Can you find distinct positive integers $a_1, a_2, \ldots, a_{n-1}, a_n$ for any $n$ such that

$$-\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_{n-1}}+\frac{1}{a_n} = -\frac{1}{a_1} \cdot \frac{1}{a_2} \cdot \ldots \cdot \frac{1}{a_{n-1}} \cdot \frac{1}{a_n}$$

The previous puzzle shows a solution for $n=4$. Computers are allowed this time.


1 Answer 1


The generalization of Retudin's answer to the previous puzzle works:

Repeatedly use the formula $-\frac{1}{x} + \frac{1}{x+1} = -\frac{1}{x(x+1)}$ to get the next term.

Let $a_1$ be any positive integer. Let $a_2 = a_1 + 1$: $$ -\frac{1}{a_1} + \frac{1}{a_2} = -\frac{1}{a_1} + \frac{1}{a_1+1} = -\frac{1}{a_1(a_1+1)} = -\frac{1}{a_1 a_2} $$

Since $-\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} = -\frac{1}{a_1 a_2} + \frac{1}{a_3}$, applying the same formula leads to a possible value of $a_3 = a_1 a_2 + 1$.

We can apply the same formula until we get the value of $a_n$. This gives the recurrence equation $$ a_n = a_1 a_2 \cdots a_{n-1} + 1 $$ which works for arbitrary $n$ and $a_1$, and all the integers are distinct since this sequence is strictly increasing for any positive integer $a_1$.

One concrete example of this construction would be

OEIS A129871: 1, 2, 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443, 12864938683278671740537145998360961546653259485195807, ...

  • $\begingroup$ Excellent work - just what I was looking for! The OEIS sequence is where I got the puzzle from. $\endgroup$ Mar 4 at 1:13
  • $\begingroup$ Interestingly $\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_{n-1}}+\frac{1}{a_n}$ for the OEIS values leads to the closest approximation to 1 for $n$ Egyptian fractions. $\endgroup$ Mar 4 at 2:12

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