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2022-01-13, 03:12   #276
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

244338 Posts

Quote:
 Originally Posted by sweety439 I either post on this thread or update my article (or both).
Maybe just over on your google doc. Then people can fetch it at their leisure. When you post here, you generate notifications for people that there is something new. While it is new, it is not worth noticing.

2022-01-13, 06:31   #277
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

22×5×132 Posts

This is the smallest GFN primes and the smallest GRU primes in bases b<=64

Note 1: we do not include the case where the base of the GFNs is perfect odd power and the case where the base of the GRUs is either perfect power or of the form -4*m^4 with integer m, since such numbers have algebra factors and are composite for all n or are prime only for very small n, such families for bases 2<=b<=64 are:

Code:
base         GFN family         GRU family
4                               {1}
8            1{0}1              {1}
9                               {1}
16                              {1}, 1{5}, {C}D
25                              {1}
27           {D}E               {1}
32           1{0}1              {1}
36                              {1}
49                              {1}
64           1{0}1              {1}, 1{L}, 5{L}, 1{9}, {u}v
Such small primes are: 11 in base 4, 111 in base 8, 11 in base 16, 111 in base 27, 11 in base 36, 19 in base 64 (they are 5, 73, 17, 757, 37, 73, respectively, in decimal)

Note 2: All GFN base b and all GRU base b are strong-probable-primes (primes and strong pseudoprimes) to base b, since they are over-probable-primes (primes and overpseudoprimes) to base b (references: https://oeis.org/A141232 http://arxiv.org/abs/0806.3412 http://arxiv.org/abs/0807.2332 http://arxiv.org/abs/1412.5226 https://cs.uwaterloo.ca/journals/JIS...shevelev19.pdf), and all overpseudoprimes are strong pseudoprimes to the same base b, all strong pseudoprimes are Euler–Jacobi pseudoprimes to the same base b, all Euler–Jacobi pseudoprimes are Euler pseudoprimes to the same base b, all Euler pseudoprimes are Fermat pseudoprimes to the same base b, so don't test with this base (see https://mersenneforum.org/showthread.php?t=10476&page=2, https://oeis.org/A171381, https://oeis.org/A028491, also see https://oeis.org/A210454, https://oeis.org/A210461, https://oeis.org/A216170, https://oeis.org/A217841, https://oeis.org/A243292, https://oeis.org/A217853, https://oeis.org/A293626, https://oeis.org/A210454/a210454.pdf, https://cs.uwaterloo.ca/journals/JIS...hamahata44.pdf, all generalized repunits in base b^2 with length p (where p is prime not dividing b*(b^2-1)) are Fermat pseudoprimes to base b, thus there are infinitely many pseudoprimes to every base b), note that there are also (but very few) numbers in the simple families which are neither GFN families nor GRU families, which are pseudoprimes, e.g. for the family {5}25 in base 8 (which have the smallest prime 555555555555525, corresponding to the second-largest base 8 minimal prime (start with b+1)), a smaller number 525 is 341 in decimal, which is Fermat pseudoprime and Euler pseudoprime (although not strong pseudoprime, but there are many examples of strong pseudoprimes to base 2 and/or base 3, e.g. the smallest composite number which is strong pseudoprime to both base 2 and base 3 is 1373653 (see https://oeis.org/A072276), which has no proper subsequence which is prime > base (b) in bases b = 55, 58, 59, 65, 66, 70, 79, 82, 95, 103, 112, 113, 115, 116, 117, 121, 127, 130, 133, 134, 135, 136, 137, 138, 139, 141, 146, 147, 149, 151, 152, 155, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 177, 179, 183, 184, 185, 187, 188, 189, 191, 192, 193, 195, 196, 197, 199, 200, ..., and the second-smallest composite number which is strong pseudoprime to both base 2 and base 3 is 1530787, which has no proper subsequence which is prime > base (b) in bases b = 77, 91, 95, 98, 109, 113, 120, 123, 125, 127, 129, 131, 132, 135, 136, 139, 141, 142, 143, 144, 147, 151, 155, 159, 160, 161, 162, 169, 170, 173, 176, 177, 179, 181, 183, 184, 187, 188, 189, 190, 191, 192, 194, 197, 199, 200, ..., and if we assume a number which has passed the Miller–Rabin primality tests to both base 2 and base 3 is in fact prime, our data will be wrong for these bases b) to base 2 (and thus to base 8, since pseudoprimes to base b are always (the same type) pseudoprimes to base b^r for all r>1, and 8=2^3).
Attached Files
 smallest GFN and smallest GRU.txt (17.0 KB, 32 views)

