- Loser66

Conjecture the formula for the nth term of {a_n} if the first ten terms of it is
1,0,0,1,0,0,0,0,1,0
Please, help

- katieb

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- Loser66

I want to use the floor function to put it in neat. Obviously,
|dw:1440875326682:dw|

- Loser66

But how?

- freckles

what do you mean how if you say it is obvious?

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## More answers

- freckles

is the piecewise part obvious but not how to write in the floor format? is that what you?

- freckles

is that what you mean*?

- ikram002p

oh that remind me of Legendre function

- Loser66

When I say "obvious" , it means we can see it obviously.

- freckles

ok then what do you need help on then

- Loser66

I want the floor format.

- freckles

oh okay

- Loser66

\(\lfloor{\sqrt n}\rfloor\)

- Loser66

How about this? \(1<\sqrt 2 < 2\), hence \(\lfloor{\sqrt 2}\rfloor =1\)
and \(\lfloor{\dfrac{\lfloor{\sqrt 2}\rfloor}{\sqrt2}}\rfloor=0\)

- Loser66

It works well for n =3,4, 5, 6...
but how can I just list them out like this? I want a general logic to lead me there.

- ikram002p

so sqrt(n^2) is integer, then [sqrt(n^2) ] would be same integer n
else u would have an additional decimals
there are two lovely things about floor function
1-[sqrt(n^2) ]<= sqrt(n^2)
2-[r]=0 if 0

- ikram002p

and you can use n =i^2 instead n^2 as the general term lol

- Loser66

Hey, friend , I don't get why \(\sqrt n^2 \). Is it not that it is always = n and n >0 since it is the order of the term.

- Loser66

If we consider \(\lfloor{\sqrt n^2}\rfloor = \lfloor{n}\rfloor\) and it is = n itself.

- Loser66

since \(n\in \mathbb N\)

- ikram002p

yes u are correct , its just a variable :D sorry to confuse u ;)

- ikram002p

i just used it to explain but since u were asked to make general form of n
then use n without n^2

- Loser66

oh, I got you. Thank you so much.

- ikram002p

you are the most wlc <3

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