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

[Vladimir Shevelev]

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)*
A002425((n+1)/2).

For a proof, let E_n(x) = x^n + Sum_{odd k=1..n}
 e_k(n)*x^(n-k),
then E_n(0)=e_n(n). For odd n, by my comment
 in A002425,  e_n(n)=
(-1)^((n+1)/2)*A002425((n+1)/2)/A006519(n+1). 
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)=
-N(n,0)=(-1)^((n-1)/2)*A002425((n+1)/2).
So, for k=1 the formula holds also for odd n.
Further, we do a simple induction to prove,
using (1), that 
E_n(k)=(-1)^(k-1+((n-1)/2)*E_n(1)+u(n,k)
and the formula easily follows from the above
considerations.

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,
Vladimir