<s>[INST] Is $h(n)$ independent of $n$? Let $n$ be a positive integer.
Let $f(n) = \lfloor\frac{8n+13}{25}\rfloor$ and $g(n) = \lfloor\frac{n-17}{25}\rfloor$.
Now consider $$h(n) = f(n) - \lfloor\frac{n - 12 - g(n)}{3}\rfloor$$
Then I calculated values of $h(n)$ for various values of $n$ and I'm getting $h(n)=4 $ everytime. Infact I ran a C code also and verified it till $n = 10^6$. But I am not able to prove this equality. How can I prove this?
 [/INST] Let $n=25k+r$ for some non-negative integers $k,r$ with $r<25$. We then have
$$\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor = \left\lfloor\frac{24k+r-12-\left\lfloor\frac{r-17}{25}\right\rfloor}{3}\right\rfloor = 8k-4+\left\lfloor\frac{r-\left\lfloor\frac{r-17}{25}\right\rfloor}{3}\right\rfloor.$$
We also have
$$\left\lfloor\frac{8n+13}{25}\right\rfloor=8k+\left\lfloor\frac{8r+13}{25}\right\rfloor.$$
So it suffices to show that for any non-negative integer $r<25$,
$$\left\lfloor\frac{8r+13}{25}\right\rfloor - \left\lfloor\frac{r-\left\lfloor\frac{r-17}{25}\right\rfloor}{3}\right\rfloor=0.$$
Try showing that this is true.
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