20201218, 16:50  #431 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}·3^{6} Posts 
There is no possible 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 Last fiddled with by sweety439 on 20201218 at 16:54 
20201218, 19:21  #432 
Feb 2017
Nowhere
11073_{8} Posts 
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. Last fiddled with by Dr Sardonicus on 20201218 at 19:31 Reason: xingif optsy 
20201219, 05:08  #433  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}·3^{6} Posts 
Quote:
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 afile for A243147 is not right, n=64 should be "0" instead of "unknown" 

20201219, 22:43  #434 
Sep 2010
Weston, Ontario
2^{3}·5^{2} Posts 
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. 
20201220, 09:14  #435 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}·3^{6} Posts 
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
Last fiddled with by sweety439 on 20201220 at 09:17 
20201220, 12:15  #436  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·47·101 Posts 
Quote:
What do these have to do with this thread? Better start your own thread. 

20201220, 16:56  #437 
Sep 2010
Weston, Ontario
2^{3}×5^{2} Posts 
While I have you here, I've been wondering if you might prefer Sergey over Serge in my Leyland primes indexing effort. It's an easy fix.

20201220, 17:11  #438  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10010100010110_{2} Posts 
Quote:
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^{y}+y^{x}, so the changes to sieve code are evident. 

20201224, 21:47  #439 
"Norbert"
Jul 2014
Budapest
109 Posts 
Another new PRP:
27496^27577+27577^27496, 122422 digits. 
20201231, 12:17  #440 
Sep 2010
Weston, Ontario
2^{3}·5^{2} Posts 
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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