# Uporabniški pogovor:Anton Mravcek~slwiki

Dejavni pogovori

For 2004 user talk, please see Uporabniški pogovor:Anton Mravcek/Pogovor2004.

For 2005 user talk, please see Uporabniški pogovor:Anton Mravcek/Pogovor2005.

## 89

89 is the smallest Fibonacci number that is not also Chen prime /(OEIS A005478)/.   I also unfortunely do not know which one is next.   I guess it is just a matter of some calculation. To show if next such number - 1597 is also Chen prime. But this takes time. And also they grow very fast. The next candidate is already 28657 ... --xJaM 05:07, 28 april 2006 (CEST)

I've checked: 1597, 28657 are not Chen numbers. Now the next candidate is 514229. --xJaM 06:33, 28 april 2006 (CEST)

514229 is such number,
433494437,
2971215073,
99194853094755497,
475420437734698220747368027166749382927701417016557193662268716376935476241 ? So, 89 and 514229 are at least two numbers with this property. --xJaM 06:43, 28 april 2006 (CEST)

Uh, I've made some stupidity here. Once again - prime Fibonacci numbers which are also Chen numbers are:

2, 3, 5, 13, 89, 233, 514229,

and Chen primes which are not prime Fibonacci numbers:

7, 11, 17, 19, 23, 29, 31, 37, 41, 47, 53, 59, 67, 71, ...

And then finally (prime) Fibonacci numbers that are not Chen primes:

1597, 28657, 433494437, 2971215073, 99194853094755497, 1066340417491710595814572169, 19134702400093278081449423917,

so 1597 is that smallest number and not 89. 1597 is prime, 1597+2 is not prime or semiprime, since it is a product of three (not necessarily distinct) and not two prime numbers: 1599 = 3 · 13 · 41. --xJaM 07:19, 28 april 2006 (CEST)

With the semiPrimeQ function defined in A109611 I was able to doublecheck your results in Mathematica in a matter of seconds. They all check out, and are in the OEIS, A11770, A117743, A117745. Anton Mravcek 00:20, 5 maj 2006 (CEST)
Yes I've also made a function ts_chen(n) in Maple (ts_ stands for teorija števil (number theory in Slovene :-) and checked them out. I've send my results to Sloane's Encyclopedia (so A117743 and A117745 are my results :-) Unfortunately I've made another error and I've sent a wrong sequence (which still can be seen in A117739). A117740 and A117743 are already corrected. I have to send also my Maple code. Is A11770 a typo? Which sequence did you mean in fact? A011770, A117700 or which one? Somehow a 15th article (475420437734698220747~) in A005478 also have to be checked if it is a Chen prime. Maple after some hours still didn't give any results. I guess another strategy should be implemented for this case. --xJaM 00:43, 5 maj 2006 (CEST)
I'm sorry, it is a typo, I meant A117740.
Mathematica tells me that the factorization (with timing) of 475420437734698220747368027166749382927701417016557193662268716376935476243 is {0.329 Second, {{3, 1}, {8689, 1}, {422453, 1}, {8175789237238547574551461093, 1}, {5280544535667472291277149119296546201, 1}}}. (Though sometimes it crashes on much simpler calculations). Also, the number's μ is -1^5. If you can verify this by multiplying, then it probably means the next Fibonacci Chen prime candidate is too large to write out explicitly in the OEIS. Anton Mravcek 21:34, 5 maj 2006 (CEST)
Is this result (with timing) also an explicit one? I've check it with multiplication and it checked okay, so it is really (and of course Ω(pF+2)=5):
pF+2 = 3 · 8689 · 422453 · 8175789237238547574551461093 · 5280544535667472291277149119296546201
Another three candidates from prime Fibonacci numbers pF, which are not listed in OEIS, are:
• 529892711006095621792039556787784670197112759029534506620905162834769955134424689676262369,
Timing[FactorInteger[\

529892711006095621792039556787784670197112759029534506620905162834769955134424\ 689676262371]]

