**Another variance and sample size question** by jfs Oct 8Hi, I know that the variance halves when the sample size doubles and this makes sense to me when using the variance to compute a confidence interval for the mean. But I presume that relationship doesn't mean that you can adjust data with it?
We measure combustions stability in an engine with stdev(IMEP) where IMEP is ~ the average pressure in the cylinder. We normally do stdev(IMEP) over 300 cycles (n = 300). My colleague collected a lot of data with only 100 cycles and has asked me if this can be converted to 300 cycle equivalent. I don't think so but not completely sure about this so would like to ask the maths community?
Any help much appreciated.>

[P] **sketch proof** by akdevaraj Oct 8Attention: Pahio and others. Many feel that my terminology is
not comprehensible. I therefore propose to develop the sketch
proof, step by step, using only commmonly understood terminology.
After each step I will give a break for a day. Is this ok?

[P] **failure functions - another example** by akdevaraj Oct 7Let our definition of a failure be a non-Carmichael number.
Let the parent function be the polynomial f(x) = x^2 + x + 9.
When x = 23 we get the Carmichael number 561. However x =
8 + 81k ( k belongs to Z ) is a failure function since
when we substitute a value of x generated by this failure function we
get only failures ( non-Carmichael numbers ).

[P] **Random thoughts on proofs** by akdevaraj Oct 61) Perhaps some conjectures cannot be proved without collaboration
with a programmer - unless the mathematician himself happens to
be a gifted programmer 2) What is the value of alternate
proofs? For example suppose someone comes up with an alternate
proof of Fermat's last theorem - who is going to pay
attention even if the latter is simpler?

[P] **conjecture - a fewpoints (contd)** by akdevaraj Oct 53) Bonus: when a value of x, say x', is generated by
a failure function we not only know that f(x') is composite-
we also know what f(x') is a multiple of. For example
let us take the failure function x = 14 + 211k; when k =1,
x' = 225. f(14+ 211) is a composite and it is a multiple
211. Needless to say this bonus aspect is not important
for application of failure functions in proving the
infinitude of primes os shape f(x).

[P] **conjecture - a fewpoints** by akdevaraj Oct 51) I double checked my computations - I am convinced that
a value of x, say x', if generated by one or more of the
relevant failure functions then f(x') is a failure (composite).
If not f(x') is prime and it need not be tested for primality.
2) in the demonstration sample the largest uncovered value of
x in the interval chosen is 167; f(167) = 28057 and the second
interval to be considered is 167, 28224. Obviously this cannnot
be dealt with manually.

[P] **conjecture - three points** by akdevaraj Oct 31) Carl Pomerance has just pointed out an exception to
my generalisation.However I stand by my message on x^2 +x +1.
2) A minor correction -second last line - shape 3^q.....
(p_(i-1)^z.
3) I ran the program {p(n) = factorint( n^2+n+1) }
from n=1 to n=1000. I could not find a single exponent
greater than 2.

[P] **Conjecture - clarification** by akdevaraj Oct 3Sorry I misunderstood Pomerance; he has clarified that
it has not yet been proved.

[P] **Conjecture - sketch proof - concluding message** by akdevaraj Sep 30Note 1) p_0 less than p_1 less than p_2........
Hence the intervals get progressively larger; even a small percentage of xs not covered
by the relevant failure functions means a sizable number of candidates for initiating the next
iteration. However we need only one x not covered by any failure function
to go to iteration i + 1. This is the largest of the xs not covered.
2) Only a competent programmer can program the failure functions and perform
the iteration. 3) For the sake of demonstration I have selected x = 12 and the
interval x = 12, 12 + 157 (12^2 + 12 +1 =157 ). The iteration is done as follows:
I make a list of xs starting with 12 and ending with 169. Which are the
relevant failure functions? 1+2k, 2 + 7k, 3 + 13k, 4 + 7k, 5 +31k, 6 +43k........ending
with 47 + 61k. These are relevant because these cover the interval chosen.
Next I circle the xs covered by each failure function. For example 1 + 3k
generates 4, 7, 10, ...After performing the same procedure with all the relevant
failure functions I found I was left with 37 unmarked xs. The percentage of
unmarked xs is 24%. Needless to say f(x), where x is any these 37 is prime.
The list of 37 unmarked xs: 14, 15, 17, 20, 24, 27, 33,38, 41, 50...ending with
167.
3) What are the implications of the iteration coming to an end after the ith
iteration? p_(i-1) is the largest prime with with shape x^2 + x+1.
This means x_(i-1) is the largest uncovered x; all subsequent xs are
such that f(x) are composite having shape 3^q7^m13^n.....x_(i-1)^z ( here
q,m,n...z are exponents belonging to N). This is highly improbable,
perhaps impossible.

