Leyland Primes (x^y+y^x primes)

[pxp]

Originally Posted by sweety439

The smallest k such that n^k+k^n is prime (A243147)

47 is 40182, known since Nov 2014. Why is 27 a zero?

[sweety439]

Originally Posted by pxp

47 is 40182, known since Nov 2014. Why is 27 a zero?

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^2-a*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

[Dr Sardonicus]

Originally Posted by sweety439

Conjecture: There are infinitely many primes for all other n

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.

[sweety439]

Originally Posted by Dr Sardonicus

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.

Well, x = 4 (and y>1) proven composite by partial algebra factors (like 25*12^n-1, 27*12^n-1, 4*24^n-1, 6*24^n-1 in Riesel conjectures)

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 a-file for A243147 is not right, n=64 should be "0" instead of "unknown"

[pxp]

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.

[sweety439]

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

[Batalov]

Originally Posted by sweety439

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

Looking for fish in the meat market?
What do these have to do with this thread?

Better start your own thread.

[pxp]

Originally Posted by Batalov

Looking for fish in the meat market?

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.

[Batalov]

Originally Posted by pxp

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.

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^y+y^-x, so the changes to sieve code are evident.

[NorbSchneider]

Another new PRP:
27496^27577+27577^27496, 122422 digits.

[pxp]

I have examined all Leyland numbers in the gap between L(49205,532) <134129> and L(49413,580) <136550> and found 42 new primes.