## best T-Sequence in B-A-S-I-C

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• not giving up... I took the decision tree out for L here s the T-Sequence in its most raw form, and it s more efficient like David Broadhurst wanted. the t-
Message 1 of 17 , Oct 27, 2010
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not giving up... I took the decision tree out for L
here's the T-Sequence in its most raw form, and it's
more efficient like David Broadhurst wanted. the t-
sequence works, but I couldn't display it before...

I'm only bringing it to you again, because you math
guys have the brains to figure out where it might go
wrong; It must have an L (-) type and one L (+) type.

I noticed that you were able to translate it into the
Pari/GP (language/platform) to check larger numbers.

100 CLS : C = 1
102 DIM A(1001), LT(1), T(1001)
104 FOR N = 3 TO 7919 STEP 2
106 IF N MOD 3 <> 0 AND N MOD 5 <> 0 AND N <> 7 THEN
108 A(0) = N: A(1) = (N + 1) / 2: A(2) = A(1) - 1
110 FOR I = 3 TO 1001 STEP 2
112 IF A(I - 1) < 3 THEN
114 IF A(I - 1) = 2 THEN
116 A(I) = 1: A(I + 1) = 0
118 M = I + 1: I = 1001
120 ELSE
122 A(I) = 0: M = I: I = 1001
124 END IF
126 ELSE
128 IF A(I - 1) MOD 2 = 1 THEN
130 A(I) = A(I - 2) / 2
132 ELSE
134 A(I) = (A(I - 2) + 1) / 2
136 END IF
138 A(I + 1) = A(I) - 1
140 END IF
142 NEXT I
144 D = 0: L = 3: LT(0) = 2: LT(1) = 2
146 WHILE (L < N - 2)
148 T(M) = 2: T(M - 1) = L: T(M - 2) = (T(M - 1) ^ 2 - 2) MOD N
150 IF T(M - 2) < 2 THEN T(M - 2) = T(M - 2) + N
152 K = 0: Z1 = 0: Z2 = 0
154 FOR J = M - 3 TO 0 STEP -1
156 IF A(J) MOD 2 = 1 THEN
158 IF A(J) = A(J + 1) + A(J + 2) THEN K = 0 ELSE K = 1
160 T(J) = (T(J + 1 + K) * T(J + 2 + K) - L) MOD N
162 ELSE
164 T(J) = (T(J + 2) ^ 2 - 2) MOD N
166 END IF
168 IF T(J) < 2 THEN T(J) = T(J) + N
170 NEXT J
172 Z1 = (T(2) ^ 2 - 2) MOD N
174 IF Z1 < 2 THEN Z1 = Z1 + N
176 Z2 = (T(1) ^ 2 - 2) MOD N
178 IF Z2 < 2 THEN Z2 = Z2 + N
180 IF T(0) = L THEN
182 IF Z1 = 2 AND Z2 = T(M - 2) THEN
184 LT(D) = -1
186 ELSE
188 IF Z1 = T(M - 2) AND Z2 = 2 THEN
190 LT(D) = 1
192 END IF
194 END IF
196 ELSE
198 PRINT N; "is composite!": L = N - 2
200 END IF
202 IF LT(0) = -LT(1) THEN
204 PRINT N; "is prime!": L = N - 2: C = C + 1
206 END IF
208 D = 1: L = L + 1
210 WEND
212 ELSE
214 IF N = 3 OR N = 5 OR N = 7 THEN
216 PRINT N; "is prime!": C = C + 1
218 ELSE
220 PRINT N; "is composite!"
222 END IF
224 END IF
226 NEXT N
228 PRINT C
230 END

the only thing that would not make it work according
to the co-authors that want to patent it... is that
it must have [T(M-2) mod N <> 2]; it must have a per-
iod larger than k(p) = 2.

65 lines of code; let me know if you like it... Bill
the maximum L is ever got to for N <= 7919 was L= 17
• ... At last you have discovered the difference between a weak and a strong Lucas test. By demanding that one of your Lucas tests is strong, you are on the way
Message 2 of 17 , Oct 27, 2010
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leavemsg1" <leavemsg1@...> wrote:

> must have an L (-) type and one L (+) type

At last you have discovered the difference between a weak
and a strong Lucas test. By demanding that one of your Lucas
tests is strong, you are on the way to discovering what BPSW
did and should no longer be tortured by small gremlins.

