[P] **No questions.** by pahio Nov 10No questions.

[P] **sketch proof - in commonly understood terminology** by akdevaraj Nov 10I would like to conclude my message with one point:
The fact that any value of n skiped is such that
f(n) is prime can be proved: f(n)=n^2+n+1 is such that
whenever f(n) is composite one of its factors
is smaller than n. This implies that it that the relevant
n has to satisfy a prior failure function.
Any question, Jussi?

[P] **sketch proof - in commonly understood terminology** by akdevaraj Nov 9Note a) p_0 less than p_1 less than p_2 less than p_i
b) This means the intervals become succeedingly bigger
c)As the fraction of primes in each interval becomes
suceedingly smaller it may appear that the number of primes
i.e. the number of ns not overed by any failure function
becomes smaller- actually it may turn out that the number
of primes become bigger since the intervals become bigger
d) there is no need to test any f(n) where n is
skiped by the failure functions for primality - they are
automatically prime e) my conjecture is that the value
of beta is 0.03 - even if beta is 0.01 it makes no
difference since the perpetuality of the iteration is
established f) only a competent programmer can program
the failure functions and the iteration g)to go from
iteration i to iteration i+1 we need only one value of
n not covered by any falure function h) the iteration
can come to an end only if all the values of n
in an interval are covered by one or more of the failure
functions ( this is highly improbable )
(to be continued)

[P] **sketch proof - in commonly understood terminology** by akdevaraj Nov 9Since the fraction of values of n in any interval which are
not covered by any failure function decreases asymptotically
to beta the iteration is perpetual. This is because the
fraction of primes in any interval never reaches zero - hence
there are infinitely many primes of shape n^2+n+1 (proved-
subject to confirmation by a programmer.)
(to be continued).

[P] **sketch proof - in commonly understood terminology** by akdevaraj Nov 8Before proceeding let me emphasise that the values of
n skiped by the failure functions i.e. n_0+kf(n_0) are
such that f(n) are prime and they need not be tested for
primality.
We now set up an iteration as follows: Let p_0 be
the largest known prime having shape n^2+n+1. Let n_0
be the corresponding value of n i.e. n_0^2+n_0+1 =
p_0. Consider the interval n_0, n_0+p_0. I am considering
this interval because f(n_0+kf(n_0)) is congruent to 0
(mod f(n_0)).
Note the fraction of this interval not covered by the
failure functions( arithmetic progressions) 1+3k, 2+7k, 3+13k
4+7k............ . Let n_1 be the largest of these values
of n not covered by any failure function. f(n_1) is prime
which need not be tested for primality. Let this prime
be p_1. 2nd iteration: consider the interval n_1, n_1 +p_1.
proceed as in the 1st iteration i.e. mark the fraction of
values of n in this interval not covered by any failure function.
Let n_2 be the largest value of n in this interval not
covered by any failure function. f(n_2) is prime; call this
p_2. Consider the interval n_2, n_2 + p_2. Mark the
fraction of values of n in this interval not covered by
any failure function. The fraction of values of n not
covered by any failure function in any interval decreases
asymptotically from iteration to iteration to a rational number- call this beta
(to be continued ).

[P] **sketch proof - in commonly understood terminology** by akdevaraj Nov 8This means we have a case of indirect primality testing:
f(n) is composite if n = n_0 +kf(n_0). Otherwise
f(n) is prime. Note we are not testing f(n)
directly for primality; we are only testing
whether is n is generated by n_0+kf(n_0); if so
f(n) is composite-if not f(n) is prime.
(to be continued)

[P] **sketch proof - in commonly understood terminology** by akdevaraj Nov 6Let me be very clear as to what counter example you
should try to find; I will do this by taking
a specific interval - n = 5 to 36.This interval
is spaned by the failure functions( arithmetic
progressions ) 1+3k, 2+7k, 3 +13k, 4 +7k and
5+31k, 7+19k and 9+13k. However the values of n
skiped by the above are 5,6, 8, 12, 14, 15, 17, 20, 21,
24,27 and 33. Now all these skiped values of n are such
that f(n) is prime. So you have to find a value of
n skiped by the relevant failure functions such that
f(n) is composite. Such a counter example may be very
difficult to find.

[P] **sketch proof - in commonly understood terminology** by akdevaraj Nov 6I forgot to mention that k belongs to Z.
Now I will take a break for two days to enable you to
find a counter-example.
(to be continued)

[P] **sketch proof - in commonly understood terminology** by akdevaraj Nov 6Let me summarise up to the point I had already contributed:
let f(n) = n^2 + n + 1 ( n belongs to N). Let n_0 be a
specific value of n. Then n = n_0 + kf(n_0) generates values
of n such that f(n) is composite.
Note: please ignore trivial cases like k = 0.
This is easily proved. All other values of n are such
that f(n) is prime.
(to be continued).

[P] **Sketch proof -the proposition** by akdevaraj Nov 2Thank you, Jussi.
Proposition: There are infinitely many prime numbers with shape
n^2+n+1.

[P] **sketch proof** by pahio Nov 1Yes. First the proposition!

[P] **sketch proof - in commonly understood terminology** by akdevaraj Nov 1Dear Jussi, would you like me to continue?

[P] **Kind attention Jussi and others who are interested ** by akdevaraj Oct 19Any doubt so far pl?

[P] **Please show the proposition** by pahio Oct 19Please show the proposition you want to prove. Then give its proof in a concise and clear form.
Regards,
Jussi