... I'm still working on the "why". I suspect it has something to do with how quickly iterates of eulerphi for a prime hit a power of 2. The phenomenon doesn't...
... In line 50, use the modulus m3 = eulerphi(P-1). In line 40, use the modulus m4 = eulerphi(m3). In line 30, use the modulus m5 = eulerphi(m4). Below is a...
... Let's use the shorthand notation (like you did) of "p^^n+k" to mean "p^p^..^p+k", where there are n p's. Using Jens's pari code, it's quite quick (c.20...
... Let f(n) = 4^n + n^4. If n is even, then so is f(n). If n is odd, then f(n) = (A(n) + B(n))*(A(n) - B(n)) with A(n) = 2^n + n^2 B(n) = 2^((n+1)/2)*n Hence...
... I can shed some light on this behaviour. For any prime q, define b(q,n) by the recurrence relation b(q,n+1)=q^b(q,n), with initial condition b(q,0)=1. [So...
... There is no need to restrict this to q /prime/, it can be any integer. And the Theorem can be generalized to Theorem: For k any integer, if a prime p...
... Sebastian Martin Ruiz <s_m_ruiz@...> wrote: ... On the contrary, the PNT makes it probable (but not yet provable) that there is an *infinite* number of...
correcting a typo, which does not seem to me to affect the argument: The probability that ((2*(n!))^8)/2+1 is prime is (heuristically) asymptotic to...
... Very nice! Can you give an example where the arithmetic progression is different from the so-far-observed case of "every powering after the first...
... Err, Mike, how does the second line follow from the first? If x = y mod (p-1) then q^x = q^y mod p, in my book. Has one of us made a big boob? David...
... Right... of course! Bad example - that's not what I meant to say. Maybe, Is 8*n!^8+1 composite for all n>4 ? Probably not. Actually I don't mind at all....
... Indeed there is a nonsequitor, David :-( While I try and mend the proof (I still believe the Theorem to be true - HELP, anyone?), here are the results of a...
Here's a Lemma that should be both true and useful. For any non-zero integer q, define b(q,n) by the recurrence relation b(q,n+1)=q^b(q,n), with initial...
... For a prime p to occur twice with a gap in between then p must divide x^m^n+c and x+c but not x^n+c. It must therefore also divide the difference x^m^n-x...
... Not partucularly useful, I suggest. b(q.n) is a power of q. b(q,n+1) is a bigger power of q. So it is rather trivial that b(q,n)|b(q,n+1). Dressing that up...
... Isn't this equivalent to the not very interesting observation that x divides x^y when x and y are nonzero integers? If you were to define b(q,c,n) as...
Maybe Sum Of Digits Is Prime - SODIP or SODIPS for more than one. So we have another name for "additive" primes in The Encyclopedia of Integer Sequences. An...
... No, it's not actually. If you set x=b(q,n), then b(q,n+1)=q^x, NOT x^y for some y, so your observation does not apply. But I do agree I didn't need an...
... Cool findings, thanks David and others. After some reflection, I could tell everyone why we shouldn't be surprised. It's from a very simple and more...
... Here is a case with q=79: q=79;p=5591987;c=3385011; for(k=2,10,a=vector(k,j,q);print([lift(pmod(a,p)+c),a])) [4909930, [79, 79]] [0, [79, 79, 79]] [578606,...
... The stakes have risen, Mark. You have to explain 3 features of the data that Richard Heylen has exposed: 1) some primes stay as divisors at successive...
... Feature number one, where some primes stay forever as divisors, is a natural consequence. Specifically, if a factor appears twice in a row at two...
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