Problems & Puzzles: Puzzles Puzzle 85.
A puzzle about the Generalized Mills' theorem The
Generalized theorem of Mills says that
exists couples of values A & B such that [A^(B^n)]
is prime for all n=>1. The cost of having a formula to produce only
primes is that A is (probably) an irrational constant such that usually
you need a great quantity of decimal digits in A, in order to produce only
a few primes. For
example if B=3, and we use the corresponding approximation to A with only
10 decimal digits  A=1.3063778838  then you will produce only two prime
numbers (2 for n=1 & 11 for n=2) and a composite 1360 for n=3. Maybe
this
is why, despite to the correctness of the statement, it's said that this
formulas are useless. "Hardy
and Wright[5] wisely pointed out that formulas like Mills' formula
(and many related formulas listed in [2,4]) are not very useful. One would
need to know C correctly to many places in order to compute only a few
primes. To make matters worse, there does not seem to be any way of
estimating C except via the primes q(1), q(2), q(3), ... i.e., the
reasoning becomes circular. The only manner in which Mills' formula could
be useful is if an exact value for C were to somehow become available,
which no one has conjectured might ever happen."
(S. Finch) "Though
amusing, this type of formula is useless for determining primes because we
need to know the primes determined before we find A (and the subsequence
of primes represented is so small!) (Chris
Caldwell) After
all this discredit about the importance of the calculation of this kind of
(A & B) constants for calculating primes, only very brave people would
walk around this subject. Luis Rodríguez,
from Venezuela, is that kind of men, and he asks if there are couples of A
& B values such that all the decimal digits used in both A and in B,
are less than the number of distinct primes obtained for
consecutive n values for n=1 to N; in
other words he asks for A & B values such that D(A)+D(B)< N He
has found a couple of such values, A=2.0845511122
& B=1.221, D(A)+D(B)=15, that
produces 17 prime numbers, one for each n=1 to 17 (BTW only 16 of
them are distinct prime numbers). I'll give to this result the Qualification Q = P/D = 16/15 *100= 106.66%, where P are the quantity of distinct primes produced and D are the decimal digits in A & B. Can you improve it? Hints: Solution Felice Russo has improved (13/03/2000) the qualification with the following results: A=4.4, B=1.179, primes = 5, 7, 11, 17, 29, 53, 109; Q=7/6=116.66% *** Luis Rodríguez noticed that shortening his previous A value to 6 digits (2.08455) and using the same B (1.221) 12 distinct primes are produced getting a qualification of 120%, and suggests that maybe we should change the qualification formula to stimulate mainly the production of the quantity of distinct primes. Rodríguez propose the following formula: Q=P^(3/2)/D. What do you think about this subject? Any better/simpler formula? *** Luis Rodríguez has gotten a qualification of 175% with A = 3 y C =1.22 obtaining P=7 and D=4. *** One suggestion from David Terr (1/11/02): I think a better qualification would be the total number of digits of the distinct primes divided by the total number of digits of A and B. If this is greater than 1, it means you're getting out more than you put in, which means the formula is more useful in some sense than the resulting list of primes. *** On July 01, 2016, Arkadiusz Wesolowski wrote:
***






