[QUOTE=sweety439;566447]The smallest k such that n^k+k^n is prime ([URL="https://oeis.org/A243147"]A243147[/URL])[/QUOTE]
47 is 40182, known since Nov 2014. Why is 27 a zero? 
[QUOTE=pxp;566573]47 is 40182, known since Nov 2014. Why is 27 a zero?[/QUOTE]
There is [I]no possible[/I] prime for 27 27^n+n^27 = (3^n)^3+(n^9)^3 and a^3+b^3 can be factored as (a+b)*(a^2a*b+b^2) Similarly, no possible prime for n = (3*k)^3, (5*k)^5, (7*k)^7, (9*k)^9, (11*k)^11, ... (r*k)^r with odd r>1 Conjecture: There are infinitely many primes for all other n 
[QUOTE=sweety439;566575]Conjecture: There are infinitely many primes for all other n[/QUOTE]Counterexample: If x = 4, the only integer y > 0 for which 4^y + y^4 is prime is y = 1.
Proof: If y > 1, and y is even, 4^y and y^4 are both divisible by 16, so the sum is divisible by 16. If y > 1 and odd, say y = 2*k + 1, then 4^y = 4*(2^k)^4, whence 4^y + y^4 = (2*4^k  2*y*2^k + y^2)* (2*4^k + 2*y*2^k + y^2) [and remember, y = 2*k + 1] Both factors are greater than 1 for odd y, except for k = 0, y = 1. Similarly with x = 4*m^4, x^y + y^x is either divisible by 16 or has an algebraic factorization. 
[QUOTE=Dr Sardonicus;566589]Counterexample: If x = 4, the only integer y > 0 for which 4^y + y^4 is prime is y = 1.
Proof: If y > 1, and y is even, 4^y and y^4 are both divisible by 16, so the sum is divisible by 16. If y > 1 and odd, say y = 2*k + 1, then 4^y = 4*(2^k)^4, whence 4^y + y^4 = (2*4^k  2*y*2^k + y^2)* (2*4^k + 2*y*2^k + y^2) [and remember, y = 2*k + 1] Both factors are greater than 1 for odd y, except for k = 0, y = 1. Similarly with x = 4*m^4, x^y + y^x is either divisible by 16 or has an algebraic factorization.[/QUOTE] Well, x = 4 (and y>1) proven composite by partial algebra factors (like 25*12^n1, 27*12^n1, 4*24^n1, 6*24^n1 in [URL="http://www.noprimeleftbehind.net/crus/Rieselconjectures.htm"]Riesel conjectures[/URL]) Also for x = 64, 324, 1024, 2500, 5184, ... 4*m^4, since even y are divisible by 2 and odd y factored as a^4+4*b^4 (x^y is of the form 4*b^4 if x = 4*m^4, y is odd, and y^x is 4th power, since x is divisible by 4) So the [URL="https://oeis.org/A243147/a243147.txt"]afile[/URL] for A243147 is not right, n=64 should be "0" instead of "unknown" 
I have examined all Leyland numbers in the gap between L(145999,10) <146000> and L(146999,10) <147000> and found 17 new primes.
I am going to be doing intervals #23 and #24, postponing #18 until late January. 
Are there any test limit of y, for x=6 (6^y*y^6+1), x=10 (10^y*y^10+1), and x=13 (13^y*y^13+1)? There are no known primes for x=13, and the only known primes for x=6 and x=10 are 6^1*1^6+1 and 10^1*1^10+1

[QUOTE=sweety439;566753]Are there any test limit of y, for x=6 (6^y*y^6+1), x=10 (10^y*y^10+1), and x=13 (13^y*y^13+1)? There are no known primes for x=13, and the only known primes for x=6 and x=10 are 6^1*1^6+1 and 10^1*1^10+1[/QUOTE]
Looking for fish in the meat market? What do these have to do with this thread? Better start your own thread. 
[QUOTE=Batalov;566759]Looking for fish in the meat market?[/QUOTE]
While I have you here, I've been wondering if you might prefer Sergey over Serge in my [URL="http://chesswanks.com/num/a094133.txt"]Leyland primes[/URL] indexing effort. It's an easy fix. 
[QUOTE=pxp;566773]While I have you here, I've been wondering if you might prefer Sergey over Serge in my [URL="http://chesswanks.com/num/a094133.txt"]Leyland primes[/URL] indexing effort. It's an easy fix.[/QUOTE]
Sergey, thanks. A word or two about that other prime class; one can modify existing sieves for that; just think of them as denominators of the x[SUP]y[/SUP]+y[SUP]x[/SUP], so the changes to sieve code are evident. 
Another new PRP:
27496^27577+27577^27496, 122422 digits. 
I have examined all Leyland numbers in the gap between L(49205,532) <134129> and L(49413,580) <136550> and found 42 new primes.

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