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How do I interpret the formula for OEIS A334742 (https://oeis.org/A334742), which is given as:

$$a(A033638(n)) = a(A002620(n)) \,\,\mathrm{for}\,\, n > 1.$$

Since

$$A002620(n)= \Bigl\lfloor\frac{n^2}{4}\Bigr\rfloor$$

and

$$A033638(n) = A002620(n)+1 = \Bigl\lfloor\frac{n^2}{4}\Bigr\rfloor+1,$$

it seems we have:

$$a( \Bigl\lfloor\frac{n^2}{4}\Bigr\rfloor+1) = a( \Bigl\lfloor\frac{n^2}{4}\Bigr\rfloor) \,\,\mathrm{for}\,\, n > 1.$$

theorist
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  • For $n>1$ , we have $$\underbrace{a\left( \lfloor \frac{2^2}{4} \rfloor +1\right)}_{a(2)} = a\left( \lfloor \frac{2^2}{4}\rfloor \right) = a(1)=1$$ Again $$a(3)=a\left( \lfloor \frac{3^2}{4} \rfloor+1\right)=a\left( \lfloor \frac{3^2}{4} \rfloor\right) = a(2)=1 $$ and so on – Amrit Awasthi Dec 09 '21 at 04:11

1 Answers1

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The sequence $\left\{\left\lfloor\frac{n^2}{4}\right\rfloor\right\}=\{0,1,2,4,6,9,12,16,20,25,\ldots\}$

So your final line tells us: $$\begin{align} &a(1)=a(0)\\ &a(2)=a(1)=a(0)\\ &a(3)=a(2)=a(1)=a(0)\\ &a(5)=a(4)\\ &a(7)=a(6)\\ &a(10)=a(9)\\ &a(13)=a(12)\\ &a(17)=a(16)\\ &a(21)=a(20)\\ &a(26)=a(25) \end{align}$$

That is, it tells us some members of $a(n)$ are equal to other members.

Note that it says nothing about certain members like $a(8)$. And it doesn't say anything about how $a(3)$ and $a(4)$ relate to each other (for example).

2'5 9'2
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