Re: [math-fun] Propp's prime factor factory
My intuition is that F(x) = x^2 + 1 is supercritical. Jim On Sunday, June 17, 2018, Warren D Smith <warren.wds@gmail.com> wrote:
If in your process instead of doubling & add 1, i.e. the map 2x+1, do the map "F(x)" for integer polynomials F I would guess for fast enough growing F(x) the process ought to blow to create an infinite set of primes while for slow enough F it will not.
I.e. I suspect there is a "critical mass" phenomenon where at some point you are breeding new primes fast enough to create exponential population explosion, but below that point it self-limits.
So what sort of growth for F constitutes that "critical mass"? Interesting & likely delicate question.
Just as an initial guess, perhaps F(X) = 1 + X^floor(lnlnX) is supercritical, but F = any polynomial(X) is subcritical.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
if x is an odd prime and k is a positive integer then x^k+1 is not prime. On Sun, Jun 17, 2018 at 5:59 PM, James Propp <jamespropp@gmail.com> wrote:
My intuition is that F(x) = x^2 + 1 is supercritical.
Jim
On Sunday, June 17, 2018, Warren D Smith <warren.wds@gmail.com> wrote:
If in your process instead of doubling & add 1, i.e. the map 2x+1, do the map "F(x)" for integer polynomials F I would guess for fast enough growing F(x) the process ought to blow to create an infinite set of primes while for slow enough F it will not.
I.e. I suspect there is a "critical mass" phenomenon where at some point you are breeding new primes fast enough to create exponential population explosion, but below that point it self-limits.
So what sort of growth for F constitutes that "critical mass"? Interesting & likely delicate question.
Just as an initial guess, perhaps F(X) = 1 + X^floor(lnlnX) is supercritical, but F = any polynomial(X) is subcritical.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
Going back a ways in this discussion, the map 3x+2 produces a different set from 2x+1, but seems to have the same sort of behavior. Starting with 7, you get 5 7 13 17 23 41 43 53 71 79 83 127 131 151 239 251 383 691 719 1151 and starting with 59 you get 5 7 11 13 17 23 41 43 53 59 71 79 83 127 131 151 179 239 251 383 691 719 1151 and so on. Here is a terrible probabilistic argument for why this sort of process will always be finite. Primes become increasingly sparse as n increases, so the probability that 2x+1 is prime goes down as x increases, so you have to be lucky to get a very large number that is also prime. Eventually, the run of chance means you won’t be able to go any higher. Yes, I know numbers are prime or not regardless of probabilities related to the prime number theorem, but it does seem back of the envelope reasonable. Steve On Jun 17, 2018, at 7:08 PM, W. Edwin Clark <wclark@mail.usf.edu<mailto:wclark@mail.usf.edu>> wrote: if x is an odd prime and k is a positive integer then x^k+1 is not prime. On Sun, Jun 17, 2018 at 5:59 PM, James Propp <jamespropp@gmail.com<mailto:jamespropp@gmail.com>> wrote: My intuition is that F(x) = x^2 + 1 is supercritical. Jim On Sunday, June 17, 2018, Warren D Smith <warren.wds@gmail.com<mailto:warren.wds@gmail.com>> wrote: If in your process instead of doubling & add 1, i.e. the map 2x+1, do the map "F(x)" for integer polynomials F I would guess for fast enough growing F(x) the process ought to blow to create an infinite set of primes while for slow enough F it will not. I.e. I suspect there is a "critical mass" phenomenon where at some point you are breeding new primes fast enough to create exponential population explosion, but below that point it self-limits. So what sort of growth for F constitutes that "critical mass"? Interesting & likely delicate question. Just as an initial guess, perhaps F(X) = 1 + X^floor(lnlnX) is supercritical, but F = any polynomial(X) is subcritical. -- Warren D. Smith https://urldefense.proofpoint.com/v2/url?u=http-3A__RangeVoting.org&d=DwICAg... <-- add your endorsement (by clicking "endorse" as 1st step) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com<mailto:math-fun@mailman.xmission.com> https://urldefense.proofpoint.com/v2/url?u=https-3A__mailman.xmission.com_cg...