> it's more efficient

You were still very inefficient, because you did not compute
kronecker(L^2-4,N), before each Lucas test on V(L,1,N).

Consider the prime p = 18539153100230615161, given in
Michael Jacobson's M.Sc. thesis, "Computational Techniques
in Quadratic Fields" [*]. Then L^2-4 is a square, modulo p,
for L = 3 to 260, and so you waste a lot time by computing
everything up to L = 261, to find a negative sign.
In this case, you spend 2*(261-2) = 518 selfridges
to achieve what anyone else could do in 3 selfridges,
simply by performing a Fermat and a strong Lucas test.

If you will learn just a very small bit of maths,
namely how to compute a kronecker symbol, then you
should be able to write a 3-selfridge test that is
secure against the absolute Pinchs to which you had
laid yourself wide open in the days of your weakness.

[*] To find this thesis, simply google 18539153100230615161

David
• ... With the help of a good programmer, Vincent Diepeven, and some computation collaborators, Carlos Pinho and Jeff Gilchrist, we have pushed up the tested
Message 3 of 17 , Oct 27, 2010
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>
>
>
> leavemsg1" <leavemsg1@> wrote:
>
> > must have an L (-) type and one L (+) type
>
> At last you have discovered the difference between a weak
> and a strong Lucas test. By demanding that one of your Lucas
> tests is strong, you are on the way to discovering what BPSW
> did and should no longer be tortured by small gremlins.
>
> > it's more efficient
>
> You were still very inefficient, because you did not compute
> kronecker(L^2-4,N), before each Lucas test on V(L,1,N).
>
> Consider the prime p = 18539153100230615161, given in
> Michael Jacobson's M.Sc. thesis, "Computational Techniques
> in Quadratic Fields" [*]. Then L^2-4 is a square, modulo p,
> for L = 3 to 260, and so you waste a lot time by computing
> everything up to L = 261, to find a negative sign.
> In this case, you spend 2*(261-2) = 518 selfridges
> to achieve what anyone else could do in 3 selfridges,
> simply by performing a Fermat and a strong Lucas test.
>
> If you will learn just a very small bit of maths,
> namely how to compute a kronecker symbol, then you
> should be able to write a 3-selfridge test that is
> secure against the absolute Pinchs to which you had
> laid yourself wide open in the days of your weakness.
>
> [*] To find this thesis, simply google 18539153100230615161
>

With the help of a good programmer, Vincent Diepeven, and some computation collaborators, Carlos Pinho and Jeff Gilchrist, we have pushed up the tested range to around n=3.9*10^8, without counterexamples, for the second 6-selfridge "double" test ("a-1 and "a+1") given in my paper.
More help with testing is always welcome.

I found this pseudocode very useful for calculating the Jacobi symbol:
http://primes.utm.edu/glossary/xpage/JacobiSymbol.html

Paul
• ... That s an impressive number of tests. How many times was it necessary to invoke your (seemingly) ad hoc restrictions that the target is coprime to 5 and 7
Message 4 of 17 , Oct 28, 2010
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"paulunderwooduk" <paulunderwood@...> wrote:

> With the help of a good programmer, Vincent Diepeven,
> and some computation collaborators, Carlos Pinho and
> Jeff Gilchrist, we have pushed up the tested range to
> around n=3.9*10^8, without counterexamples, for the
> second 6-selfridge "double" test ("a-1 and "a+1")
> given in my paper.

That's an impressive number of tests.
How many times was it necessary to invoke your (seemingly) ad hoc
restrictions that the target is coprime to 5 and 7 ?
No reason was given in your article why a large odd square-free
target, coprime to 6*a, should be denied these factors,
in order to find a "suitable" pair (a-1,a+1). Perhaps some
pseudoprimes, might be informative?

> I found this pseudocode very useful for calculating
> the Jacobi symbol:
> http://primes.utm.edu/glossary/xpage/JacobiSymbol.html

Thank for providing Bill with that link,
which is indeed very clear. Note however
> Suppose n is odd and 0 < a < n
whereas Kronecker's extension
http://mathworld.wolfram.com/KroneckerSymbol.html
of Jacobi's extension of the Legendre symbol allows
for all pairs of integers.

