20200816, 11:58  #1 
"Jeppe"
Jan 2016
Denmark
2×3^{4} Posts 
3 as Leyland prime?
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 = (n1)^1 + 1^(n1)). (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. 
20200816, 23:26  #2  
Sep 2010
Weston, Ontario
2·7·13 Posts 
Quote:
Last fiddled with by pxp on 20200816 at 23:35 Reason: pluralized 'sequence' 

20200817, 09:26  #3  
"Jeppe"
Jan 2016
Denmark
10100010_{2} Posts 
Quote:
So it is n^k + k^n where either 1 < k ≤ n or 1 = k = n1. 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 b2 = m = 0. /JeppeSN 

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