[seqfan] Re: Surprising Patterns in Tangent and Secant Numbers
maxale at gmail.com
Thu Nov 12 23:06:27 CET 2009
Let h(x) = serlaplace(g(x)).
With the help of Lagrange inversion theorem, we have
serlaplace(exp(x)*g(x)) = h(x/(1-x)) / (1-x)
and the required identity follows trivially.
On Tue, Nov 10, 2009 at 9:39 AM, Max Alekseyev <maxale at gmail.com> wrote:
> It appears that 1/cosh(x) can be replaced by an arbitrary function g(x):
> serreverse(x*serlaplace(exp(x)*g(x))) = 1/( 1 +
> 1/serreverse(x*serlaplace(g(x))) )
> On Tue, Nov 10, 2009 at 2:23 AM, Paul D Hanna <pauldhanna at juno.com> wrote:
>> I suppose that the proofs of the main observations in my prior email
>> could start with establishing the following identity (in PARI notation):
>> = x + x/serreverse(x*serlaplace(1/cosh(x)))
>> = 1 + x - x^2 + 3*x^4 - 38*x^6 + 947*x^8 - 37394*x^10 +...
>> Is this identity easy to prove?
>> Cf. A157308.
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