So instead of x^2+1, which will be composite when x is prime, how about x^2+2? Steve On Jun 17, 2018, at 7:08 PM, W. Edwin Clark <wclark@mail.usf.edu<mailto:wclark@mail.usf.edu>> wrote: if x is an odd prime and k is a positive integer then x^k+1 is not prime. On Sun, Jun 17, 2018 at 5:59 PM, James Propp <jamespropp@gmail.com<mailto:jamespropp@gmail.com>> wrote: My intuition is that F(x) = x^2 + 1 is supercritical. Jim On Sunday, June 17, 2018, Warren D Smith <warren.wds@gmail.com<mailto:warren.wds@gmail.com>> wrote: If in your process instead of doubling & add 1, i.e. the map 2x+1, do the map "F(x)" for integer polynomials F I would guess for fast enough growing F(x) the process ought to blow to create an infinite set of primes while for slow enough F it will not. I.e. I suspect there is a "critical mass" phenomenon where at some point you are breeding new primes fast enough to create exponential population explosion, but below that point it self-limits. So what sort of growth for F constitutes that "critical mass"? Interesting & likely delicate question. Just as an initial guess, perhaps F(X) = 1 + X^floor(lnlnX) is supercritical, but F = any polynomial(X) is subcritical. -- Warren D. Smith https://urldefense.proofpoint.com/v2/url?u=http-3A__RangeVoting.org&d=DwICAg... <-- add your endorsement (by clicking "endorse" as 1st step) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com<mailto:math-fun@mailman.xmission.com> https://urldefense.proofpoint.com/v2/url?u=https-3A__mailman.xmission.com_cg...
If we start with a nonempty collection of odd primes, x^2+2 seems to give all the primes that are 1 or 3 mod 8, whereas x^2-2 seems to give all the primes that are 1 or 7 mod 8. These are all the odd primes we could hope for. Jim Propp On Tue, Jun 19, 2018 at 10:43 AM, Lucas, Stephen K - lucassk < lucassk@jmu.edu> wrote:
So instead of x^2+1, which will be composite when x is prime, how about x^2+2?
Steve
On Jun 17, 2018, at 7:08 PM, W. Edwin Clark <wclark@mail.usf.edu> wrote:
if x is an odd prime and k is a positive integer then x^k+1 is not prime.
On Sun, Jun 17, 2018 at 5:59 PM, James Propp <jamespropp@gmail.com> wrote:
My intuition is that F(x) = x^2 + 1 is supercritical.
Jim
On Sunday, June 17, 2018, Warren D Smith <warren.wds@gmail.com> wrote:
If in your process instead of doubling & add 1, i.e. the map 2x+1, do the map "F(x)" for integer polynomials F I would guess for fast enough growing F(x) the process ought to blow to create an infinite set of primes while for slow enough F it will not.
I.e. I suspect there is a "critical mass" phenomenon where at some point you are breeding new primes fast enough to create exponential population explosion, but below that point it self-limits.
So what sort of growth for F constitutes that "critical mass"? Interesting & likely delicate question.
Just as an initial guess, perhaps F(X) = 1 + X^floor(lnlnX) is supercritical, but F = any polynomial(X) is subcritical.
-- Warren D. Smith https://urldefense.proofpoint.com/v2/url?u=http-3A__ RangeVoting.org&d=DwICAg&c=eLbWYnpnzycBCgmb7vCI4uqNEB9RSjOdn_5nBEmmeq0&r= vge6KOo90zMf7Wx14WFtiQ&m=84TqKdy-ioBA6N-oUuiXkit5YkR2Fsf44Kl9gXw0H_o& s=-VM_bTQLnt6isGej8y5_Wonl9N5GmAg0PUQRrqPzIbU&e= <-- add your endorsement (by clicking "endorse" as 1st step)
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Something interesting happens if instead of starting with a finite set of primes, we start with the set of all primes (call it P). In that situation, the factor reactor that uses f(p) = 2p-1, f(p) = 2p+1, or f(p) = p^2-1 gives back P again, while the factor reactor that uses f(p) = p^2+1 gives just the prime 2 and the primes that are 1 mod 4. That is, the three rules that tend to make big finite sets of primes get smaller leaves the set P alone, while the rule that tends to make nonempty finite sets get bigger makes P get smaller. These assertions are not hard to prove. To show that every prime odd prime q divides some number of the form 2p-1, find k such that 2k is 1 mod q, and note that by Dirichlet's theorem, there are infinitely many primes p (and hence at least one!) such that p is k mod q; then 2p-1 is 0 mod q. The same argument shows that q divides some number of the form 2p+1 and that q divides some number of the form p-1 (which in turn divides p^2-1). The assertion about f(p) = p^2+1 follows from that fact that for every prime q that's 1 mod 4, there exists k such that k^2 is -1 mod q, combined with one more application of Dirichlet's theorem. Jim Propp
With the polynomial F(x) = x^2 + 1, the seed 1 appears to yield all the primes it for which -1 is a quadratic residue, i.e., the prime 2 and all primes congruent to 1 mod 4. (Note that these are the only primes we could hope to get.) Can anyone prove this? I've obtained all the 1-mod-4 primes up to 1000. Jim On Sunday, June 17, 2018, James Propp <jamespropp@gmail.com> wrote:
My intuition is that F(x) = x^2 + 1 is supercritical.
