Puzzle 726. Completing the OEIS sequence A235031

Here we produce a sequence for a starting integer using recursively the following rules:

1) the next term of the sequence is equal to the sum of the product of the non-zero digits plus the sum of the digits, of the previous term in both operations.

2) The sequence stops when the last term repeats some term already produced, discarding this last term.

Examples

a) Lets start with the integer = 1. The sequence has exactly 18 distinct terms:

1, 2, 4, 8, 16, 13, 7, 14, 9, 18, 17, 15, 11, 3, 6, 12, 5, 10, [2]

b) Lets start with the integer = 26. The sequence has exactly 19 distinct terms:

26, 20, 4, 8, 16, 13, 7, 14, 9, 18, 17, 15, 11, 3, 6, 12, 5, 10, 2, [4]

As a matter of fact on Jan 2, 2014, I sent the following smallest initial terms for the OEIS seq A235031:

a(n): 1, 26, 28, 66, 289, 579, 3468, 23889, 2366688, for n= 18, 19, ... 26 terms.

Almost immediately Giovanni Resta produced five more terms: a(17) and
a(27)-a(30).

Later he also produced twelve more terms: a(1) to a(12).

Accordingly, the sequence A235031 looks currently something like this, for n=1 to 30.

19, 34, 46, 177, 458, 2699, 279999, 4557888, 23366667799, 456667788889999, 246666666666666667888999, 23777777777777777888888888899999999, X, X, X, X, 2, 1, 26, 28, 66, 289, 579, 3468, 23889, 2366688, 45579999, 356688888888, 35888888888888889, 2455566666777777999999999999999

That is to say, we don't know what are the four a(n) values for n= 13, 14, 15 & 16.

Q. Can you find the missing terms in this sequence?