Last fiddled with by sweety439 on 2022-05-06 at 14:57

2022-01-13, 06:36   #278
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

D3416 Posts

Quote:
 Originally Posted by Uncwilly it is not worth noticing.
but it is my newer researching result, and I think that it is important, sometimes I update my old posts, such as #208 and #215, also there are posts which are lists of references: #140 and #154

Last fiddled with by sweety439 on 2022-01-13 at 06:37

2022-01-13, 11:42   #279
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

22×5×132 Posts

Quote:
 Originally Posted by sweety439 done to base b=31 the original text file for base 31 is too large (1546 KB) to upload, thus zipped it
all bases b<=36 are done.

the original text file for base 35 is too large (1743 KB) to upload, thus zipped it
Attached Files
 kernel32.txt (539.7 KB, 25 views) kernel33.txt (871.2 KB, 25 views) kernel34.txt (550.9 KB, 35 views) kernel35.zip (694.7 KB, 30 views) kernel36.txt (161.0 KB, 22 views)

Last fiddled with by sweety439 on 2022-01-13 at 11:43

 2022-01-15, 07:11 #280 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 22·5·132 Posts Conjecture: There is no base b such that the largest minimal prime (start with b+1) and the second-largest minimal prime (start with b+1) have the same number of digits in base b, note that in the original minimal prime (i.e. prime > base is not required), the three largest minimal primes in decimal (base 10) all have the same number of digits (60000049, 66000049, 66600049, all have 8 digits), and in base 2 the largest (and the only) two minimal primes are 10 and 11, both have 2 digits, also, in base 5, the largest two minimal primes are 14444 and 44441, both have 5 digits. For the problem in this project (i.e. the minimal primes (start with b+1)), the largest and the second-largest minimal primes (start with b+1) have the numbers of digits: (combine with the third-largest and the fourth-largest minimal primes (start with b+1), see the table below): Code: base 1st largest 2nd largest 3rd largest 4th largest 2 2 N/A N/A N/A 3 3 2 2 N/A 4 3 2 2 2 5 96 6 5 5 6 5 4 4 2 7 17 10 8 7 8 221 15 13 11 9 1161 689 331 38 (conjectured) 10 31 12 8 8 12 42 30 9 8 and (the number of digits of 1st largest) / (the number of digits of 2nd largest) getting large very quickly if (b-1)*eulerphi(b) gets large, thus I do not think such base can exist. Another conjecture: For any number n>=2, there exists a minimal primes (start with b+1) with exactly n digits in base b, for every enough large b Clearly, all 2-digit primes (except "10" (i.e. = b) when b itself is prime) are minimal primes (start with b+1) base b, I conjectured that all bases b != 2, 6 have a 3-digit minimal prime (start with b+1), also all bases b>4 have a 4-digit minimal primes (start with b+1), all bases b>4 have a 5-digit minimal primes (start with b+1), etc. (note that in the original minimal prime (i.e. prime > base is not required), all single-digit primes are minimal primes, and I conjectured that all bases b != 8 have a 2-digit minimal prime, all bases b != 2, 4, 6, 7 have a 3-digit minimal prime, all bases b != 2, 3, 4, 5, 7 have a 4-digit minimal prime, all bases b != 2, 3, 4, 9 have a 5-digit minimal prime, etc. (the bases with no n-digit minimal prime for given n is more complex, thus the problem in this project (i.e. the minimal primes (start with b+1)) is really better)) (for more data, see post 145) we can research: * the possible length of the minimal primes (start with b+1) * the possible (first,last) combo of the minimal primes (start with b+1) * for these minimal primes (start with b+1), the digit which appears the most times in this minimal prime (start with b+1) * the length such that there are the most minimal primes (start with b+1) * the (first,last) combo such that there are the most minimal primes (start with b+1) Last fiddled with by sweety439 on 2022-01-15 at 15:54
 2022-01-17, 06:17 #281 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 22×5×132 Posts find the set of the minimal primes (start with b+1) base b for various bases b (2<=b<=36) is the target of the project in this thread. minimal prime (start with b+1) base b is always minimal prime (start with b'+1) base b' = b^n, if it is > b', for any integer n>1 original minimal prime (i.e. prime > b is not required) base b is always minimal prime (start with b+1) base b, if it is > b
2022-01-21, 15:44   #282
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

1101001101002 Posts

zipped file for the minimal primes (start with b+1) in bases 17<=b<=36
Attached Files
 minimal primes.zip (3.16 MB, 24 views)

