Re: 2*k+3 = p
- From: Danny <fasttrack2a@xxxxxxxxxxxxx>
- Date: 11 May 2007 10:18:35 -0700
On 4 May, 09:29, Danny <fasttrac...@xxxxxxxxxxxxx> wrote:
On 4 May, 04:34, Vincenzo Librandi <vincenzo.librandw...@xxxxxxxx>
wrote:
Danny wrote:
Where all (n) which includes 0(mod 3), only appears
just once giving (n) of the form2n+3= composite.
Well, the divisors:
If n=59, will2n+3 = 121; integer n= 59 into the matrix
A(5,5), therefore divisors: 5*2+1 = 11; 5*2+1 = 11.
If n = 70, will2n+3=143; integer n=70 into the matrix
A(6,5), therefore divisors: 6*2+1=13; 5*2+1=11.
If n=103, will2n+3=209; integer 103 in the matrix
A(9,5), therefore divisors: 9*2+1=19; 5*2+1=11.
If n=448, will2n*3=899; integer 448 in the matrix
A(15,14), therefore divisors: 15*2+1=31; 14*2+1=29.
Never a prime but all interers (n) not in this table
will be of the form2n+3 = prime
Very good.
Note all 0(mod 3) will only appear in the left most >column
and only appear once because of the doubling pattern.
In the non trivial rows where +(1) is represented as
a single add and +(2) is represented as a double add so
then the pattern is--+-- 1,1,2,1,2,1,2,1,2... but >ending with
the appropreate number of terms for that row.
Row 1 beginning with (3) has 1 term.
Row 2 beginning with (6) has 2 terms.
Row 3 beginning with (9) has 3 terms.
Row 4 beginning with (12) has 1 term.
Row 5 beginning with (15) has 4 terms.
Row 6 beginning with (18) has 5 terms.
Row 7 beginning with (21) has 1 term.
Row 8 beginning with (24) has 6 terms.
Row 9 beginning with (27) has 7 terms.
Row 10 beginning with (30) has 1 term.
Row 11 beginning with (33) has 8 terms.
Row 12 beginning with (36) has 9 terms.
Row 12 beginning with (36) has 2 terms
Row 13 beginning with (39) has 1 term.
Row 14 beginning with (42) has 10 terms.
Row 14 beginning with (42) has 9 terms.
Row 15 beginning with (45) has 10 terms.
Row 16 beginning with (48) has 1 term.
Row 17 beginning with (51) has 2 terms.
Row 18 beginning with (54) has 11 terms
etc.
Ok, I screwed up, many will repeat in my table starting
with (n) = 86.
So there is no simplified matrix or table that can be uniform
and still represent all the integers of the form2n+3 = composite
only once?
etc.
3
6,11
9,16,23
12-- trivial with no summations
15,26,37,59 =(15+11+11+22)
18,31,44,70,83 =(18+13+13+26+13)
21--trivial -- ditto.
24,41,58,92,109,143 =(24+17+17+34+17+34)
27,46,65,103,122,160,179 =(27+19+19+38+19+38+19)
30--trivial --ditto.
33,56,79,125,148,194,217,263 =(33+23+23+46+23+46+23+46)
36,61,86,136,161,211,236,286,311 =(36+25+25+50+25+50+25+50+25)
36,61 =(36+25)
39--trivial --ditto.
42,71,100,158,187,245,274,332,361,419 =(42+29+29+58+29+58+29+58+29+58)
42,71,100,158,187,245,274,332,419 =(42+29+29+58+29+58+29+58+87)
45,76,107,169,200,262,293,355,448,479 =(45+31+31+62+31+62+31+62+93+31)
etc.
etc.
So each (n) of the form2n+3 = composite
where each (n) only appears once in this table.
This is quite remarkable because all (n) that
do not appear in this table are of the form
2n+3 = prime. Representing all the primes.
Very well.
Can anyone explain this?
Da più tempo sto facendo la stessa domanda !
The sequence of integers (n) in this table.
2n+3 = composite.
n =
3,6,9,11,12,15,16,18,21,23,24,26,27,30,31,33,36,37,
39,41,42,44,45,46,48,51,54,56,57,58,59,60,61,63,65,
66,69,70,71,72,75,76,78,79,81...