Complete pseudo-code for the Kronecker symbol is
given by Cohen in CCANT Algorithm 1.4.10
with notes on how to use a "bitwise and"
and cautions in Exercise 24 about possible
exponential blow-up of the computation time if we
do not clean out powers of 2 at every step of the
Euclidean loop.

David
• ... The number tests, taking into consideration the symmetry, is of the proportion (2-1)/2*(3-1)/3*(5-1)/5*(7-1)/7/2 of squared n That is 0.114285714*n^2. For
Message 5 of 17 , Oct 28, 2010
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>
>
>
> "paulunderwooduk" <paulunderwood@> wrote:
>
> > With the help of a good programmer, Vincent Diepeven,
> > and some computation collaborators, Carlos Pinho and
> > Jeff Gilchrist, we have pushed up the tested range to
> > around n=3.9*10^8, without counterexamples, for the
> > second 6-selfridge "double" test ("a-1 and "a+1")
> > given in my paper.
>
> That's an impressive number of tests.
> How many times was it necessary to invoke your (seemingly) ad hoc
> restrictions that the target is coprime to 5 and 7 ?

The number tests, taking into consideration the symmetry,
is of the proportion
(2-1)/2*(3-1)/3*(5-1)/5*(7-1)/7/2 of squared n

That is 0.114285714*n^2.
For n<3.9*10^8 the estimate for the number of tests is therefore
1.73828571*10^16

> No reason was given in your article why a large odd square-free
> target, coprime to 6*a, should be denied these factors,
> in order to find a "suitable" pair (a-1,a+1).

Note that the discriminants are (a+1)*(a-3) and (a-1)*(a+3) and so 3,5, and 7 can be eliminated by deduction.

> Perhaps some
> pseudoprimes, might be informative?
>

I have thousands of cases where the gcd was necessary, mostly with a common gcd for a given n, but a few which might be considered interesting, for example n=35285489 which needed for "a" the gcds:
n,a,gcd 35285489 357865 71573
n,a,gcd 35285489 1574606 71573
n,a,gcd 35285489 2075617 2075617
n,a,gcd 35285489 2433482 1216741
n,a,gcd 35285489 3650223 1216741
n,a,gcd 35285489 4509099 71573
n,a,gcd 35285489 5725840 71573

Here is another, but "uninteresting":
n,a,gcd 36780239 87781 87781
n,a,gcd 36780239 438905 87781
n,a,gcd 36780239 2194525 87781
n,a,gcd 36780239 2457868 87781
n,a,gcd 36780239 2808992 87781
n,a,gcd 36780239 3160116 87781
n,a,gcd 36780239 3511240 87781
n,a,gcd 36780239 3686802 87781
n,a,gcd 36780239 4037926 87781
n,a,gcd 36780239 4301269 87781
n,a,gcd 36780239 4389050 87781
n,a,gcd 36780239 4476831 87781
n,a,gcd 36780239 4652393 87781
n,a,gcd 36780239 5003517 87781
n,a,gcd 36780239 5266860 87781
n,a,gcd 36780239 5354641 87781
n,a,gcd 36780239 5442422 87781
n,a,gcd 36780239 5530203 87781
n,a,gcd 36780239 5617984 87781
n,a,gcd 36780239 5705765 87781
n,a,gcd 36780239 5793546 87781
n,a,gcd 36780239 5969108 87781
n,a,gcd 36780239 6056889 87781
n,a,gcd 36780239 6408013 87781
n,a,gcd 36780239 6759137 87781
n,a,gcd 36780239 6846918 87781
n,a,gcd 36780239 7812509 87781
n,a,gcd 36780239 8163633 87781
n,a,gcd 36780239 8778100 87781
n,a,gcd 36780239 9217005 87781
n,a,gcd 36780239 9743691 87781
n,a,gcd 36780239 10094815 87781
n,a,gcd 36780239 10358158 87781
n,a,gcd 36780239 11235968 87781
n,a,gcd 36780239 11762654 87781
n,a,gcd 36780239 12991588 87781
n,a,gcd 36780239 13079369 87781
n,a,gcd 36780239 13430493 87781
n,a,gcd 36780239 13957179 87781
n,a,gcd 36780239 14044960 87781
n,a,gcd 36780239 14396084 87781
n,a,gcd 36780239 14747208 87781
n,a,gcd 36780239 15888361 87781
n,a,gcd 36780239 16239485 87781
n,a,gcd 36780239 16415047 87781
n,a,gcd 36780239 16502828 87781
n,a,gcd 36780239 17380638 87781
n,a,gcd 36780239 17731762 87781
n,a,gcd 36780239 18170667 87781
n,a,gcd 36780239 18258448 87781

