My favorite integer sequence

[jaydfox]

As for J(x) <=> J ' (x) = J(x/2)
I think I know a way to prove lim x-> +oo A(x)/J(x) = 1.083...

But what really intrests me is a closed form for 1.083...
Havent I seen this number before ??

Well, I think a good place to start is to treat f(x) (or are we calling it J(x) now?) as a limit:

Define a step size h, and set x=hk
\(
J_h(x) = J_h(hk) = A_h(k) \\
A_h(2k+1) = A_h(2k) + h A_h(k)
\)

Then, the standard A(x) becomes A_1(x), and J(x) becomes:
\(
J(x) = \lim_{h \to 0} J_h(x) = \lim_{h \to 0} A_h(\lfloor x/h \rfloor)
\)

This form presents a couple of nice benfits. For starters, we can derive imaginary iterates without having to find a taylor series and plug in imaginary arguments. We simply use a complex/imaginary step size. The sequence itself is still linear. (Not linear as in degree 1, but linear in the sense that we follow a line in the complex plane, like a line integral, but with a sequence that simply has an integer index.)

The second benefit is that we can draw a direct analogy with Euler's constant e = 2.718... Let's redefine exponentiation using a similar form:

\(
\exp_h(x) = \exp_h(hk) = E_h(k) \\
E_h(k+1) = E_h(k) + h E_h(k)
\)

Then e is given by the limit:
\(
e = \lim_{h \to 0} \exp_h(1) = \lim_{h \to 0} E_h(\lfloor 1/h \rfloor)
\)

So what's the analogy to 1.083...?

1.083... is J_0(oo) / J_1(oo)

\(
a_1 = \left(\lim_{h \to 0} \lim_{x \to \infty} \frac{J_h(x)}{J_1(x)}\right) \\
\Rightarrow a_1 = \lim_{k \to \infty} \frac{J(k)}{A_1(k)} \\
\Rightarrow a_1 \approx 1.08306...
\)

Therefore, a_1 is analogous to e_1, where e_h = exp_0(1) / exp_h(1), with exp_0(1) = e. Numerically, e_1 = e/2 = 1.35914...