Comparing this sequence above with A067076 --
where (n) is of the form --2n+3 = prime
n =
0,1,2,4,5,7,8,10,13,14,17,19,20,22,25,28,29,32,34,35,
38,40,43,47,49,50,52,53,55,62,64,67,68,73,74,77,80,
82...
La sequenza è apparsa su A067076 molto tempo
dopo i miei post su sci math.
No integer match and not any missing integers
between the two sequences.
A nice find by Vincenzo but with a few
modifications to elliminate douplicate
0(mod 3)'s.
Stabilito ciò, chiedo, è possibile determinare, data una
N che appartiene alla matrice, per esempio 109, il numero degli elementi che la precedono ?
Esempio:
prima di 23, nella matrice compaiono i termini
(3,6,9,11,12,15,16,18,21), cioè 9 termini;
il primo termine utile, prima di 23, per determinare un numero primo è 22 che porta al primo 47; che occupa il posto (23-9 =14)quattordicesimo numero primo dispari.
Parleremo, se vuoi, più in là, degli elementi che compaiono nella diagonale della matrice (3,11,23,59,83,143,179,263,419,479,683 etc,)
e i numeri primi di Sophie Germain.
A risenterci,
Vincenzo Librandi
vincenzo.librabdw...@xxxxxxxxx Hide quoted text -
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Your matrix is interesting in as much that
it does not satisfy for n*2+3 = prime but
what it does satisfy is n*2+3 = composite.
There are many interesting things about this
matrix and I show one below.
The first column is the index
and the next column is the
odd integers starting with 3.
The third column is the first left column
in the matrix with the odd in the second
column added to it and throughout the
sequence (row).
I = index
O=all odd >1
T=all 0(mod 3) and first left column of matrix.
I--O--T
1--3--3
2--5--6,11----------------------------11*2+3 = 5^2
3--7--9,16,23------------------------23*2+3 = 7^2
4--9--12,-----trivial.
5--11--15,26,37,48,59-------------59*2+3 = 11^2
6--13--18,31,44,57,70,83---------83*2+3 = 13^2
7--15--21,-----trivial.
8--17--24,41,58,75,92,109,126,143--143*2+3 = 17^2
9--19--27,46,65,84,103,122,141160,179--179*2+3 = 19^2
10-21-30,-----trivial.
11-23-33,56,79,102,125,148,171,194,217,240,263--263*2+3 = 23^2
12-25-36,61,86,111,136,161,186,211,236,261,286,311--311*2+3 = 25^2
13-27-39,-----trivial.
14-29-42,71,100,129,158,187,216,245,274,303,332,361,390,419-- etc..
15-31-45,76,107,138,169,200,231,262,293,324,355,386,417,448,479
16-33-48,-----trivial.
17-35-51,86,121,156,191,226,261,296,331,366,401,436,471506,541,576,611
18...
etc..
Say you have a very large integer (n) (index) that you wish
to test the row of that index for any possible 2n+3= p.
Where (n) = index then -- (n*2)+1 * n +(n-1) = the last
integer for that row.
First the index integer should not be 1(mod 3).
So by the formula the integer 1287950 (index number).
(1287950*2) +1 * 1287950 + (1287950 - 1) = 3317632980899
a composite but also the last digit in the row for that index.
To check this I take your formula for the right diagonal integer--
3317632980899*2 + 3 = 2575901^2 where 2575901 = the odd
component add throughout the sequence.
Now to find the first term (column) in your matrix it has
to be a 0(mod 3).
The first term = 3*1287950=(3863850) + 2575901 + 2575901 ...
for a total of 1287950 terms for that row.
Checking index # 1287950--
index-----odd integer------0(mod 3)
1287950---2575901--------3863850 + 2575901 + 2575901 + 2575901 +
2575901 +.. = 3317632980899
Now you can calculate any larger row in the matrix without
all the previous rows.
The OP may have proposed this very same property about
calculating larger rows without showing any previous rows
if so I apologize to the OP.
I ask the same question as before, can anyone explain
this matrix where it contains all (n) of the form
2n+3 = composite and all (n) not in this matrix is
of the form 2n+3= prime?
The only way this matrix can be refuted is with just one
counterexample.
Dan
.
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- Re: 2*k+3 = p
- From: Vincenzo Librandi
- Re: 2*k+3 = p
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