Puzzle 1182 Another sequence by Paolo Lava

Paolo Lava sent the following puzzle:

Let us consider this sequence: a(1) = 1; a(n) counts the number of digits (n mod 10) already present in the sequence. (this one is not yet in OEIS)

Here below the first terms:

1, 0, 0, 0, 0, 0, 0, 0, 0, 8, 1, 0, 0, 0, 0, 0, 0, 1, 0, 15, 4, 0, 0, 1, 1, 0, 0, 1, 0, 20, 7, 1, 0, 1, 1, 0, 1, 1, 0, 24, 12, 3, 1, 2, 1, 0, 1, 1, 0, 26, 17, 5, 1, 2, 2, 1, 2, 1, 0, 27, 21, 10, 1, 2, 2, 1, 3, 1, 0, 29, 26, 14, 2, 3, 2, 2, 3, 1, 1, 29, 29, 19, 4, ...

E.g. a(1)=1, then a(2) tells the number of (2 mod 10) = 2 already present in the sequence, that are 0. The same up to a(9). For a(10) we have 8 because the number of (10 mod 10) = 0 already present are 8. Again, for instance, a(21) = 4 because the number of (21 mod 10) = 1 already present are 4 (a(1) = 1, a(11) = 1, a(18) = 1 and there is a "1" in a(20) =15).

I tested up to 2*10^4. The distribution of digits should follow Benford's law: the sequence is mostly populated by digit "1", with digit "0" to the other extreme.

Here below is a graph of the number of different digits for the first 2*10^4 terms.

The first consecutive different primes for exactly k values are:

k = 1 -> [7] for n = 31.
k = 2 -> [17,5] for n = 51, 52.
k = 3 -> [7, 3, 5] for n = 104, 105, 106.
k = 4 -> [2351, 2467, 2477, 2437] for n = 8677, 8678, 8679, 8680.
k = 5 -> [421, 467, 443, 401, 433] for n = 2096, 2097, 2098, 2099, 2100.

Q1. Can you extend the list for k>5?

And what about the same prime in consecutive positions? I found only primes in two consecutive positions. Here below the list:

Prime	n
2	54, 55
2	64, 65
2	75, 76
29	80, 81
2	85, 86
5	116, 117
11	165, 166
113	552, 553
109	695, 696
223	972, 973
223	1317, 1318
251	1386, 1387
1109	3602, 3603
1481	5268, 5269
2767	9837, 9838
3517	12278, 12279

Q2.Are there primes in 3 or more consecutive positions?