Problems & Puzzles: Puzzles Puzzle 83. The 6k+1 and 6k1 boxes As
a result of a discussion about the distribution of the twin prime numbers
sent to me by Alberto Zelaya (that will deserve later another puzzle than this) I
made a numerical empirical search that conveyed to the following other
issues: Knowing
that all
the prime numbers can be dropped in one of the following boxes: the 6k+1
box or the 6k1 box: a)
Are the prime numbers equally distributed in these two boxes? Let
P and Q to be the total (accumulated) number of prime numbers in the 6k+1
and 6k1 boxes, respectively, for k=1 to certain K value. Empirically can
be observed that it seems to be that P is never greater than Q. For
k<315589197, I got that P=Q
only in the following few cases: K=1, 2, 3, 6, 7, 13, 27, other wise
P<Q. I
stopped the search when the status went at: b)
Exist other K values such that P=Q? Solution Luis Rodríguez
wrote (26/02/2000): "The
puzzle related to Zelaya search is very interesting; Paulo Ribenboim
explains its details in the p. 275 of "The new book of prime
number records". According with this, the answers to the posted
questions are:
The
explanation about the large series of numbers needed to the inequalities
to be inverted is related to the same random characteristic distribution
of the primes which produces a kind of 'random walk', needing large
quantity of steps for the required inversion . Please see also: "An
Introduction to Probability Theory and Its Applications", Feller
p. 83." I
(CR) have not the Feller's book, but at the p. 285 of the Ribenboim
book we can read this: "Tschebycheff
noted already in 1853, that p_{3,1}(x)<
p_{3,2}(x)
and p_{4,1}(x)<
p_{4,1}(x)
for small values of x; in other words, there are more primes of the form
3k+2 than of the form 3k+1 (resp. more primes 4k+3 than primes 4k+1) up to
x (for x not too large). Are these inequalities true for every x?…it may
be shown that these inequalities are reversed infinitely often. Thus Leech
computed in 1957 that x1=26861 is the smallest prime for which p_{4,1}(x)>
p_{4,1}(x);
see also Bays and Hudson (1978) who found that
x1=608981813029 is the smallest prime for which p_{3,1}(x)>
p_{3,2}(x)…On
the average p_{3,2}(x)
 p_{3,1}(x)
is asymptotically (x^1/2)/(log x). However, Hudson (with the help
of Schinzel) showed in 1985 that limit
as xàinfinite
(p_{3,2}(x)
 p_{3,1}(x))/((x^1/2)/(log
x)) does not exist (in particular, it is not equal to
1)" So, after this precious paragraphs pointed out kindly by Rodríguez,
we know now that for sure P will be equal to Q again infinitely times,
remaining only the following question: when, for the first time, P will be
Q again after k=27 in x=6*k+/1?





