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Question Number 202306 by hardmath last updated on 24/Dec/23

$$$$Find the 2023rd term in the sequence 2,3,5,6,7,8,10,11,12,13,14,15,17,18,... obtained by subtracting integer squares from natural numbers.

Answered by deleteduser1 last updated on 24/Dec/23

$$1,2,3,...,2023contains44squares$$$$\Rightarrow 2023isthe(2023-44=1979)thterm$$$$2023-1979thterm$$$$2023+44-2022ndterm;since{45}^2=2025$$$$\Rightarrow 2023rdterm=2023+45=2068$$

Commented by hardmath last updated on 24/Dec/23

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}$$

Answered by mr W last updated on 24/Dec/23

$$1,2,3,4,5,...,n$$$$1^2,2^2,3^2,...,k^{2}with{k}^2<n,i.e.k=\lfloor \sqrt{n}\rfloor$$$$m=n-k=n-\lfloor \sqrt{n}\rfloor$$$$thatmeansthe{m}^{th}termisn.$$$$\lfloor \sqrt{n}\rfloor ⩽\sqrt{n}$$$$m=n-\lfloor \sqrt{n}\rfloor ⩾n-\sqrt{n}$$$$n-\sqrt{n}-m⩽0$$$$\Rightarrow \sqrt{n}⩽\frac{1+\sqrt{1+4m}}{2}$$$$\Rightarrow n⩽\frac{{(\sqrt{4m+1}+1)}^2}{4}$$$$i.e.the{m}^{th}termis$$$$a_m=\lfloor \frac{{(\sqrt{4m+1}+1)}^2}{4}\rfloor$$$$examples:$$$$m=2023:$$$$a_{2023}=\lfloor \frac{{(\sqrt{4\times 2023+1}+1)}^2}{4}\rfloor =2068$$$$m=2025:$$$$a_{2025}=\lfloor \frac{{(\sqrt{4\times 2025+1}+1)}^2}{4}\rfloor =2070$$$$m=1000:$$$$a_{1000}=\lfloor \frac{{(\sqrt{4\times 1000+1}+1)}^2}{4}\rfloor =1032$$$$m=3000:$$$$a_{3000}=\lfloor \frac{{(\sqrt{4\times 3000+1}+1)}^2}{4}\rfloor =3055$$

Commented by hardmath last updated on 24/Dec/23

$$\mathrm{perfect}\mathrm{solution}\mathrm{tahnkyou}\mathrm{dear}\mathrm{professor}$$

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