3 as Leyland prime?

[JeppeSN]

Sorry if this has been asked before.

OEIS has the following two sequences:

(A076980) Leyland numbers: 3, together with numbers expressible as n^k + k^n nontrivially, i.e., n,k > 1 (to avoid n = (n-1)^1 + 1^(n-1)).

(A094133) Leyland primes: 3, together with primes of form x^y + y^x, for x > y > 1.

Does anyone know why 3 is specifically included in these sequences? Historically, it was not. The last version of each which did not have it: A076980, version 12; A094133, version 26.

/JeppeSN

PS! There is a third sequence without the 3, which also excludes the case 2*x^x:

(A173054) Numbers of the form a^b+b^a, a > 1, b > a.

[pxp]

Originally Posted by JeppeSN

Does anyone know why 3 is specifically included in these sequences?

The inclusion may well be my fault! With the sole intention of generating a discussion, on 6 April 2015 I wrote to the Sequence Fanatics discussion list: "The triviality condition excludes 3 (= 2^1 + 1^2), which strikes me as a useful initial term. For example, because 3 is also excluded from the Leyland primes (A094133), the comment therein that A094133 'contains A061119 as a subsequence' isn't really correct because A061119 includes 3." Unilaterally, Neil Sloane added 3 to the sequences some fifteen minutes later.

[JeppeSN]

Originally Posted by pxp

The inclusion may well be my fault! With the sole intention of generating a discussion, on 6 April 2015 I wrote to the Sequence Fanatics discussion list: "The triviality condition excludes 3 (= 2^1 + 1^2), which strikes me as a useful initial term. For example, because 3 is also excluded from the Leyland primes (A094133), the comment therein that A094133 'contains A061119 as a subsequence' isn't really correct because A061119 includes 3." Unilaterally, Neil Sloane added 3 to the sequences some fifteen minutes later.

Thanks, that sounds like the explanation!

So it is n^k + k^n where either 1 < k ≤ n or 1 = k = n-1.

It is not really important if we include that exceptional case, or not.

There are similar situations for other definition, for example a generalized Fermat is b^(2^m) + 1. If you take m ≥ 0, then all numbers are generalized Fermat (which we do not want). If you take m > 0, then the classical Fermat prime F_0 is not a generalized Fermat. Finally, you can make the "mixed" criterion m > 0 or b-2 = m = 0.

/JeppeSN