I am doing the past Putnam problems and one is
Prove that is an integer .
I make a generalization , which is :
is an integer iff , where is the minimun prime factor of .
If it is true , then the problem from Putnam is just a special case . However , I couldn't trust my proof so i beg for help here .
Thanks
p.s. I make use of floor function to prove it .
EDIT: I find some errors in my statement , iff should be changed to if and the equality can hold . i.e.
Thats false,
take n=6 k=5. We have that n+k = 11, so x = 11. Clearly 5 <= 11/2, but the result is 216/5 as per n = 6 k =5, 1/(n+k) *binom( 2*n, n ) - Wolfram|Alpha.
I think thats a counter example ....
Hello,
actually there is a very simple way to generalize it :
But obviously, so :
And . So we are left with :
Thus :
Now we know that has got to divide the top part of the fraction iff . But we also know that each factor of (not prime factor) will divide its double (which is present in the top part of the fraction since ).
This leads to the conclusion that is an integer for any .
Am I right ?
I think you may have something wrong here.
Correct me if I am wrong but you are using the contrapositive to try to find a counter example.
So let me restate the original proposition:
the OP claims
where
is the logical statement
is the logical statement
is the logical statement
thus the contrapositive is
thus we search for and for a counter example. I think....