$m$ 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. ## Not the answer you're spring for? Browse other questions tagged number-theory or ask your own question. If each each the two herbal numbers$a$and$b$is a amount of two squares climate$ab$is likewise a sum of two squares uncover the quantity of natural numbers that deserve to be composed as$x^2$,$x^3$and also$x^5$that are smaller or equal 보다$2^30\$ site design / logo © 2021 stack Exchange Inc; user contributions license is granted under cc by-sa. Rev2021.9.24.40305

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