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Old 2022-05-12, 22:58   #1
sweety439
 
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"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

22×5×173 Posts
Default Curious of generalized repunit primes

Mersenne conjectured that for primes p <= 257, 2^p-1 is prime if and only if p = {2, 3, 5, 7, 13, 17, 19, 31, 67, 127, 257}, but this is not true, the correct one is: for primes p <= 257, 2^p-1 is prime if and only if p = {2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127}, Mersenne incorrectly included the primes p = 67 and p = 257 (and missed the primes p = 61 and p = 89 and p = 107), note that 67 and 257 are irregular primes (https://oeis.org/A000928), and all primes p <= 257 such that 2^p-1 is prime are regular primes (the smallest prime p such that 2^p-1 is prime is 607), in general, for irregular prime p, the generalized repunit (b^p-1)/(b-1) has less probability to be a prime, e.g. for the smallest irregular prime, p = 37, the smallest base b such that (b^37-1)/(b-1) is prime is 61 (much larger than the excepted value), and for the prime 67 which is mentioned above, the smallest base b such that (b^67-1)/(b-1) is prime is 46 (also large), indeed, 61 is the smallest base b such that (b^41+1)/(b+1) is prime (41 is the next prime after 37), and 46 is the smallest base b such that (b^71+1)/(b+1) is prime (71 is the next prime after 67)

No matter positive or negative base, the generalized repunit (b^p-1)/(b-1) or (b^p+1)/(b+1) has much less probability to be a prime if 2*p+1 is prime (i.e. p is Sophie Germain prime), since there is about 1/2 probability that 2*p+1 divides the generalized repunit (b^p-1)/(b-1) or (b^p+1)/(b+1) if the base b is random chosen, similarly, for 4*p+1 (1/4 probability), 6*p+1 (1/6 probability), ..., e.g. for the prime 233, with a combination the fact that it is an irregular prime and 2*233+1 = 467 is prime, the smallest base for (b^p-1)/(b-1) and (b^p+1)/(b+1) are both large (602 and 489, respectively), also the prime 491, the smallest base such that (b^491-1)/(b-1) is prime is very large (514), conversely, 61 and 127 are regular primes, and neither 2*61+1 nor 2*127+1 is prime, there are many small bases b such that (b^p-1)/(b-1) is prime for b = 61 and 127

For base b = 3, the smallest irregular prime p such that (b^p-1)/(b-1) is 103, and for base b = 6, this prime is 271, and for base b = 10, this prime is (probably) 86453, and for b = 12, this prime is 353

A curious sequence: For given prime p --> finding the smallest base b such that (b^p-1)/(b-1) is prime --> finding the smallest prime p such that (b^p-1)/(b-1) is prime --> finding the smallest base b such that (b^p-1)/(b-1) is prime --> ..., which prime sets the record of number of terms in this sequence to terminate (until a cycle)?

(i.e. A085398(p) --> A084740(b) --> A085398(p) --> A084740(b) --> ...)

e.g. choose the prime p = 271639 --> the smallest base is 92 --> the smallest prime is 439 --> the smallest base is 76 --> the smallest prime is 41 --> the smallest base is 14 --> the smallest prime is 3 --> the smallest base is 2 --> the smallest prime is 2 (terminate), this sequence has 8 terms

There are other cycles (other than (2,2)): e.g. (prime,base) = (313,35) and (349,39) and (599,124), it is unknown whether there are infinitely many such cycles.

Last fiddled with by sweety439 on 2022-05-12 at 22:59
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Old 2022-05-12, 23:20   #2
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"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

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Default Strictly generalized repunit primes

Like the sense ...

Generalized Cullen primes base b: primes of the form n*b^n+1 (https://oeis.org/A240234)
Generalized Woodall primes base b: primes of the form n*b^n-1 (https://oeis.org/A240235)
Strictly generalized Cullen primes base b: primes of the form n*b^n+1 with n>=b-1 (https://oeis.org/A327660)
Strictly generalized Woodall primes base b: primes of the form n*b^n-1 with n>=b-1 (https://oeis.org/A327661)

We use this sense ...

https://oeis.org/A085104 (all generalized repunit primes with length >=3, Brazilian primes)
https://oeis.org/A179625 (more strictly, with p > b/5, but why 5? Not 4 or 6? Only because we use decimal? It is not reasonable!)

