2, p$ ; $p > 3 \in \mathbb{P}$ $\longrightarrow$ $5 + p^2 \equiv 0 \pmod{6}$ , but $3$ is not a divisor.
Case Five: smallest three positive divisors are $1, 3, p$ ; $p > 3 \in \mathbb{P}$ $\longrightarrow$ $10 + p^2 \equiv 5 \pmod{6}$ , but $3$ is a divisor.
Case Six: smallest three positive divisors are $1, p, q$ ; $p, q > 3 \in \mathbb{P}$ $\longrightarrow$ $1 + p^2 + q^2 \equiv 3 \pmod{6}$ , but $3$ is not a divisor.

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#4 h Jun 22, 2009, 4:47 PM • 2 Y

Y by Adventure10, Mango247

The only lemur numbers less than 100000000 are:

1 in $L_1$
130 in $L_4$
1860 in $L_{11}$
148480 in $L_{19}$
3039520 in $L_{31}$

through C++. I am currently running it up to 1000000000.

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#5 h Jan 8, 2023, 5:13 PM

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Sequence of Lemur numbers in the OEIS: https://oeis.org/A185584

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