$m$ doesn't come in the succession $a_n=$ iff m is a perfect square (1 answer)

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define which herbal numbers do not belong to the set $$E = \left\\left \mid n \in \tasiilaq.netbbN\right\.$$

The price is the set of hopeful perfect squares. I am not sure just how to prove it. We understand that $\left = n+\left<\sqrtn+\frac12 \right>$, but what carry out I execute from here?


Let $f(n):=n+\sqrtn+\frac12$ and also $g(n):=\big\lfloor f(n)\big\rfloor$ because that each $n=1,2,\ldots$. Observe the $f(n)$ is no an creature for any type of positive integer $n$. We have $1k+2\,.$$Hence, $$\left(\sqrtn+\frac12\right)^2k+\frac74\,.$$Therefore,$$\sqrtn\sqrtk+\frac74-\frac12\,.$$Ergo,$$k+1-\sqrtk+\frac74l^2+l+1+\sqrtl^2+l+\frac14+\frac12=l^2+2l+2\,.$$Therefore, $g\left(l^2+l\right)=l^2+2l=(l+1)^2-1$ and $g\left(l^2+l+1\right)=l^2+2l+2=(l+1)^2+1$. Hence, $(l+1)^2\notin E$ because that every $l=1,2,\ldots$. The proof is now complete.


when walk $\lfloor \sqrt n + \frac 12\rfloor$ jump?

$(k+0.5)^2 = k^2 + k + 0.25$

when $n$ can be composed as $k^2 + k$ you are on the low side that the next step, and also when $n$ deserve to be composed as $k^2 + k + 1$ you space on the high next of a step.

$\lfloor k^2 + k +\sqrt k^2 + k + \frac 12\rfloor = k^2 + 2k\\\lfloor k^2 + k + 1+\sqrt k^2 + k + 1 + \frac 12\rfloor = k^2 + 2k + 2\\$

And, $(k^2 + 2k + 1) = (k+1)^2$ gets skipped.


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