Q1. The first 10 non-boring (9...90...01) ones seem
to be:

811451682377384625400019885321

666809750614092756440997454321145572124297
9487975174553917765709752598913886733902249807
616702241273241464228741455123758372745732277149
482405967216190677658368860828837029782991706641
451758720286129809463302229118977816556275026247
45568777448898926771505932475441338201823020604851
758751408135177632584329442563587578403802457203988171

379845109286996981286201548908130030187137984510928699698129

333333333266666666666666666666333333333333333333326666666667

Q2. The two smallest titanic prime [a+1]:[a] numbers
are (only the first 10 initial and 10 ending digits shown, CR; the whole
primes can be sent on
request)

a) 4715395621...2730905317 (1000 digits)

b) 4895358452...2412532529 (1000 digits)

No other numbers of exactly 1000 digits exist, if my
code worked properly.

My method id this one:

We have: p^2-n^2 = a(10^l+1) where n^2 = 1 or 10^l
depending on the puzzle. Therefore _all_ factors of 10^l+1 must be either
factors of p-n or p+n.

Fix l. In the n^2=10^l case, l must be even. Take a
list of all prime power factors of 10^l+1. Then for each partition of that
list between p-n and p+n (so there are 2^f numbers to look at where f is
the number of prime divisors of 10^l+1) solve the Chinese Remainder
Theorem.

Throw away any answer less than 10^l/sqrt(10), and
then check for primality. My method could be improved massively if I
didn't solve the CRT from scratch _every time_ (all 2^f combinations).
There's no reason why solutions with many hundreds of digits shouldn't be
found almost instantly with sensibly-written code.

Earlier this week I received from Jean-Claude Rosa
the following nice pattern that yield infinitely many sibling numbers of
the second kind :

326666333267 ^ 2 = 106710893290_106710893289

332666666333332667 ^ 2 = 110667110889332890_110667110889332889

333266666666333333326667 ^ 2 =
111066671110888933328890_111066671110888933328889

With the next general form I turned to PFGW to start
looking for 3-PRP's :

(3[k])(2)(6[2.k+2])(3[2.k+1])(2)(6[k])(7) ^ 2 =

(1[k])(0)(6[k])(7)(1[k])(0)(8[k])(9)(3[k])(2)(8[k]+1)(0)_(1[k])(0)(6[k])

(7)(1[k])(0)(8[k])(9)(3[k])(2)(8[k])(9)

p^2 = (a+1).a

--------------

p = (3[k])(2)(6[2.k+2])(3[2.k+1])(2)(6[k])(7)

(a+1) =
(1[k])(0)(6[k])(7)(1[k])(0)(8[k])(9)(3[k])(2)(8[k]+1)(0)

a =
(1[k])(0)(6[k])(7)(1[k])(0)(8[k])(9)(3[k])(2)(8[k])(9)

with k = 9, 13, 40, 363 and 6305 and lengths
respectively (6.k+6) 60, 84, 246, 2184 and 37836

I have thus found the first five probable primes.
Also, there it is, with k = 6305, my first gigantic sibling-II number of
the form p^2 = (a+1)_(a) with prime length 37836 ! The fourth one is an
earliest titanic solution for your puzzle 258 where parts 'p', 'a+1' and
'a' all have a length of 2184 digits. For Question Q2 this is my only
personal record. Of course, I have to wait and see if other puzzlers find
even smaller titanic ones [**Carmody** did it, CR]...