20211005, 12:07  #188 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
BE9_{16} Posts 
Quote:

20211006, 23:52  #189 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
101111101001_{2} Posts 
New minimal prime (start with b+1) in base b is found for b=908: 8(0^243438)1, see post https://mersenneforum.org/showpost.p...&postcount=992
File https://docs.google.com/spreadsheets...RwmKME/pubhtml updated. 
20211016, 13:38  #190 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3049_{10} Posts 
Conjecture: If sequence (a*b^n+c)/gcd(a+c,b1) (a>=1 is integer, b>=2 is integer, c is (positive or negative) integer, c>=1, gcd(a,c) = 1, gcd(b,c) = 1) does not have covering set (full numerical covering set, full algebraic covering set, or partial algebraic/partial numerical covering set), then the sum of the reciprocals of the positive integers n such that (a*b^n+c)/gcd(a+c,b1) is prime is converge (i.e. not infinity) and transcendental number. (of course, this conjecture will imply that there are infinitely many such n)
For the examples of (a,b,c) triples (a>=1 is integer, b>=2 is integer, c is (positive or negative) integer, c>=1, gcd(a,c) = 1, gcd(b,c) = 1) such that (a*b^n+c)/gcd(a+c,b1) have covering set (full numerical covering set, full algebraic covering set, or partial algebraic/partial numerical covering set), see post https://mersenneforum.org/showpost.p...&postcount=678 
20211016, 13:45  #191 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3,049 Posts 
Another conjecture (seems to already be proven, but I am not sure that): If all but finitely many primes p divide (a*b^n+c)/gcd(a+c,b1) (a>=1 is integer, b>=2 is integer, c is (positive or negative) integer, c>=1, gcd(a,c) = 1, gcd(b,c) = 1) for some n>=1, then a=1 and c=1, i.e. (a*b^n+c)/gcd(a+c,b1) is generalized repunit number (b^n1)/(b1)
The factor tables have many examples for the special case that b=10, e.g. {2}1 in base 10 is (a,b,c) = (2,10,11), the section Prime factors that appear periodically lists the primes that divide (a*b^n+c)/gcd(a+c,b1) = (2*10^n11)/9 for some n, and we note that the primes 2, 5, 11, 31, 37, 41, 43, ... divides no numbers of the form (a*b^n+c)/gcd(a+c,b1) = (2*10^n11)/9, and this sequence of primes seems to be infinite. 
20211016, 13:54  #192 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
5751_{8} Posts 
The smallest generalized nearrepdigit primes (i.e. of the form x{y} or {x}y) base b is always minimal primes (start with b+1) in base b unless the repeating digit (i.e. y for x{y}, or x for {x}y) is 1, since the generalized repunit numbers base b may be prime unless b is 9, 25, 32, 49, 64, 81, 121, 125, 144, ... (A096059) bases without any generalized repunit primes, and for a repdigit to be prime, it must be a repunit (i.e. the repeating digit is 1) and have a prime number of digits in its base (except trivial singledigit numbers), since, for example, the repdigit 77777 is divisible by 7, in any base > 7, thus (i.e. of the form x{y} or {x}y) base b is always minimal primes (start with b+1) in base b if the repeating digit (i.e. y for x{y}, or x for {x}y) is not 1, thus, the families A{1} in base 22 and 8{1} in base 33 and 4{1} in base 40 are not unsolved families in this problem (i.e. finding all minimal primes (start with b+1) in base b) although all they are nearrepdigit families and all they have no known primes or PRPs and none of them can be ruled out as only contain composites (only count numbers > base), since their repeating digit are 1, and the prime F(1^957) in base 24 (its value is (346*24^9571)/23) is not minimal prime (start with b+1) in base b=24, since its repeating digit is 1
Last fiddled with by sweety439 on 20211017 at 12:55 
20211017, 17:00  #193 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3,049 Posts 
The algebra form ((a*b^n+c)/d) of the unsolved families are:
Code:
base unsolved family algebra form 11 5(7^n) (57*11^n7)/10 13 9(5^n) (113*13^n5)/12 13 A(3^n)A (41*13^(n+1)+27)/4 16 (3^n)AF (16^(n+2)+619)/5 16 (4^n)DD (4*16^(n+2)+2291)/15 
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