Lorentz and Doppler and signal pick up

[Stephanus]

B will reach A at 115.4701 years.

I think you mean by this that A's clock will read 215.47years when B reaches A.
Because γ=2, then B will reach A according to B clock for 57.7350 years...

Very smart and meticulous of you, very careless of me! Yes that's right.
When A's clock read 215.47 year or 115.47 year from A's clock read 100 years which is the first time A receive signal from B.

B will receive a signal from A every γ(1−v/c) years as per wiki. There are several alternate equivalent formulae. I get this as a period of .2679 years, close to your figure.

Yes, thanks 0.2679 actually, just copied only 2 dec points.

A will send out a total of 216 plus a fraction signals that B will receive, the first signal A0 will be received by B when B starts the trip at B's time of 100. (Note that the set {0,1,...,215} contains 216 elements.) The first signal is numbered A0, the last A215. The spacing between signals is .2679 years. The last signal from A, A215, will be received at a time of 215*.2679 = 57.6 years according to B's clock, near the end of B's trip.

B will send out a total of 58 signals on the trip, the first signal B100, the last signal B157. A will receive the first signal, B100, at 200 years on A's clock due to propagation delays, the same time at which A will see B's rocket flare through a telescope (if A is looking).

A will receive signals from that point at the rate of 1 signal every .2679 years, so the last signal, B57, will arrive at 200+.269*57 = 215.27 on A's clock, shortly before B arrives.

Thank you, thank you. I learn much from this.