You seem to have ignored the possibility of A=1:

k=1 -> 1+1=2

k=2 -> 1+2=3

k=3 -> 1+5=6

k=4 -> 1+14=15

k=7 -> 1+714=715

For k=13, you missed **
1235+495726=496961**

There are none for k=14,15,16,17,18,19,20,21. There
are almost certainly no more solutions for puzzle 284:

The probability that a random integer N is of the
form AB(A+B) is no more than 1/N^(1/6):

For a given A, the probability that N is of the form
AB(A+B) is no more than 1/N^(1/2), and there are N^(1/3) possibilities for
A given that A<=B.

Therefore the expected number of solutions with
k>=13 is less than the sum of the reciprocals of the 6th roots of the
primorials for k>=13.

...

Choose a random integer N. The probability that N is
of the form AB(A+B) is less than 1/N^(1/3).

Let s_n be the nth primordial. So s_1=2, s_2=2*3=6,
s_3=2*3*5=30 and so on. Let p_n be the nth prime.

Hand-waving claim: "The probability that a random
large integer N is of the form AB(A+B) is no more than 1/N^(1/6)"

Hand-waving proof: "For a fixed A, the probability
that N is of the form

AB(A+B) is no more than 1/N^(1/2)." Further, with
A<=B, there are fewer than N^(1/3) possible values of A. QED.

Now, s_19=7858321551080267055879090>=10^24. Further,
for n>=19, p_n>=64. Therefore for n>=19, s_n>=(10^24)*64^(n-19).
Therefore, s_n^(1/6)>=(10^4)*(2^(n-19)).

Therefore "the expected number of solutions to
AB(A+B)=s_n with n>=19 is smaller than
1/(10^4)+1/(10^4)*2+...=2/10^4=0.0002.

Also, there are no new solutions with n<100 and
A<=1000000.