Jim
On Sunday, June 17, 2018, Warren D Smith <warren.wds@gmail.com> wrote:
If in your process instead of doubling & add 1, i.e. the map 2x+1, do the map "F(x)" for integer polynomials F I would guess for fast enough growing F(x) the process ought to blow to create an infinite set of primes while for slow enough F it will not.
I.e. I suspect there is a "critical mass" phenomenon where at some point you are breeding new primes fast enough to create exponential population explosion, but below that point it self-limits.
So what sort of growth for F constitutes that "critical mass"? Interesting & likely delicate question.
Just as an initial guess, perhaps F(X) = 1 + X^floor(lnlnX) is supercritical, but F = any polynomial(X) is subcritical.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
I think "prime factor reactor" (along with the subcritical/supercritical distinction) is more apt than "prime factor factory". Jim Propp On Mon, Jun 18, 2018 at 9:57 AM, James Propp <jamespropp@gmail.com> wrote:
With the polynomial F(x) = x^2 + 1, the seed 1 appears to yield all the primes it for which -1 is a quadratic residue, i.e., the prime 2 and all primes congruent to 1 mod 4. (Note that these are the only primes we could hope to get.)
Can anyone prove this?
I've obtained all the 1-mod-4 primes up to 1000.
Jim
On Sunday, June 17, 2018, James Propp <jamespropp@gmail.com> wrote:
My intuition is that F(x) = x^2 + 1 is supercritical.
Jim
On Sunday, June 17, 2018, Warren D Smith <warren.wds@gmail.com> wrote:
If in your process instead of doubling & add 1, i.e. the map 2x+1, do the map "F(x)" for integer polynomials F I would guess for fast enough growing F(x) the process ought to blow to create an infinite set of primes while for slow enough F it will not.
I.e. I suspect there is a "critical mass" phenomenon where at some point you are breeding new primes fast enough to create exponential population explosion, but below that point it self-limits.
So what sort of growth for F constitutes that "critical mass"? Interesting & likely delicate question.
Just as an initial guess, perhaps F(X) = 1 + X^floor(lnlnX) is supercritical, but F = any polynomial(X) is subcritical.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
Here's a related result: You can't have everything except a finite number of missing primes. In fact, you can't have 100% - epsilon, with epsilon->0. Why not? Suppose 17 is missing. Since sqrt(-1) = +-4 (mod 17), all the primes 17k+-4 must also be missing. This mean 2/16 = 12.5% (at least) are missing. This contradicts the 100%-epsilon hypothesis. [QED] Moreover, these missing primes in turn create another set of missing primes. This suggests that the only possibilities for the final set are All Primes, or some much smaller percentage, maybe 0%. This would still permit an infinite number of primes, but they'd need to be sparse. Notice that F(x) = x^2 - 1 behaves very differently. I did a couple of experiments with simply iterating F(x) to make a sequence, rather than building out the whole tree. F(x) = largest-prime-factor(x^2+1) seems to head for infinity. The ride is bumpy, but I'm up to 70 digit numbers. I did a small experiment with F(x) = LPF(floor(x^1.5)); all x<101 peter out into small numbers, although I did get some intermediate 10 & 15-digit values. 157-281-157 is a small loop. Rich ---- Quoting James Propp <jamespropp@gmail.com>:
With the polynomial F(x) = x^2 + 1, the seed 1 appears to yield all the primes it for which -1 is a quadratic residue, i.e., the prime 2 and all primes congruent to 1 mod 4. (Note that these are the only primes we could hope to get.)
