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September 23rd, 2015, 05:04 AM   #11
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Yes, that's one way to do it. Another way would be a function of the form f(x) = ax + b + c*(-1)^x + d*i^x for appropriate constants a, b, c, and d. (This is from memory; check the gf/recurrence for the right form.)
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September 23rd, 2015, 07:16 AM   #12
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Thank you all.

I have found something interesting :

U(5)=((5*6)/2)-3=12

U(5)*10=5!

How many U(n) are of that form?

U(n)*(10^k)=m!

m could be distinct from n

Ps : I think that 25+8*(n!) could be perfect square fro some value of n. I`m building the proof.
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September 23rd, 2015, 08:43 AM   #13
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Quote:
Originally Posted by mobel View Post
I have found something interesting :

U(5)=((5*6)/2)-3=12

U(5)*10=5!

How many U(n) are of that form?

U(n)*(10^k)=m!
I don't expect very many at all -- factorials are very sparse. U(5) * 10, U(317) * 10^-1, U(893) * 10^2.
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September 23rd, 2015, 11:26 AM   #14
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I think that 25+8*(n!) could be perfect square fro some value of n. I`m building the proof.
I'd be surprised, but let me know if you find something.
Thanks from mobel
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September 23rd, 2015, 11:44 AM   #15
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I'd be surprised, but let me know if you find something.
We have an infinite number of factorials.
U(n) is growing with almost random differences between k! and U(n)

7 is one away from 3!
25 is one away from 4!
and so on ...
If you look at the minimal distance between k! and U(n) you will see that this value is somewhat random.
The question is : what will stop U(n) not to be equal to k!

We will find it by analyzing how the difference between U(n+d)-U(n) behave.
It depends mainly on the value of d (and in second on n).

There is a way to give at least the size of n is such equation U(n)=k! holds.
The size of n is around int(sqrt(n!) but we can find more precise value.
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