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I was doing some problems in Arthur Engel's "Problem-Solving Strategies" when I came across this problem in the Induction section:

Find a closed formula for the sequence $a_1 = 1,$ $$a_{n+1} = \frac{1}{16}(4a_n + 1 + \sqrt{24a_n + 1}).$$

Now the problem itself is not so interesting, but what I found interesting was a comment by the author at the end of his solution. He leaves the following remark:

"The sequence $a_n$ converges to $\int_0^1 x^2dx$. The recursion is a "duplication formula" for the parabola $y = x^2$. This is the way I discovered it."

Obviously the statement that the sequence converges to the same value as the integral ($1/3$) is not hard to verify. But I am at a loss as to how the recursion formula is in any way related to the parabola $y=x^2$ and what the author means by "duplication" formula in this case. Typically I would think of a duplication formula being an expression for $f(2x)$ in terms of $f(x)$, which in the case of the stated parabola would simply lead to the recursion $b_{n+1} = 4b_n$. It is not clear to me how this could be reflected in a definite integral.

Any insight would be greatly appreciated as I am curious how the author connected this recursive sequence to the definite integral.

ThreeDee
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    I'm guessing that $a_n$ represents a particular Riemann sum that approximates the given integral, and that $a_{n+1}$ represents perhaps the same type of Riemann sum but with twice as many intervals. – Greg Martin Aug 14 '22 at 05:56
  • It would be good to see the mentioned closed formula, that is, $a_n={(2^n+1)(2^n+2)\over3\cdot2^{2n}}$. – Ivan Neretin Aug 14 '22 at 22:01

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