Alternating series of primes

Question

Consider the infinite alternating series: $2-3+5-7+11-13+17...$ taken over all primes. Partial sums at odd terms gives: $$2-3+5=2^2\\ 2-3+5-7+11=2^3\\ 2-3+5-7+11-13+\dotsb+23=2^4\\ \vdots$$ Is there a proof that there are infinite partial sums that give as a result a number of the form $2^{k}$?

Answer 1

Effectively you are accumulating alternate prime gaps. $2+(5-3) + (11-7)+(17-13)+ \cdots $. The downstep values are irrelevant because they are odd. The alternating even values are monotonically increasing.

Prime gaps are fairly small compared to the primes themselves but are very difficult to put strict limits on analytically. I would not be surprised to continue to find occasional hits on powers of two indefinitely, but they evidently become rarer.

Tabulating the results for primes out to $500$ million for the increasing power of two, the prime $p_k$ where the series reaches that value and the actual series value $S_k$ at that point:

\begin{array}{|c|c|} \text{power of $2$} & p_k & S_k & \text{hit?} \\ \hline 2 & 2 & 2 & \checkmark \\ 4 & 5 & 4 & \checkmark \\ 8 & 11 & 8 & \checkmark \\ 16 & 23 & 16 & \checkmark \\ 32 & 59 & 32 & \checkmark \\ 64 & 127 & 70 & \times \\ 128 & 211 & 128 & \checkmark \\ 256 & 449 & 258 & \times \\ 512 & 977 & 512 & \checkmark \\ 1024 & 2087 & 1026 & \times \\ 2048 & 4091 & 2052 & \times \\ 4096 & 8329 & 4104 & \times \\ 8192 & 16649 & 8194 & \times \\ 16384 & 33107 & 16386 & \times \\ 32768 & 64997 & 32788 & \times \\ 65536 & 131009 & 65556 & \times \\ 131072 & 264949 & 131084 & \times \\ 262144 & 525359 & 262148 & \times \\ 524288 & 1051747 & 524306 & \times \\ 1048576 & 2107319 & 1048594 & \times \\ 2097152 & 4204223 & 2097198 & \times \\ 4194304 & 8408747 & 4194312 & \times \\ 8388608 & 16780681 & 8388614 & \times \\ 16777216 & 33563741 & 16777218 & \times \\ 33554432 & 67113811 & 33554438 & \times \\ 67108864 & 134255887 & 67108866 & \times \\ 134217728 & 268466503 & 134217778 & \times \\ \end{array}