October 5th, 2017, 11:42 AM  #1 
Senior Member Joined: Dec 2013 Posts: 1,112 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 
Senior Member Joined: Dec 2013 Posts: 1,112 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: 18,036 Thanks: 1394 
Why do you think there is a next one?

October 5th, 2017, 12:27 PM  #4 
Senior Member Joined: Dec 2013 Posts: 1,112 Thanks: 41  
October 6th, 2017, 02:10 AM  #5 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,068 Thanks: 692 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
There are no new ones for n up to 2000 (using a python script).

October 6th, 2017, 02:37 AM  #6 
Senior Member Joined: Dec 2013 Posts: 1,112 Thanks: 41  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 
Senior Member Joined: Dec 2013 Posts: 1,112 Thanks: 41 
Little up for this one too.


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