The only pattern I have spotted so far is that when a gcd is required, n/gcd is a product of primes of the form 6*k-1,

Paul
• ... I think I am off by a factor 4... Paul
Message 6 of 17 , Oct 28, 2010
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--- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
>
>
> The number tests, taking into consideration the symmetry,
> is of the proportion
> (2-1)/2*(3-1)/3*(5-1)/5*(7-1)/7/2 of squared n
>
> That is 0.114285714*n^2.
> For n<3.9*10^8 the estimate for the number of tests is therefore
> 1.73828571*10^16
>

I think I am off by a factor 4...

Paul
• ... Drats. It is only n
Message 7 of 17 , Oct 28, 2010
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>
>
>
> "paulunderwooduk" <paulunderwood@> wrote:
>
> > With the help of a good programmer, Vincent Diepeven,
> > and some computation collaborators, Carlos Pinho and
> > Jeff Gilchrist, we have pushed up the tested range to
> > around n=3.9*10^8, without counterexamples, for the
> > second 6-selfridge "double" test ("a-1 and "a+1")
> > given in my paper.
>
> That's an impressive number of tests.

Drats. It is only n<3.9*10^7 (39 million). So the number of individual tests is about:
0.228571429*(n/2)^2 ~= 0,87*10^14

Paul
• ... Whenever you re implementing this algorithm, you need to be warned that different programming languages implement different conventions for the mod
Message 8 of 17 , Oct 28, 2010
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On 10/27/2010 10:52 PM, paulunderwooduk wrote:
> I found this pseudocode very useful for calculating the Jacobi symbol:
> http://primes.utm.edu/glossary/xpage/JacobiSymbol.html

Whenever you're implementing this algorithm, you need to be warned
that different programming languages implement different conventions for
the "mod" operator when working with negative numbers. Since the page
doesn't indicate which convention is expected, your results may be
wrong. Some languages round towards zero, others round towards negative
infinity.

Actually, some languages round *both* ways. For example, Java's "%"
operator (used for ints) and its BigDecimal.mod method (used for bigger
numbers) use *different* sign conventions! Augh! This was actually the
source of a very nasty bug in Java for about 3 years. Their
implementation of Jacobi symbol returned incorrect values for negative
BigIntegers because of this difference in how mod worked. This caused
silent failures of its primality testing routines, which were especially
nasty because it failed relatively often and in a way that Rabin-Miller
tests could *not* fail:

http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=4624738

I used a different algorithm, implemented from similar pseudocode in
the implementation of the JacobiSymbol in my programming language Frink
( http://futureboy.us/frinkdocs/ ) and had this same problem for a
specific negative number range because of the difference in the
implementation of % and mod in Java. (To my credit, I released a fix
later that day, as opposed to Java, which had primality testing failures
for 3 years.)

--
Alan Eliasen
eliasen@...
http://futureboy.us/
• ... Oops. That was intended to read BigInteger. -- Alan Eliasen eliasen@mindspring.com http://futureboy.us/
Message 9 of 17 , Oct 29, 2010
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On 10/28/2010 09:38 PM, Alan Eliasen wrote:
> Actually, some languages round *both* ways. For example, Java's "%"
> operator (used for ints) and its BigDecimal.mod method (used for bigger
> numbers)

Oops. That was intended to read BigInteger.

--
Alan Eliasen
eliasen@...
http://futureboy.us/
• ... Faulty deduction , it seems to me. Consider n = 7*373*383 = 1000013 Then we may satisfy all three of these conditions {cond(a,n) = gcd(6*a,n) == 1 &&
Message 10 of 17 , Oct 29, 2010
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"paulunderwooduk" <paulunderwood@...> wrote:

> Note that the discriminants are (a+1)*(a-3) and (a-1)*(a+3)
> and so 3, 5, and 7 can be eliminated by deduction.

Faulty "deduction", it seems to me.