{0.562 Second, {{3, 1}, {13, 1}, {433, 1}, {557, 1}, {2417, 1}, {21401, 1}, {44269, 1}, {374929, 1}, {217221773, 1}, {226981241, 1}, {2191174861, 1}, {126192465881, 1}, {6274653314021, 1}, {767056342442009, 1}}}
• 1387277127804783827114186103186246392258450358171783690079918032136025225954602593712568353,
This number + 2 must is clearly (5^1)x, but even dividing it by 5 before asking for factorization it's taking a while.
Here's a case of overlooking the obvious: What if the number has a prime factor smaller than 5? It doesn't take a sophisticated algorithm to see that this number's digital root is 3. Therefore, the factorization begins {{3, 1}, {5, 1} ... Even if the divisor 92485141853652255140945740212416426150563357211452246005327868809068348396973506247504557 is prime, the number is not a semiprime. Anton Mravcek 23:15, 11 maj 2006 (CEST)
• 3061719992484545030554313848083717208111285432353738497131674799321571238149015933442805665949
I guess here Mathematica uses better factorisation method then Maple, since its function ifactor() again stops for quite a while ... (I'm just saying this by heart). bigomega() is a little bit faster, but this does not help much. --xJaM 02:13, 6 maj 2006 (CEST)
According to the Mathematica Help Browser, "FactorInteger switches between removing small primes by trial division and using the Pollard a, Pollard rho and quadratic sieve algorithm."
For the timing, I can either wrap Timing[] around a command or put //Timing at the end. I have the habit of using it for almost any calculation involving numbers greater than 10^20 or so. If I'm running some other app besides a Web browser, I've noticed that it slows Mathematica down. Anton Mravcek 19:45, 6 maj 2006 (CEST)

Yes, it is really obvious that Fibonacci number (433)+2 (OEIS A001605) is divisible (firstly) by 5 and 3 and that this prime Fibonacci number is not also a Chen prime. I've forgotten to say that Maple's function ifactor uses 4 additional parameters, which indicate a method of integer factorisation. The last one easy does no further work and just expresses factors (if any) for a given number with simple division. By default (without parameters) it uses Morrison-Brillhart algorithm. All prime Fibonacci numbers for indices to 571 are not Chen primes. As Maple with parameter easy shows:

(433): p433+2 = 3 · 5 · 1307 · 433494437 · 2607553541 · _c68, where _c68 is not prime (and what is not relevant here, as you've said too) and it is 62600761515871045554937593876108427840854574001297205916753298831503,
(449): p449+2 = 3 · 4013 · 6329 · 108377 · 2151521 · _c75, where _c75 is not prime and it is 172328554634949009455574417787337750177268989791629527260033946176980328313
(509): p509+2 = 89 · 28657 · 34303 · _c96, _c96 is not prime and it is 121135392661744712953539295588450167176115330163162249034648716629286109147628362145470574079169
(569): p569+2 = 3 · 43 · 307 · 5147 · 10753 · 90481 · 825229 · 15791401 · _c89, _c89 is not prime and it is 14194426980814777224114603417290004530224892241407393349470934981595711987611002876636323
(571): p571+2 = 7 · 13 · 2789 · 59369 · _c109, _c109 is again not prime and it is 6373939547491682249021949229434880396923405072788863313833695805810685450770017881472463223581020731890207801. According to  the largest known prime Fibonacci number is F604711, but it is just a probable prime.

We know that there are an infinite number of Chen primes, but we do not know if there are infinite many Fibonacci numbers that are primes, and we also do not know a 'general' term for Chen primes. I guess these kind of questions wait for another Gauss or Ramanujan. --xJaM 18:29, 12 maj 2006 (CEST)

The famous Green-Tao paper of 2005 might shed some light on this. They've proven that there are infinitely many primes in arithmetic progressions. Perhaps they might be able to extend this result to Fibonacci and related sequences. Anton Mravcek 21:50, 12 maj 2006 (CEST)

## Vaš uporabniški račun bo preimenovan

09:16, 20. marec 2015 (CET)

## Renamed

13:14, 19. april 2015 (CEST)