After over three days of CPU time, I found the first group of six
consecutive integers whose prime factorizations have the same exact
pattern of digit lengths (in base 10, of course).

That may be a little confusing, so let me give the result, which
should clear things up a bit:

853173502025 == 5 * 5 * 13 * 2625149237
853173502026 == 2 * 3 * 37 * 3843123883
853173502027 == 7 * 7 * 11 * 1582882193
853173502028 == 2 * 2 * 29 * 7354943983
853173502029 == 3 * 3 * 59 * 1606729759
853173502030 == 2 * 5 * 17 * 5018667659

Each number is a one-digit-prime times a one-digit-prime times a
two-digit prime times a ten-digit-prime. The actual pattern is
not important, just the fact that six consecutive integers all
share the same pattern.

Smaller records include:

(two consecutive numbers)
2 == 2
3 == 3

(three consecutive numbers)
85 == 5 * 17
86 == 2 * 43
87 == 3 * 29

(four consecutive numbers)
34172 == 2 * 2 * 8543
34173 == 3 * 3 * 3797
34174 == 2 * 7 * 2441
34175 == 5 * 5 * 1367

(five consecutive numbers)
779930 == 2 * 5 * 23 * 3391
779931 == 3 * 3 * 19 * 4561
779932 == 2 * 2 * 73 * 2671
779933 == 7 * 7 * 11 * 1447
779934 == 2 * 3 * 43 * 3023


Before somebody attempts to "best" this result by finding a group
of seven consecutive integers with the same pattern, let me warn
you... No such group exists. The proof is left to the reader. :)

If anybody would like to find some more examples of six, perhaps
to justify inclusion in the OEIS, note that the first integer of
the six must be congruent to either 3770 or 84425 (mod 88200).
My crude method of factoring every such group of six completely
and then testing for compatibility could be greatly improved upon
by using sieve methods.