2022-01-21, 15:52   #283
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

22·5·132 Posts

Quote:
 Originally Posted by sweety439 e.g. proving the set of the minimal primes (start with b+1) in base b = 10 is {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 227, 251, 257, 277, 281, 349, 409, 449, 499, 521, 557, 577, 587, 727, 757, 787, 821, 827, 857, 877, 881, 887, 991, 2087, 2221, 5051, 5081, 5501, 5581, 5801, 5851, 6469, 6949, 8501, 9001, 9049, 9221, 9551, 9649, 9851, 9949, 20021, 20201, 50207, 60649, 80051, 666649, 946669, 5200007, 22000001, 60000049, 66000049, 66600049, 80555551, 555555555551, 5000000000000000000000000000027}, is equivalent to: * Prove that all of 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 227, 251, 257, 277, 281, 349, 409, 449, 499, 521, 557, 577, 587, 727, 757, 787, 821, 827, 857, 877, 881, 887, 991, 2087, 2221, 5051, 5081, 5501, 5581, 5801, 5851, 6469, 6949, 8501, 9001, 9049, 9221, 9551, 9649, 9851, 9949, 20021, 20201, 50207, 60649, 80051, 666649, 946669, 5200007, 22000001, 60000049, 66000049, 66600049, 80555551, 555555555551, 5000000000000000000000000000027 are primes > 10. * Prove that all proper subsequence of all elements in {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 227, 251, 257, 277, 281, 349, 409, 449, 499, 521, 557, 577, 587, 727, 757, 787, 821, 827, 857, 877, 881, 887, 991, 2087, 2221, 5051, 5081, 5501, 5581, 5801, 5851, 6469, 6949, 8501, 9001, 9049, 9221, 9551, 9649, 9851, 9949, 20021, 20201, 50207, 60649, 80051, 666649, 946669, 5200007, 22000001, 60000049, 66000049, 66600049, 80555551, 555555555551, 5000000000000000000000000000027} which are > 10 are composite. * Prove that all primes > 10 contain at least one element in {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 227, 251, 257, 277, 281, 349, 409, 449, 499, 521, 557, 577, 587, 727, 757, 787, 821, 827, 857, 877, 881, 887, 991, 2087, 2221, 5051, 5081, 5501, 5581, 5801, 5851, 6469, 6949, 8501, 9001, 9049, 9221, 9551, 9649, 9851, 9949, 20021, 20201, 50207, 60649, 80051, 666649, 946669, 5200007, 22000001, 60000049, 66000049, 66600049, 80555551, 555555555551, 5000000000000000000000000000027} as subsequence (equivalently, prove that all numbers > 10 not containing any element in {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 227, 251, 257, 277, 281, 349, 409, 449, 499, 521, 557, 577, 587, 727, 757, 787, 821, 827, 857, 877, 881, 887, 991, 2087, 2221, 5051, 5081, 5501, 5581, 5801, 5851, 6469, 6949, 8501, 9001, 9049, 9221, 9551, 9649, 9851, 9949, 20021, 20201, 50207, 60649, 80051, 666649, 946669, 5200007, 22000001, 60000049, 66000049, 66600049, 80555551, 555555555551, 5000000000000000000000000000027} as subsequence are composite).
the last (i.e. Proving that all primes > 10 contain at least one element in {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 227, 251, 257, 277, 281, 349, 409, 449, 499, 521, 557, 577, 587, 727, 757, 787, 821, 827, 857, 877, 881, 887, 991, 2087, 2221, 5051, 5081, 5501, 5581, 5801, 5851, 6469, 6949, 8501, 9001, 9049, 9221, 9551, 9649, 9851, 9949, 20021, 20201, 50207, 60649, 80051, 666649, 946669, 5200007, 22000001, 60000049, 66000049, 66600049, 80555551, 555555555551, 5000000000000000000000000000027} as subsequence) is equivalent to:

* If n is an integer which is > 10, and the base 10 representation of n contains none of {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 227, 251, 257, 277, 281, 349, 409, 449, 499, 521, 557, 577, 587, 727, 757, 787, 821, 827, 857, 877, 881, 887, 991, 2087, 2221, 5051, 5081, 5501, 5581, 5801, 5851, 6469, 6949, 8501, 9001, 9049, 9221, 9551, 9649, 9851, 9949, 20021, 20201, 50207, 60649, 80051, 666649, 946669, 5200007, 22000001, 60000049, 66000049, 66600049, 80555551, 555555555551, 5000000000000000000000000000027} as subsequences, prove that n is composite.