Can anyone prove this?
I've obtained all the 1-mod-4 primes up to 1000.
Jim
On Sunday, June 17, 2018, James Propp <jamespropp@gmail.com> wrote:
My intuition is that F(x) = x^2 + 1 is supercritical.
Jim
On Sunday, June 17, 2018, Warren D Smith <warren.wds@gmail.com> wrote:
If in your process instead of doubling & add 1, i.e. the map 2x+1, do the map "F(x)" for integer polynomials F I would guess for fast enough growing F(x) the process ought to blow to create an infinite set of primes while for slow enough F it will not.
I.e. I suspect there is a "critical mass" phenomenon where at some point you are breeding new primes fast enough to create exponential population explosion, but below that point it self-limits.
So what sort of growth for F constitutes that "critical mass"? Interesting & likely delicate question.
Just as an initial guess, perhaps F(X) = 1 + X^floor(lnlnX) is supercritical, but F = any polynomial(X) is subcritical.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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What happens with x^2-1? Jim Propp On Monday, June 18, 2018, <rcs@xmission.com> wrote:
Here's a related result: You can't have everything except a finite number of missing primes. In fact, you can't have 100% - epsilon, with epsilon->0.
Why not? Suppose 17 is missing. Since sqrt(-1) = +-4 (mod 17), all the primes 17k+-4 must also be missing. This mean 2/16 = 12.5% (at least) are missing. This contradicts the 100%-epsilon hypothesis. [QED]
Moreover, these missing primes in turn create another set of missing primes. This suggests that the only possibilities for the final set are All Primes, or some much smaller percentage, maybe 0%. This would still permit an infinite number of primes, but they'd need to be sparse.
Notice that F(x) = x^2 - 1 behaves very differently.
I did a couple of experiments with simply iterating F(x) to make a sequence, rather than building out the whole tree. F(x) = largest-prime-factor(x^2+1) seems to head for infinity. The ride is bumpy, but I'm up to 70 digit numbers. I did a small experiment with F(x) = LPF(floor(x^1.5)); all x<101 peter out into small numbers, although I did get some intermediate 10 & 15-digit values. 157-281-157 is a small loop.
Rich
---- Quoting James Propp <jamespropp@gmail.com>:
With the polynomial F(x) = x^2 + 1, the seed 1 appears to yield all the primes it for which -1 is a quadratic residue, i.e., the prime 2 and all primes congruent to 1 mod 4. (Note that these are the only primes we could hope to get.)
Can anyone prove this?
I've obtained all the 1-mod-4 primes up to 1000.
Jim
On Sunday, June 17, 2018, James Propp <jamespropp@gmail.com> wrote:
My intuition is that F(x) = x^2 + 1 is supercritical.
Jim
On Sunday, June 17, 2018, Warren D Smith <warren.wds@gmail.com> wrote:
If in your process instead of doubling & add 1, i.e. the map 2x+1,
do the map "F(x)" for integer polynomials F I would guess for fast enough growing F(x) the process ought to blow to create an infinite set of primes while for slow enough F it will not.
I.e. I suspect there is a "critical mass" phenomenon where at some point you are breeding new primes fast enough to create exponential population explosion, but below that point it self-limits.
So what sort of growth for F constitutes that "critical mass"? Interesting & likely delicate question.
Just as an initial guess, perhaps F(X) = 1 + X^floor(lnlnX) is supercritical, but F = any polynomial(X) is subcritical.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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I did a few experiments with x^2-1 in my head, and in every case, the reaction was subcritical. Is it possible that x^2-1 is subcritical for every initial seed, whereas x^2+1 Is supercritical for every non-empty initial seed? Jim On Monday, June 18, 2018, James Propp <jamespropp@gmail.com> wrote:
What happens with x^2-1?
Jim Propp
On Monday, June 18, 2018, <rcs@xmission.com> wrote:
Here's a related result: You can't have everything except a finite number of missing primes. In fact, you can't have 100% - epsilon, with epsilon->0.
