two further puzzles

[Jack Brennen]

> Can anyone figure them out?

> %S A112037 2,3,5,11,7,23,13,29,41

> %S A112038 5,7,11,23,29,47,53,59,83

Here's the PARI/GP code I threw together.  I get these same sequences,
except that I think the second sequence rightfully begins with 3,
rather than 5.

g=1;forprime(p=2,299,f=factorint(p-1)[,1];z=factorback(f);
             r=z/gcd(z,g);g*=r;if(r>1,print(r," ",p)));


Essentially, go through all of the primes p, and for each one, factor p-1
into primes.  List the primes in the order of their first appearance in
the p-1 factorizations -- that gives you A112037.  For each member r of
A112037, give the smallest prime p which is congruent to 1 (mod r) --
that gives you A112038.  (Or you could think of it as the "p" values
which produced new members of A112037.)

  Jack