A proof that A341214 has no terms beyond A341214(5):

Suppose a term p = a(n) exists for some n > 5. Then tau(p)=2, 
tau(p-1)=4, tau(p-2)=6, tau(p-3)=8, tau(p-4)=10, and tau(p-5)=12, 
where tau(k) is the number of divisors function A000005.

Since tau(p)=2 (and p > 2),

     p is an odd prime.

Since tau(p-1)=4, p-1 is a number of the form q^3 or q * r, 
where q and r are distinct primes. But p is odd, so p-1 is even, 
and it's not 2^3 = 8, so

     p-1 = 2 * q (where q is an odd prime).

Neither p nor p-1 is divisible by 3 (p can't be 3, and p-1 can't 
be 2*3), so 3 divides p-2. Since tau(p-2)=6, p-2 is of the form 
r^5 or r^2 * s (with r, s distinct primes). Since p-2 isn't r^5 
(it would have to be 3^5 = 243, which wouldn't work; e.g., it 
would make p = 245, which is not prime), p-2 must be either of 
the form 3^2 * r = 9 * r or 3 * r^2.

Suppose p-2 = 3 * r^2; then (p-2)/3 = r^2. For every prime r > 5, 
r^2 == (1 or 49) (mod 120), and since r is a prime > 5, we would 
have p-2 = 3 * r^2 == (3 or 147) (mod 360). But that would mean 
p-1 = (p-2) + 1 == (4 or 148) (mod 360), so 4 would divide p-1: 
impossible. So

     p-2 = 9 * r (where r is a prime > 3).

Since p-1 = 2 * q (with q an odd prime), 4 divides p-3. But 
tau(p-3)=8, which is not a multiple of 3, so p-3 cannot be 4 
times an odd number, so 8 divides p-3. Thus, since tau(p-3)=8, 
p-3 must be either 2^7 or a number of the form 2^3 * s (where s 
is an odd prime). But 2^7=128 won't work (e.g., if p-3=128 then 
tau(p-1) = tau(130) = tau(2*5*13) = 8, not 4), so

     p-3 = 8 * s (where s is a prime > 3).

Since 5 cannot divide p (p=5 wouldn't work), p-1 (p-1 = 2 * q = 
2*5 = 10 wouldn't work), p-2 (p-2 = 9 * r = 9*5 = 45 wouldn't 
work), or p-3 (p-3 = 8 * s = 8*5 = 40 wouldn't work), 5 must 
divide p-4, and since tau(p-4)=10, p-4 must be of the form t^9 
or t^4*u (with t, u distinct primes). But p-4 cannot be 5^9 = 
1953125 (that would mean p-3 = 1953126, which is not divisible 
by 8), so p-4 is either of the form 5^4*t = 625*t or of the form 
5*t^4 (where t is a prime > 5).

Suppose p-4 = 5*t^4; then (p-4)/5 = t^4. For every prime t > 5,
t^4 == 1 (mod 240), and since t is a prime > 5, we would have
p-4 == 5 * t^4 == 5 (mod 1200). But that would mean p-3 = (p-4) 
+ 1 == 6 (mod 1200), so 4 would not divide p-3: impossible. So

     p-4 = 625*t (where t is a prime > 5).

Since tau(p-5)=12, p-5 must be of the form u^11, u^5 * v, 
u^3 * v^2, or u^2 * v * w (with u, v, w distinct primes). But 
since p-3 is a multiple of 8, p-5 is a multiple of 2, but not of 
2^2. Similarly, since p-2 is a multiple of 9, p-5 is a multiple 
of 3, but not of 3^2. As a result, p-5 must be of the form 
2 * 3 * u^2, i.e.,

     p-5 = 6 * u^2 (where u is a prime > 5).

Since u is not 7 (p-5 = 6*7^2 = 294 wouldn't work), u is a 
prime > 7. For any prime u > 7, u^2 == (1, 2, or 4) (mod 7); 
as a result, p-5 = 6 * u^2 == (3, 5, or 6) (mod 7). But all 
three cases are impossible:

- if p-5 == 3 (mod 7), then 7 | p-1,
  so p-1 = 2 * q = 2*7 = 14 (which won't work)

- if p-5 == 5 (mod 7), then 7 | p-3, 
  so p-3 = 8 * s = 8*7 = 56 (which won't work)

- if p-5 == 6 (mod 7), then 7 | p-4, 
  so p-4 = 625 * t = 625*7 = 4375 (which won't work).

So there exists no prime p (nor any composite number p, since a 
composite p would not satisfy the requirement that tau(p)=2) 
such that p, p-1, p-2, p-3, p-4, and p-5 have 2, 4, 6, 8, 10, 
and 12 divisors, respectively.  Thus, A341214 has no terms 
beyond A341214(5).