Why not? Suppose 17 is missing. Since sqrt(-1) = +-4 (mod 17), all the primes 17k+-4 must also be missing. This mean 2/16 = 12.5% (at least) are missing. This contradicts the 100%-epsilon hypothesis. [QED]
Moreover, these missing primes in turn create another set of missing primes. This suggests that the only possibilities for the final set are All Primes, or some much smaller percentage, maybe 0%. This would still permit an infinite number of primes, but they'd need to be sparse.
Notice that F(x) = x^2 - 1 behaves very differently.
I did a couple of experiments with simply iterating F(x) to make a sequence, rather than building out the whole tree. F(x) = largest-prime-factor(x^2+1) seems to head for infinity. The ride is bumpy, but I'm up to 70 digit numbers. I did a small experiment with F(x) = LPF(floor(x^1.5)); all x<101 peter out into small numbers, although I did get some intermediate 10 & 15-digit values. 157-281-157 is a small loop.
Rich
---- Quoting James Propp <jamespropp@gmail.com>:
With the polynomial F(x) = x^2 + 1, the seed 1 appears to yield all the primes it for which -1 is a quadratic residue, i.e., the prime 2 and all primes congruent to 1 mod 4. (Note that these are the only primes we could hope to get.)
Can anyone prove this?
I've obtained all the 1-mod-4 primes up to 1000.
Jim
On Sunday, June 17, 2018, James Propp <jamespropp@gmail.com> wrote:
My intuition is that F(x) = x^2 + 1 is supercritical.
Jim
On Sunday, June 17, 2018, Warren D Smith <warren.wds@gmail.com> wrote:
If in your process instead of doubling & add 1, i.e. the map 2x+1,
do the map "F(x)" for integer polynomials F I would guess for fast enough growing F(x) the process ought to blow to create an infinite set of primes while for slow enough F it will not.
I.e. I suspect there is a "critical mass" phenomenon where at some point you are breeding new primes fast enough to create exponential population explosion, but below that point it self-limits.
So what sort of growth for F constitutes that "critical mass"? Interesting & likely delicate question.
Just as an initial guess, perhaps F(X) = 1 + X^floor(lnlnX) is supercritical, but F = any polynomial(X) is subcritical.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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Well certainly x^2-1 can never go critical because, because it can never even increase! If p is prime, then p^2-1 = (p+1)(p-1), both of which must be composite themselves (once p>3), so you never create a prime larger than the one you started with. --Michael On Tue, Jun 19, 2018 at 8:54 AM James Propp <jamespropp@gmail.com> wrote:
I did a few experiments with x^2-1 in my head, and in every case, the reaction was subcritical.
Is it possible that x^2-1 is subcritical for every initial seed, whereas x^2+1 Is supercritical for every non-empty initial seed?
Jim
On Monday, June 18, 2018, James Propp <jamespropp@gmail.com> wrote:
What happens with x^2-1?
Jim Propp
On Monday, June 18, 2018, <rcs@xmission.com> wrote:
Here's a related result: You can't have everything except a finite number of missing primes. In fact, you can't have 100% - epsilon, with epsilon->0.
Why not? Suppose 17 is missing. Since sqrt(-1) = +-4 (mod 17), all the primes 17k+-4 must also be missing. This mean 2/16 = 12.5% (at least) are missing. This contradicts the 100%-epsilon hypothesis. [QED]
Moreover, these missing primes in turn create another set of missing primes. This suggests that the only possibilities for the final set are All Primes, or some much smaller percentage, maybe 0%. This would still permit an infinite number of primes, but they'd need to be sparse.
Notice that F(x) = x^2 - 1 behaves very differently.
I did a couple of experiments with simply iterating F(x) to make a sequence, rather than building out the whole tree. F(x) = largest-prime-factor(x^2+1) seems to head for infinity. The ride is bumpy, but I'm up to 70 digit numbers. I did a small experiment with F(x) = LPF(floor(x^1.5)); all x<101 peter out into small numbers, although I did get some intermediate 10 & 15-digit values. 157-281-157 is a small loop.
Rich
---- Quoting James Propp <jamespropp@gmail.com>:
With the polynomial F(x) = x^2 + 1, the seed 1 appears to yield all the primes it for which -1 is a quadratic residue, i.e., the prime 2 and all primes congruent to 1 mod 4. (Note that these are the only primes we could hope to get.)
Can anyone prove this?
I've obtained all the 1-mod-4 primes up to 1000.
