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 October 5th, 2017, 11:42 AM #1 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 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?
 October 5th, 2017, 12:12 PM #2 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 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?
 October 5th, 2017, 12:23 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,731 Thanks: 1809 Why do you think there is a next one?
October 5th, 2017, 12:27 PM   #4
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Quote:
 Originally Posted by skipjack 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?

 October 6th, 2017, 02:10 AM #5 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,131 Thanks: 716 Math Focus: Physics, mathematical modelling, numerical and computational solutions There are no new ones for n up to 2000 (using a python script). Thanks from mobel
October 6th, 2017, 02:37 AM   #6
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Quote:
 Originally Posted by Benit13 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.

 October 9th, 2017, 06:02 AM #7 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 Little up for this one too.

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