special relativity - Do relativistic events need to match if accounted for time dilation and length contraction?

To explain the question let me give you a short example. In the scenario there are two references frames A and B.

A consists of a x'=1 Ls (lightsecond) long pole in the positive x direction. At t=0 a flash is generated at its origin. 1s later the flash reaches the end of the pole.

B sees A moving with v=0.866c in the positive x direction. Due to length contraction, A's pole appears to only be 0.5 Ls long.

In B, 3.731s after A generated the flash the flash reaches the end of the pole, because: $$x-ct=0 \quad with \quad x=v \cdot t+x'\sqrt{1-v^2/c^2}$$ $$(v \cdot t+x'\sqrt{1-v^2/c^2})-ct=0$$ $$(0.866c \cdot 3.731s+0.5)-3.731s \cdot c=0$$

So the flash reaches the pole's end after 1s in A. But from B's point of view it takes 3.371s.

Wouldn't this require a time dilation factor of 3.371? But the actual factor is $$\frac{1}{\sqrt{1-v^2/c^2}}=2$$

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Based on a suggestion in a comment let me write out the problem more detailed:

For A: $$ct'-x'=0$$ $$t'=1s$$ $$c \cdot 1s - 1 (1c \cdot 1s) = 0$$

For B (for the formula of x see above): $$ct-x=0$$ $$t = \frac{t'}{\sqrt{1-v^2/c^2}}=1s / 0.5 = 2s$$ $$c \cdot 2s - x \neq 0$$

Answer

I think the problem is this:

In A light would move 1s to the end of the 1Ls long pole, e.g. be reflected by a mirror and take 1s back to A.

In B the light catches up with the contracted (0.5 Ls long) pole with a relative speed of 1-0.866c=0.134c, so takes 3.731s. Then it is reflected and approaches A with a relative speed of 1.866c, so that only takes 0.268s back to A.

Sum that up and we get 4s in B, 2s in A, exactly as the Lorentz factor says.

There is also a mistake in my calculations: $$t \neq \frac{t'}{\sqrt{1-v^2/c^2}}$$ instead $$t=\frac{t'+\frac{vx'}{c^2}}{\sqrt{1-v^2/c^2}}=(1s+0.866c \cdot 1(1c \cdot 1s)/c^2)/0.5=3.732s$$