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October 5th, 2017, 11:42 AM   #1
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Complete this sequence

m=4*(2n!/(n!)^2)+1

m is a perfect square for n={1,2,3,9}

n=1
m=4*(2!/((1!)^2))+1=3^2

n=2
m=5^2

n=3
m=9^2

n=9
m=441^2

I stopped at n=9 can you find the next one?
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October 5th, 2017, 12:12 PM   #2
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If you write it another way :

(1/2)*C(18,9)=T(220) where T() is a triangular number

Is it correct?

How many triangular numbers you could write such way above?
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October 5th, 2017, 12:23 PM   #3
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Why do you think there is a next one?
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October 5th, 2017, 12:27 PM   #4
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Quote:
Originally Posted by skipjack View Post
Why do you think there is a next one?
So prove there is no next one then.
Otherwise if not proven then there will be maybe another one and even an infinite number.
Who knows?
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October 6th, 2017, 02:10 AM   #5
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There are no new ones for n up to 2000 (using a python script).
Thanks from mobel
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October 6th, 2017, 02:37 AM   #6
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Quote:
Originally Posted by Benit13 View Post
There are no new ones for n up to 2000 (using a python script).
Thank you.
But that does not prove that there will be no one.
This problem is linked to the Brocard problem.
As you see the form 4*A + 1 if this number m is square then it will solve some quadratic ax^2+bx+c where a=1, b=1, c=-C(2n,n)
Now rewrite the quadratic to see what is behind.
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October 9th, 2017, 06:02 AM   #7
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Little up for this one too.
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