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For any three consecutive members of a sequence, the first and third members are near consecutive.

1 squared is 1. 2 squared is 4. So 1 and 4 are consecutive perfect squares.

1 squared is 1. 3 squared is 9. So 1 and 9 are near consecutive perfect squares.

I want to verify this formula to go from any perfect square to the next near consecutive perfect square. Let a be any perfect square. Let c be the next near consecutive perfect square. Here is the formula:

a+4(1+$\sqrt a$)=c

Example:

49+4(1+$\sqrt 49$)=81

Did I get the formula right? Is my terminology ok?

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Yes! Its right! If your square is $a$ then the number that originate it is $\sqrt{a}$ thus the next near consecutive perfect square is $(\sqrt{a}+2)^2$, but $(\sqrt{a}+2)^2 = a + 4(\sqrt{a}+1)$

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I have found there is a easier way to do this

(N+1)squared And (N-1)squared

Like for example (1+1)squared is 4 then (4-1)squared is 9 then (9-1)squared is 18

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  • $\begingroup$ I just figured out this is incorrect. $\endgroup$ – Dustin Nov 9 '19 at 21:47

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