[seqfan] A formula for numerator of Euler(n,k)

Vladimir Shevelev shevelev at bgu.ac.il
Wed Sep 6 17:19:36 CEST 2017

Dear SeqFans,

Let N(n,k) denote numerator of E_n(k)=Euler(n,k)
with integer k>=1. 
Set u(n,k) = 2*Sum_{1<=i<=k-1}(-1)^(i-1)*(k-i)^n
(such that u(n,0)=u(n,1)=0). Then,
for even n, N(n,k)= u(n,k) + (-1)^(k-1)^delta(n,0); 
for n==1(mod 4), N(n,k) = u(n,k)*A006519(n+1) +
(-1)^(k-1)*A002425((n+1)/2); for n==3(mod 4), 
N(n,k) = u(n,k)*A006519 (n+1) - (-1)^(k-1)*

For a proof, let E_n(x) = x^n + Sum_{odd k=1..n}
then E_n(0)=e_n(n). For odd n, by my comment
 in A002425,  e_n(n)=
So N(n,0)=(-1)^((n+1)/2)*A002425((n+1)/2). 
If n is even, then E_n(0)=delta(n,0),  where 
delta(n,0)=1, n=0 and delta(n,0)=0, n>0. 
Further we use the known formula
E_n(x+1)=2*x^n - E_n(x).      (1)
Since  for x=0, x^n=delta(n,0), then for even n
E_n(1)=2*delta(n,0) - E_n(0)=delta(n,0) and 
since here u(n,1)=0, for k=1 the formula is
true for even n; if n is odd, then N(n,1)=
So, for k=1 the formula holds also for odd n.
Further, we do a simple induction to prove,
using (1), that 
and the formula easily follows from the above

Our formula is relevant for A143074, A157805,
A157808, A157812, A157827, A157835, A157856,
A157864, A157875, A157886, A157907.

Remark. I noted, that in Neil's comment
in A002425 there is a misprint: instead of 
(-1)^n*a(n) is the numerator of
Euler(2n+1,1), it should be (-1)^n*a(n+1)
is the numerator of Euler(2n+1,1). 
For example, for n=3,
we have N(7,1)=-17 (not -3).

Best regards,

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