Of course, this can be generalized to other bases, such as:

* If n is an integer which is > 7, and the base 7 representation of n contains none of {14, 16, 23, 25, 32, 41, 43, 52, 56, 61, 65, 113, 115, 131, 133, 155, 212, 221, 304, 313, 335, 344, 346, 364, 445, 515, 533, 535, 544, 551, 553, 1022, 1051, 1112, 1202, 1211, 1222, 2111, 3031, 3055, 3334, 3503, 3505, 3545, 4504, 4555, 5011, 5455, 5545, 5554, 6034, 6634, 11111, 11201, 30011, 30101, 31001, 31111, 33001, 33311, 35555, 40054, 100121, 150001, 300053, 351101, 531101, 1100021, 33333301, 5100000001, 33333333333333331} as subsequences, prove that n is composite.

* If n is an integer which is > 8, and the base 8 representation of n contains none of {13, 15, 21, 23, 27, 35, 37, 45, 51, 53, 57, 65, 73, 75, 107, 111, 117, 141, 147, 161, 177, 225, 255, 301, 343, 361, 401, 407, 417, 431, 433, 463, 467, 471, 631, 643, 661, 667, 701, 711, 717, 747, 767, 3331, 3411, 4043, 4443, 4611, 5205, 6007, 6101, 6441, 6477, 6707, 6777, 7461, 7641, 47777, 60171, 60411, 60741, 444641, 500025, 505525, 3344441, 4444477, 5500525, 5550525, 55555025, 444444441, 744444441, 77774444441, 7777777777771, 555555555555525, 44444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444447} as subsequences, prove that n is composite.

* If n is an integer which is > 12, and the base 12 representation of n contains none of {11, 15, 17, 1B, 25, 27, 31, 35, 37, 3B, 45, 4B, 51, 57, 5B, 61, 67, 6B, 75, 81, 85, 87, 8B, 91, 95, A7, AB, B5, B7, 221, 241, 2A1, 2B1, 2BB, 401, 421, 447, 471, 497, 565, 655, 665, 701, 70B, 721, 747, 771, 77B, 797, 7A1, 7BB, 907, 90B, 9BB, A41, B21, B2B, 2001, 200B, 202B, 222B, 229B, 292B, 299B, 4441, 4707, 4777, 6A05, 6AA5, 729B, 7441, 7B41, 929B, 9777, 992B, 9947, 997B, 9997, A0A1, A201, A605, A6A5, AA65, B001, B0B1, BB01, BB41, 600A5, 7999B, 9999B, AAAA1, B04A1, B0B9B, BAA01, BAAA1, BB09B, BBBB1, 44AAA1, A00065, BBBAA1, AAA0001, B00099B, AA000001, BBBBBB99B, B0000000000000000000000000009B, 400000000000000000000000000000000000000077} as subsequences, prove that n is composite.

etc.

Last fiddled with by sweety439 on 2022-01-24 at 12:42

 2022-01-24, 13:06 #284 sweety439     "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 22·5·132 Posts Thus, the total proof of base 10 includes these proofs: * Prove that all of 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 227, 251, 257, 277, 281, 349, 409, 449, 499, 521, 557, 577, 587, 727, 757, 787, 821, 827, 857, 877, 881, 887, 991, 2087, 2221, 5051, 5081, 5501, 5581, 5801, 5851, 6469, 6949, 8501, 9001, 9049, 9221, 9551, 9649, 9851, 9949, 20021, 20201, 50207, 60649, 80051, 666649, 946669, 5200007, 22000001, 60000049, 66000049, 66600049, 80555551, 555555555551, 5000000000000000000000000000027 are primes. (of course, they are > 10, thus this part (i.e. all these numbers are > 10) needs no proof) (we can use ECPP (such as Primo) to prove that the largest two numbers are defined primes (i.e. not merely PRPs), in this case of base 10, the largest number has only 31 digits and can be proved primality in <1 second, but in other case, such as base 13, 14, and 16, there are numbers > 10^10000 in the set, thus ECPP (or N-1, N+1, if this prime -1 or +1 can be trivially factored, such as the case of base 14, the largest prime 5*14^19698-1 in this set) is need to prove their primality) * Prove that all proper subsequence of all elements in {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 227, 251, 257, 277, 281, 349, 409, 449, 499, 521, 557, 577, 587, 727, 757, 787, 821, 827, 857, 877, 881, 887, 991, 2087, 2221, 5051, 5081, 5501, 5581, 5801, 5851, 6469, 6949, 8501, 9001, 9049, 9221, 9551, 9649, 9851, 9949, 20021, 20201, 50207, 60649, 80051, 666649, 946669, 5200007, 22000001, 60000049, 66000049, 66600049, 80555551, 555555555551, 5000000000000000000000000000027} which are > 10 are composite. (this is the easiest part of all these parts, as we can use either trial division or Fermat test to prove their compositeness (if these numbers have small prime divisors, or these numbers fails the Fermat primality tests, then they are defined composite), unless the numbers are Fermat pseudoprimes to many bases (say bases 2, 3, 5, 7, 11) with no small divisors (say < 2^32), in this case, we need to run either Miller–Rabin primality test or Lucas primality test to prove their compositeness) * Prove that all numbers > 10 not containing any element in {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 227, 251, 257, 277, 281, 349, 409, 449, 499, 521, 557, 577, 587, 727, 757, 787, 821, 827, 857, 877, 881, 887, 991, 2087, 2221, 5051, 5081, 5501, 5581, 5801, 5851, 6469, 6949, 8501, 9001, 9049, 9221, 9551, 9649, 9851, 9949, 20021, 20201, 50207, 60649, 80051, 666649, 946669, 5200007, 22000001, 60000049, 66000049, 66600049, 80555551, 555555555551, 5000000000000000000000000000027} as subsequence are composite. (for this part, we use either covering congruences or algebraic factorization (or combine of them, such as the base 12 family {B}9B and the base 14 family 8{D}) to prove that all numbers in a given family (may be non-simple family, such as many families in base 29 and 41) are composite) Last fiddled with by sweety439 on 2022-01-24 at 20:26
2022-01-28, 23:07   #286
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

