[seqfan] Re: A conjecture about the numerator of Euler(2n-1,n)

Vladimir Shevelev shevelev at bgu.ac.il
Fri Sep 22 22:06:45 CEST 2017


Yes, Peter Luschny simplified one of my 
formula in A291897 (see there), using
a property of A002425, so you are
right. Thank you very much for the
verification of my conjecture up to
5000.

Best regards,
Vladimir


________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of rgwv at rgwv.com [rgwv at rgwv.com]
Sent: 22 September 2017 20:56
To: 'Sequence Fanatics Discussion list'
Subject: [seqfan] Re: A conjecture about the numerator of Euler(2n-1,n)

Are the log2 of the denominators of Euler(2n-1,n) the sequence https://oeis.org/A001511: The ruler function?

-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Vladimir Shevelev
Sent: Friday, 22 September, 2017 9:38 AM
To: seqfan at list.seqfan.eu
Subject: [seqfan] A conjecture about the numerator of Euler(2n-1,n)

Dear SeqFans,

Now I have just submitted a new sequence
A291897 "Numerator of Euler(2n-1,n)".
I proved that
a(n)=2(-1)^n*A006519(2n)*(1^(2n-1) - 2^(2n-1)+
...+(-1)^n*(n-1)^(2n-1)) + A002425(n).
The sequence begins 1, 9, 125, 32977,...
My observation is about an interesting property:
a(n) is divisible by (2*n-1)^2 (conjecture).
Note that sometimes, maybe, for prime n of the form 4*k+1, a(n) is divisible by (2n-1)^3 ( for example, for  n=1,5,13,17,...).
Can anyone analyze this property?

Best regards,
Vladimir

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