An Elegant Way To Define A Sequence

Answer :

I believe your sequence continues forever but grows quickly. If $n$ is large, the density of primes around $n$ is $\log n$. Since $\log n$ is so much smaller than $n$, the chance a random $n$ has $k$ arrows is about $\frac 1{(\log n)^{k+1}}$. The expected number of sequences of length $k$ above $10^{12},$ say, is then $\int_{10^{12}}^\infty \frac {dn}{(\log n)^{k+1}}$. This diverges because $(\log n)^k$ becomes less than $n$ for $n$ large enough and we know the integral of $\frac 1n$ diverges. Each subtraction is only of order $\frac n{\log n}$, which is small compared to $n$ and the log will not change much.

If we ask what length of sequence we expect to find among the $12$ digit numbers, we note that the log of these numbers is about $29$ and that $29^{8.5} \approx 3\cdot 10^{12}$. We would expect to find some sequences of $7$ arrows, maybe $8$ or $9$, and be surprised at $10$ or more. For $100$ digit numbers, the log is about $231$ and $231^{42.5} \approx 3\cdot 10^{100}$, so we would expect some sequences of length $40$ or $41$ among the $100$ digit numbers.

`my(a=0,b=0);forprime(x=1,50000,y=x;while(isprime(y-primepi(y)),y-=primepi(y);b++);if(b>a,a=b;print(x));b=0)` 

produces 14897 as the next one. Then no more below 500000. There's not too much to say except primes in the sequence will be primes at even indices after the first, simply because most primes are more than 2 away from their indices.