Jim
On Sunday, June 17, 2018, James Propp <jamespropp@gmail.com> wrote:
My intuition is that F(x) = x^2 + 1 is supercritical.
Jim
On Sunday, June 17, 2018, Warren D Smith <warren.wds@gmail.com>
wrote:
If in your process instead of doubling & add 1, i.e. the map 2x+1,
do the map "F(x)" for integer polynomials F I would guess for fast enough growing F(x) the process ought to blow to create an infinite set of primes while for slow enough F it will not.
I.e. I suspect there is a "critical mass" phenomenon where at some point you are breeding new primes fast enough to create exponential population explosion, but below that point it self-limits.
So what sort of growth for F constitutes that "critical mass"? Interesting & likely delicate question.
Just as an initial guess, perhaps F(X) = 1 + X^floor(lnlnX) is supercritical, but F = any polynomial(X) is subcritical.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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-- Forewarned is worth an octopus in the bush.
Ah! Good point. Jim On Tuesday, June 19, 2018, Michael Kleber <michael.kleber@gmail.com> wrote:
Well certainly x^2-1 can never go critical because, because it can never even increase! If p is prime, then p^2-1 = (p+1)(p-1), both of which must be composite themselves (once p>3), so you never create a prime larger than the one you started with.
--Michael
On Tue, Jun 19, 2018 at 8:54 AM James Propp <jamespropp@gmail.com> wrote:
I did a few experiments with x^2-1 in my head, and in every case, the reaction was subcritical.
Is it possible that x^2-1 is subcritical for every initial seed, whereas x^2+1 Is supercritical for every non-empty initial seed?
Jim
On Monday, June 18, 2018, James Propp <jamespropp@gmail.com> wrote:
What happens with x^2-1?
Jim Propp
On Monday, June 18, 2018, <rcs@xmission.com> wrote:
Here's a related result: You can't have everything except a finite number of missing primes. In fact, you can't have 100% - epsilon, with epsilon->0.
Why not? Suppose 17 is missing. Since sqrt(-1) = +-4 (mod 17), all the primes 17k+-4 must also be missing. This mean 2/16 = 12.5% (at least) are missing. This contradicts the 100%-epsilon hypothesis. [QED]
Moreover, these missing primes in turn create another set of missing primes. This suggests that the only possibilities for the final set are All Primes, or some much smaller percentage, maybe 0%. This would still permit an infinite number of primes, but they'd need to be sparse.
Notice that F(x) = x^2 - 1 behaves very differently.
I did a couple of experiments with simply iterating F(x) to make a sequence, rather than building out the whole tree. F(x) = largest-prime-factor(x^2+1) seems to head for infinity. The ride is bumpy, but I'm up to 70 digit numbers. I did a small experiment with F(x) = LPF(floor(x^1.5)); all x<101 peter out into small numbers, although I did get some intermediate 10 & 15-digit values. 157-281-157 is a small loop.
Rich
---- Quoting James Propp <jamespropp@gmail.com>:
With the polynomial F(x) = x^2 + 1, the seed 1 appears to yield all the primes it for which -1 is a quadratic residue, i.e., the prime 2 and all primes congruent to 1 mod 4. (Note that these are the only primes we could hope to get.)
Can anyone prove this?
I've obtained all the 1-mod-4 primes up to 1000.
Jim
On Sunday, June 17, 2018, James Propp <jamespropp@gmail.com> wrote:
My intuition is that F(x) = x^2 + 1 is supercritical.
Jim
On Sunday, June 17, 2018, Warren D Smith <warren.wds@gmail.com>
wrote:
If in your process instead of doubling & add 1, i.e. the map 2x+1, > do the map "F(x)" for integer polynomials F > I would guess for fast enough growing F(x) > the process ought to blow > to create an infinite set of primes while for slow enough F it will > not. > > I.e. I suspect there is a "critical mass" phenomenon where at some > point > you are breeding new primes fast enough to create exponential > population > explosion, but below that point it self-limits. > > So what sort of growth for F constitutes that "critical mass"? > Interesting & likely delicate question. > > Just as an initial guess, perhaps > F(X) = 1 + X^floor(lnlnX) > is supercritical, but F = any polynomial(X) is subcritical. > > -- > Warren D. Smith > http://RangeVoting.org <-- add your endorsement (by clicking > "endorse" as 1st step) > > _______________________________________________
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