Consider n = 7*373*383 = 1000013

Then we may satisfy all three of these conditions

{cond(a,n) =
gcd(6*a,n) == 1 &&
kronecker((a+1)*(a-3),n) == -1 &&
kronecker((a-1)*(a+3),n) == -1 ;}

for many values of a:

for(a=5,137,if(cond(a,1000013),print1(a", ")));

5, 33, 47, 51, 54, 68, 72, 79, 96, 117, 124, 128, 135,

Yet you wrote, in your article:

> to find a suitable pair, a-1 and a+1, we further require that
> n is non-square and gcd(210a,n) = 1

So either I am being stupid,
or you are being sloppy,
or both :-?

David (on behalf of the factor 7)
• ... Great story. Thanks, Alan. Jens has a good quote, somewhere, along the lines that failures of primality tests are not matters of life and death, whatever
Message 11 of 17 , Oct 29, 2010
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Alan Eliasen <eliasen@...> wrote:

> To my credit, I released a fix later that day,
> as opposed to Java, which had primality testing
> failures for 3 years.

Great story. Thanks, Alan.

Jens has a good quote, somewhere, along the lines
that failures of primality tests are not matters
of life and death, whatever you may read in the
grant proposals of number theorists trying to
pretend to "fight terrorism". Cf:
http://www.presstv.ir/detail/148710.html

David
• ... I observe: 373 = 2 (mod n) 383 = -2 (mod n) ... Again, all a = +-2 (mod) ... It seems that n=7^m*p^2 or n=7^m (m odd and p 7) (at least) never yield a
Message 12 of 17 , Oct 29, 2010
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>
>
>
> "paulunderwooduk" <paulunderwood@> wrote:
>
> > Note that the discriminants are (a+1)*(a-3) and (a-1)*(a+3)
> > and so 3, 5, and 7 can be eliminated by deduction.
>
> Faulty "deduction", it seems to me.
>
> Consider n = 7*373*383 = 1000013
>

I observe:
373 = 2 (mod n)
383 = -2 (mod n)

> Then we may satisfy all three of these conditions
>
> {cond(a,n) =
> gcd(6*a,n) == 1 &&
> kronecker((a+1)*(a-3),n) == -1 &&
> kronecker((a-1)*(a+3),n) == -1 ;}
>
> for many values of a:
>
> for(a=5,137,if(cond(a,1000013),print1(a", ")));
>
> 5, 33, 47, 51, 54, 68, 72, 79, 96, 117, 124, 128, 135,
>

Again, all a = +-2 (mod)

> Yet you wrote, in your article:
>
> > to find a suitable pair, a-1 and a+1, we further require that
> > n is non-square and gcd(210a,n) = 1
>
> So either I am being stupid,
> or you are being sloppy,
> or both :-?

It seems that n=7^m*p^2 or n=7^m (m odd and p>7) (at least) never yield a suitable kronecker pair.

? for(n=1,100000,if(gcd(n,30)==1&&!issquare(n)&&!isprime(n),t=0;for(a=1,(n-1)/2,x=a-1;y=a+1;if(kronecker(x^2-4,n)==-1&&kronecker(y^2-4,n)==-1,t=1));if(t==0,print(n" "factor(n)))))
343 Mat([7, 3])
847 [7, 1; 11, 2]
1183 [7, 1; 13, 2]
2023 [7, 1; 17, 2]
2527 [7, 1; 19, 2]
3703 [7, 1; 23, 2]
5887 [7, 1; 29, 2]
6727 [7, 1; 31, 2]
9583 [7, 1; 37, 2]
11767 [7, 1; 41, 2]
12943 [7, 1; 43, 2]
15463 [7, 1; 47, 2]
16807 Mat([7, 5])
19663 [7, 1; 53, 2]
24367 [7, 1; 59, 2]
26047 [7, 1; 61, 2]
31423 [7, 1; 67, 2]
35287 [7, 1; 71, 2]
37303 [7, 1; 73, 2]
41503 [7, 3; 11, 2]
43687 [7, 1; 79, 2]
48223 [7, 1; 83, 2]
55447 [7, 1; 89, 2]
57967 [7, 3; 13, 2]
65863 [7, 1; 97, 2]
71407 [7, 1; 101, 2]
74263 [7, 1; 103, 2]
80143 [7, 1; 107, 2]
83167 [7, 1; 109, 2]
89383 [7, 1; 113, 2]
99127 [7, 3; 17, 2]

