Joerg Arndt: After correction this becomes (n>=1)
0, 0, 0, 0, 0, 1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 14, 16, 19, 21, ...
which (dropping some initial terms) is A001399, A069905, and A211540.

Frank M Jackson: Can you demonstrate your correction with a(21) as an example?
From my perspective, the set of partitions counted in A218469
is a subset of the set of partitions counted in A217008 where
prime multiplicities have been excluded to give the number of
partitions of n with, at most, 3 distinct terms from {1unionPrimes}.
So A217008(21)=9 as 21 = 1+1+19 = 2+19 = 1+3+17 = 2+2+17 = 1+7+13 = 3+5+13 = 3+7+11 = 5+5+11 = 7+7+7 and
A218469(21)=5 as 21 = 2+19 = 1+3+17 = 1+7+13 = 3+5+13 = 3+7+11.
Apologies if I have not understood the point that you are making.

T. D. Noe: Joerg, this is different because only primes and 1 are in the sums.