Also, numbers k such that k concatenated with k2 gives the product of two numbers which differ by 2.
Also, numbers n such that n concatenated with n5 gives the product of two numbers which differ by 4.
Every term contains an even number of digits.  Max Alekseyev, May 14 2007
If n=(10^m1)^2+1 where m is a positive integer then n is in the sequence. Because then n has 2m digits and n concatenated with n1 is n*10^(2m)+(n1) = (10^(2m)10^m+1)^2. for example, taking m=1 we get 82, the first term of the sequence.  Farideh Firoozbakht, Aug 22 2013
As pointed out by Georg Fischer, it is very plausible that all of A054214, A116123, and A116142 contain exactly the same terms. If that is indeed true, then A054214 should be edited to mention the alternative constructions, and the other two sequences declared "dead". However, this needs careful analysis to deal with the possibilities that n, n2, and n5 may not all have the same number of digits.  N. J. A. Sloane, Oct 30 2018. Nov 05 2018: Thanks to Giovanni Resta_ (see below), this has now been done.  N. J. A. Sloane, Nov 05 2018
From Giovanni Resta, Nov 05 2018: (Start)
For n and n5 to have a different digit length, we must have n = 10^k+h with 0<=h<=4.
We want to prove that in this case the concatenation of n and n5 cannot be of the form m(m+4). The numbers m(m+4) modulo 9 can only be equal to 0, 3, 5, or 6, but it is easy to see that the concatenation of 10^k+h and 10^k+h5 can be equal to one of these values modulo 9 only if h=0.
Now, the concatenation of 10^k and 10^k5 is equal to 3 modulo 4 for every k>1, but m(m+4) modulo 4 can only be equal to 0 or 1, so A116123 is indeed equal to this sequence.
Using an identical argument (with mods 9 and 4) we can prove that the concatenation of n and n2, when n and n2 have a different number of digits, cannot be equal to m(m+2) and so A116142 is equal to this sequence. (End)
