I think that is a clever way of writing recursion. You made a typographical mistake though.
If we write: \( \mathcal{L}_s(x, x \,\,\bigtriangleup_s\,\,y) = y \)
then we have recursion as:
\( x \,\,\bigtriangleup_{s-1}\,\,y = x\,\,\bigtriangleup_s\,\,\mathcal{L}_s(x,y) + 1 \)
However. I am only concerned with when \( \mathcal{L}_s(x, y) \in \mathbb{N} \). Since going to reals requires an universal extension of hyper operators to reals.
But! I had another aha!
\( \mathcal{L}_s(x,y) = \prod_{n=0}^{\infty} \mathcal{L}_n(x,y)^{\zeta_n(s)} = (y-x)^{\zeta_0(s)} \cdot (\frac{y}{x})^{\zeta_1(s)} \cdot (\log_x(y))^{\zeta_2(s)} \cdot (\text{slog}_x(y))^{\zeta_3(s)}\cdot... \)
\( \zeta_n(s) = \frac{\sin(\pi (s-n))}{\pi (s-n)} \psi_n(s) \)
for some undetermined \( \psi \)
This would actually allow us to extend to some reals. Thanks for this formation of recursion. I think it's a more efficient formula.
If we write: \( \mathcal{L}_s(x, x \,\,\bigtriangleup_s\,\,y) = y \)
then we have recursion as:
\( x \,\,\bigtriangleup_{s-1}\,\,y = x\,\,\bigtriangleup_s\,\,\mathcal{L}_s(x,y) + 1 \)
However. I am only concerned with when \( \mathcal{L}_s(x, y) \in \mathbb{N} \). Since going to reals requires an universal extension of hyper operators to reals.
But! I had another aha!
\( \mathcal{L}_s(x,y) = \prod_{n=0}^{\infty} \mathcal{L}_n(x,y)^{\zeta_n(s)} = (y-x)^{\zeta_0(s)} \cdot (\frac{y}{x})^{\zeta_1(s)} \cdot (\log_x(y))^{\zeta_2(s)} \cdot (\text{slog}_x(y))^{\zeta_3(s)}\cdot... \)
\( \zeta_n(s) = \frac{\sin(\pi (s-n))}{\pi (s-n)} \psi_n(s) \)
for some undetermined \( \psi \)
This would actually allow us to extend to some reals. Thanks for this formation of recursion. I think it's a more efficient formula.