We use the more strictly condition: p>=b

in our sense, the "legal" generalized repunit primes < 10^100 are: {3, 7, 13, 31, 127, 1093, 8191, 19531, 55987, 131071, 524287, 797161, 12207031, 305175781, 2147483647, 16148168401, 50544702849929377, 1111111111111111111, 2305843009213693951, 6115909044841454629, 29043636306420266077, 459715689149916492091, 7369130657357778596659, 11111111111111111111111, 109912203092239643840221, 618970019642690137449562111, 162259276829213363391578010288127, 177635683940025046467781066894531, 3754733257489862401973357979128773, 26063080998214179685167270877966651, 170141183460469231731687303715884105727, 243270318891483838103593381595151809701, 568972471024107865287021434301977158534824481, 7538867501749984216983927242653776257689563451, 6957596529882152968992225251835887181478451547013, 26656068987980386414408582952871386493955339704241, 3546245297457217493590449191748546458005595187661976371, 35842614220783025524408588074144786493150233831596714503, 423622795798733187216959754496018087627393990881167960767, 70169234660105574400577005075855017842743056666917902427141, 279199061472649689615930789290784389297167871396904357110743, 615840114784814774501200690134862345946783236130283731411280186824640601, 155306613932666028670208812450645212905178047040045530562317564121001023821, 1051153199500053598403188407217590190707671147285551702341089650185945215953, 56065687629692436349945381294682921858769274981456436532996640647681369599401, 5111329463430071646630167819950683399621676569261698373582346123709742512021745954241, 94800483779652702112291995272136379042013221035884653418370569190962917425415732643821, 143798195172461138521036839345269251740737334259640879028155379795667047030720519999127, 14693679385278593849609206715278070972733319459651094018859396328480215743184089660644531, 4298038738591350241903359894361937263902417766066948735111854394786520954312466554848504069671001, 133733063818254349335501779590081460423013416258060407531857720755181857441961908284738707408499507}

Also, the smallest exponent p of generalized repunit base b for b = 2, 3, 4, ... are:

2, 3, (not exist), 7, 7, 13, (not exist), (not exist), 19, 17, 19, 137, 19, 43, (not exist), 47, 25667, 19, 1487, 43, 79, 3181, 53, (not exist), 43, (not exist), 457, 151, 163, 31, (not exist), 197, 1493, 313, (not exist), 71, 401, 349, 541, 83, 1319, 6277, 167, 53, 67, 127, 269, (not exist), 127, 4229, 103, 1571, 389, 151, 157, 109, 2333, 479, 173, 107, 163, 3067, (not exist), 631, 19973, 367, 107, 2371, 541, 157, 109, 35401, 191, 739, 157, 15361, 101, 109, (unknown, >3*10^5), (not exist), 7607, 2713, 3917, 2111, 113, 121487, 577, 109, 97, 4421, 439, 4903, 1789, 523, 3343, 1693, 2801, 383, (not exist), 337, 673, 313, 263, 389, ...

Last fiddled with by sweety439 on 2022-05-13 at 10:41
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Old 2022-05-20, 22:20   #3
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A085398(n) is much large than all A085398(k) for all k<=n for:

* n=115, and if skip n=115, then n=119 (just larger than 115) is such n
* n=545, and if skip n=545, then n=553 (just larger than 545) is such n

both 115 and 545 are 5*prime
both 119 and 553 are 7*prime

A252503 (the same as A085398, but with a restriction that k must be prime), A252503(545) and A252503(553) are both 3109

For the smallest negative k (<-1) (rather than the smallest positive k (>1))

For 545, it is A085398(1090) = 12 (related to "duodecimal" unique period)
For 553, it is A085398(1106) = 2 (related to "binary" unique period)

For all divisors d of 1090, except d = 545, Phi(d,12) are primes
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Old 2022-06-02, 09:22   #4
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Nov 2016
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317, 353, 701 are three consecutive lengths of dozenal repunits

If p is Sophie Germain prime (i.e. 2*p+1 is prime) other than 2, 3, 5, then the dozenal repunit with length p cannot be prime, since the dozenal repunit with length p is always divisible by 2*p+1

Sophie Germain prime (except 3) must == 2 mod 3, and 317, 353, 701 are also == 2 mod 3

2*317+1, 2*353+1, 2*701+1 are all semiprimes

2*317+1 is 5*prime, 2*353+1 is 7*prime
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