Paul
>
> David (on behalf of the factor 7)
>
• ... If you add a break you ll quickly see p is not necessarily prime, Paul
Message 13 of 17 , Oct 29, 2010
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--- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
>

> It seems that n=7^m*p^2 or n=7^m (m odd and p>7) (at least) never yield a suitable kronecker pair.
>
> ? for(n=1,100000,if(gcd(n,30)==1&&!issquare(n)&&!isprime(n),t=0;for(a=1,(n-1)/2,x=a-1;y=a+1;if(kronecker(x^2-4,n)==-1&&kronecker(y^2-4,n)==-1,t=1));if(t==0,print(n" "factor(n)))))

If you add a "break" you'll quickly see "p" is not necessarily prime,

Paul
• ... Obviously not: 7^m*p^2 (m odd and p 7) is equivalent to 7 as far as Kronecker is concerned. ... Obviously not: 7^m*p^2 (m odd and p 7) is equivalent to 7
Message 14 of 17 , Oct 29, 2010
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"paulunderwooduk" <paulunderwood@...> wrote:

> It seems that n=7^m*p^2 or n=7^m (m odd and p>7)
> never yield a suitable kronecker pair.

Obviously not: 7^m*p^2 (m odd and p>7) is equivalent to 7
as far as Kronecker is concerned.

> If you add a "break" you'll quickly see "p"
> is not necessarily prime,

Obviously not: 7^m*p^2 (m odd and p>7) is equivalent to 7
as far as Kronecker is concerned, for all p > 7.

But the above is merely post-hoc evasion.

since you missed out testing *oodles* of (n,a) pairs, with 7|n,
for no justifiable reason, whatsoever.

David (on behalf of the factor 7)
• ... Guilty as charged. My deduction about 7*N was flawed. The only reason for discounting 7 was oodles output from pari where a kronecker pair could not be
Message 15 of 17 , Oct 29, 2010
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>
>
>
> "paulunderwooduk" <paulunderwood@> wrote:
>
> > It seems that n=7^m*p^2 or n=7^m (m odd and p>7)
> > never yield a suitable kronecker pair.
>
> Obviously not: 7^m*p^2 (m odd and p>7) is equivalent to 7
> as far as Kronecker is concerned.
>
> > If you add a "break" you'll quickly see "p"
> > is not necessarily prime,
>
> Obviously not: 7^m*p^2 (m odd and p>7) is equivalent to 7
> as far as Kronecker is concerned, for all p > 7.
>
> But the above is merely post-hoc evasion.
>
> since you missed out testing *oodles* of (n,a) pairs, with 7|n,
> for no justifiable reason, whatsoever.
>

Guilty as charged. My deduction about 7*N was flawed.
The only reason for discounting 7 was oodles output from pari where a kronecker pair could not be found for many 7*N-- not very mathematical, I admit, but I felt it justified.

n=11 is not identified as prime by the test, because no kronecker pair is available. As much as I rely output from pari, I also disregard testing 11 as it's too small.

Paul
> David (on behalf of the factor 7)
>
• ... I did check numbers divisible by 5 or 7 for n
Message 16 of 17 , Oct 29, 2010
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>

>
> since you missed out testing *oodles* of (n,a) pairs, with 7|n,
> for no justifiable reason, whatsoever.
>

I did check numbers divisible by 5 or 7 for n<11364001, but cut them out of being tested for larger n, because of the lack of kronecker pairs for some 5*N and 7*N

Paul
> David (on behalf of the factor 7)
>
• ... Thanks, Paul. ... As official representative of the factor 7, may I cordially invite you to test (n,a) pairs with n 11364001 and 7|n ? Best regards David
Message 17 of 17 , Oct 29, 2010
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"paulunderwooduk" <paulunderwood@...> wrote:

> > since you missed out testing *oodles* of (n,a) pairs, with 7|n,
> > for no justifiable reason, whatsoever.
>
> Guilty as charged. My deduction about 7*N was flawed.

Thanks, Paul.

> I did check numbers divisible by 5 or 7 for n<11364001

As official representative of the factor 7,
may I cordially invite you to test (n,a) pairs
with n > 11364001 and 7|n ?

Best regards

David (on behalf of the factor 7)
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