[P] **Conjecture- sketch proof (contd.)** by akdevaraj Sep 29Briefly the sketch proof consists of the following steps:
1) Iteration: Let p_0 be the largest known prime with shape
x^2 + x +1. Let x_0 be the relevant value of x i.e. x_0^2 +x_0
+1 = p_0. Consider the interval x_0, x_0 + p_0.( this interval
is chosen because f(x_0 + k*f(x_0)) is congruent to 0 (mod (f(x_0)).
). In each iteration there is a percentage of values
of x not covered by the relevant failure functions - these are
such that the relevant f(x)s are prime (which need not be
tested for primality). Do the relevant failure functions cover
the whole interval? If so the iteration has come to an end
which means p_0 is the largest prime of the shape x^2 + x + 1
and that there are only a finite number of primes of this
shape. If not we go to the second iteration, i_2.: let
x_1 be the largest value of x not covered by the relevant
failure functions; f(x_1) is prime - let this be p_1. Consider
the interval x_1, x_1 + p_1. Do the relevant failure functions
cover the this interval completely? If so iteration has come
to an end and p_1 is the largest prime of this shape; this also
means there are only a finite number of primes of this shape.
If not we go to iteration, i_3. My conjecture: the percentage
of xs not covered by the relevant failure functions will
decrease from iteration to iteration progressively; however
the decrease of percentage is asymptotic to 3. i.e. it never
reaches 3. Hence the iteration is perpetural. Therefore the
infinitude of primes of shape x^2 + x + 1 is proved.
(to be continued ).

[P] **Conjecture- sketch proof** by akdevaraj Sep 29Conjecture: any irreducible quadratic polynomial, in which the
variable belongs to Z, generates an infinite set of prime
numbers. For illustration I am taking the irreducible quadratic polynomial:
f(x) = x^2+x +1. Definitions:1) failure: a composite number.
2) failure function: x = x_0 + k*x_0 (here a belongs to
N and is fixed; k belongs to Z and x_0 is a specific value of
x. In other words x is a function of x_0. When we substitute
the values of x generated by the failure function in f(x)
we get ONLY failures ( composites ). Example: x = 1 +3*k generates
4,7, 10, 13. . . Any of this infinite set, when substituted in
f(x) results in a failure (composite).
Indirect primality test: in the sketch proof I am developing I
will be testing only whether a particular value of x, say x_1, is is covered
by one or more failure functions or not. If it is covered, then f(x)
is a failure (composite ); otherwise f(x _1) is prime which need
not be tested for primality.
We do not directly test whether f(x_1) is prime or composite; that is
why this is an indirect primality test. (to be continued).

[P] **conjecture** by akdevaraj Sep 28Hi, Jussi; Sorry am unable to give the sketch proof without using the
mathematical tool viz. "failure functions ".

[P] **conjecture** by pahio Sep 26 Hi Deva, can you give the proof without failure funtions which ones I don't understand?
BTW, see the example http://planetmath.org/exampleofgcd
Jussi

[P] **Conjecture - sketch proof** by akdevaraj Sep 26Before I give the sketch proof I request members
to read my article " Failure functions " and the
the following message : Application
of failure functions - indirect primality test ( both on
planetmath ).