22·5·132 Posts

Quote:
 Originally Posted by sweety439 These are families I am interested: (of the form (a*b^n+c)/gcd(a+c,b-1) for fixed a>=1, b>=2, c != 0, gcd(a,c) = 1, gcd(b,c) = 1 and variable n) (although some of these families do not always product minimal primes (start with b+1))
These (a*b^n+c)/gcd(a+c,b-1) (a>=1, b>=2, c != 0, gcd(a,c) = 1, gcd(b,c) = 1) families are dual families (for the definition, see http://www.kurims.kyoto-u.ac.jp/EMIS...rs/i61/i61.pdf and https://oeis.org/A076336/a076336c.html):

* b^n+1 (self dual) (b == 0 mod 2)
* (b^n+1)/2 (self dual) (b == 1 mod 2)
* (b^n-1)/(b-1) (self dual)
* 2*b^n+1 and b^n+2 (b == 3, 5 mod 6)
* 2*b^n+1 and (b/2)*b^n+1 (b == 0, 2 mod 6)
* (2*b^n+1)/3 and (b^n+2)/3 (b == 1 mod 6)
* (2*b^n+1)/3 and ((b/2)*b^n+1)/3 (b == 4 mod 6)
* 2*b^n-1 and b^n-2 (b == 1 mod 2)
* 2*b^n-1 and (b/2)*b^n-1 (b == 0 mod 2)
* 3*b^n+1 and b^n+3 (b == 2, 4 mod 6)
* 3*b^n+1 and (b/3)*b^n+1 (b == 0 mod 6)
* (3*b^n+1)/2 and (b^n+3)/2 (b == 7, 11 mod 12)
* (3*b^n+1)/2 and ((b/3)*b^n+1)/2 (b == 3 mod 12)
* (3*b^n+1)/4 and (b^n+3)/4 (b == 1, 5 mod 12)
* (3*b^n+1)/4 and ((b/3)*b^n+1)/4 (b == 9 mod 12)
* 3*b^n-1 and b^n-3 (b == 2, 4 mod 6)
* 3*b^n-1 and (b/3)*b^n-1 (b == 0 mod 6)
* (3*b^n-1)/2 and (b^n-3)/2 (b == 1, 5 mod 6)
* (3*b^n-1)/2 and ((b/3)*b^n-1)/2 (b == 3 mod 6)
* (b-1)*b^n-1 and b^n-(b-1)
* (b-1)*b^n+1 and b^n+(b-1)
* (b+1)*b^n-1 and b^n-(b+1)
* (b+1)*b^n+1 and b^n+(b+1) (b == 0, 2 mod 3)
* ((b+1)*b^n+1)/3 and (b^n+(b+1))/3 (b == 1 mod 3)
* ((b-2)*b^n+1)/(b-1) and (b^n+(b-2))/(b-1) (b == 1 mod 2)
* ((b-2)*b^n+1)/(b-1) and ((1/A006519(b-2))*b^n+A000265(b-2))/(b-1) (b == 0 mod 2)
* ((2*b-1)*b^n-1)/(b-1) and (b^n-(2*b-1))/(b-1)

Note: 1/A006519(b-2) is not integer (for b == 0 mod 2), we should start with the smallest n such that (1/A006519(b-2))*b^n is integer

Last fiddled with by VBCurtis on 2022-01-29 at 01:10 Reason: stop quoting yourself

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