# Even more interesting equation with fractions

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.

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, ...

• Excellent work - just what I was looking for! The OEIS sequence is where I got the puzzle from. Commented Mar 4 at 1:13
• 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. Commented Mar 4 at 2:12