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Old 2011-01-24, 08:11   #1
allasc
 
Aug 2010
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Default integers of the form (2 ^ t +1) / (2 ^ p + x)

Рассмотрим.....
Числа удовлетворяющие условию.


Числа n таких, что
2^(n-1) = 1 (MOD n);
2^(n-1-m) = (n-2^р) (MOD n)

где m целое и 2^(m-1) <n<2^m;
р целое число, р>0 и 2^р<n


заявка в OEIS A167612, лежит на рассмотрении :)
первые элементы
3, 11, 13, 19, 41, 43, 241, 331, 683, 2113, 2731, 3277, 4033, 5419, 8321, 43691, 61681, 65281, 80581, 85489, 87211, 174763, 233017, 253241, 525313, 838861, 1016801, 1397419, 2796203, 3033169, 3605429, 4682833, 6700417, 13421773, 15790321, 16773121, 18837001, 20647621, 22253377, 22366891, 24214051, 25080101, 25781083, 30662497, 47349373, 50155733, ....

ключевое значение - показатель степени t=m+p
все элементы этой последовательности могут быть представлены в виде
n=(2^t+1) / (2^p + x)

----------------------------------------

Consider .....
Numbers satisfying the condition.

Numbers n such that
2^(n-1) = 1 (mod n);
2^(n-1-m) = (n-2^p) (mod n)
where
m is an integer and 2^(m-1) < n < 2^m;
p is an integer, p > 0 and 2^p < n

in OEIS A167612, is pending:)
the first elements
3, 11, 13, 19, 41, 43, 241, 331, 683, 2113, 2731, 3277, 4033, 5419, 8321, 43691, 61681, 65281, 80581, 85489, 87211, 174763, 233017, 253241, 525313, 838861, 1016801, 1397419, 2796203, 3033169, 3605429, 4682833, 6700417, 13421773, 15790321, 16773121, 18837001, 20647621, 22253377, 22366891, 24214051, 25080101, 25781083, 30662497, 47349373, 50155733, ....

principal value - an exponent
t = m + p
All elements of this sequence can be represented as
n=(2^t+1) / (2^p + x)
x is an integer

Last fiddled with by allasc on 2011-01-24 at 08:50
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Old 2011-01-24, 08:20   #2
allasc
 
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Свойства чисел этой последовательности

1) 2^t=1(mod n)
2) n=1(mod t) (proved user Sonic86 http://dxdy.ru/topic41309.html)

3) Если n составное число, то его можно предтавить в виде n=(a*t+1)*(b*t+1) где a и b целые числа >0

4) Если m=0(mod p), тогда n=(2^t+1)/(2^p+1), тоесть x=1
5) Если m-четное число, p=m-2, тогда n=(2^t+1)/(2^p+2^p-2^(m/2)+1) тоесть x=2^p-2^(m/2)+1

---------------------------------

Properties of the numbers in this sequence

1) 2 ^ t = 1 (mod n)
2) n = 1 (mod t) (proved user Sonic86 http://dxdy.ru/topic41309.html)

3) If n is a composite number, then it can be represented as n = (a*t +1)*(b*t +1), where a and b are integers > 0

4) If m = 0 (mod p), then n = (2^t +1) / (2^p +1), ie x = 1
5) If the m-even number and (p = m-2), then n = (2^t +1) / (2^p+2^p-2^(m/2) +1) ie x = 2^p-2^(m/2)+1

Last fiddled with by allasc on 2011-01-24 at 09:00
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Old 2011-01-24, 08:29   #3
allasc
 
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Самое интересное.

Если p=1 , тогда t=m+p - либо простое число, либо псевдопростое по основанию 2
И соответсвенно число n имеет вид n=(2^t+1)/3
Причем, если это число n является составным, то его делитель (я так понимаю больший) также принадлежит нашей последовательности.

---------------

Most interesting.

If p=1, then t=m+p - a prime number, or pseudo-prime base 2
And correspondingly the number n has the form n = (2^t +1) / 3
Moreover, if the number n is composite, then its divisor (as I understand it more) also belongs to our sequence.

Last fiddled with by allasc on 2011-01-24 at 08:54
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Old 2011-01-24, 08:38   #4
allasc
 
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(2^29+1)/3=178956971=59*3033169

178956971 belongs to our sequence, where m=28 p=1 (t=29)
3033169 belongs to our sequence, where m=22 p=7 (t=29)

likewise

(2^37+1)/3=45812984491=1777*25781083
45812984491 m=36 p=1 (t=37)
25781083 m=25 p=12 (t=37)

(2^41+1)/3=733007751851 = 83*8831418697
733007751851 m=40 p=1 (t=41)
8831418697 m=34 p=7 (t=41)

(2^47+1)/3=46912496118443 = 283*165768537521
46912496118443 m=46 p=1 (t=47)
165768537521 m=38 p=9 (t=47)

(2^53+1)/3=3002399751580331 = 107*28059810762433
3002399751580331 m=52 p=1 (t=53)
28059810762433 m=45 p=8 (t=53)

Last fiddled with by allasc on 2011-01-24 at 09:08
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Old 2011-01-24, 08:48   #5
allasc
 
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Самое главное предположение

рассматривая выражение
n=(2^t+1) / (2^p + x)
предпологая что значение x - однозначно определяется набором значений m и p
вот только как определяется - вот это главный вопрос.
Если это правило будет дастаточно простым, то нахождение делителей для
n=(2^t+1)/3 (для n, при p=1)
также будет простым

-------------

The most important assumption

Considering the expression
n = (2 ^ t +1) / (2 ^ p + x)
assuming that the value of x - is uniquely determined by the values of m and p.
How is - this is the main question.
If this rule is simple,
finding the divisors for
n = (2 ^ t +1) / 3 (for n, where p=1)
will also be a simple

Last fiddled with by allasc on 2011-01-24 at 09:19
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Old 2011-01-24, 09:25   #6
allasc
 
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Quote:
Originally Posted by allasc View Post

Properties of the numbers in this sequence

1) 2 ^ t = 1 (mod n)
извините ошибся
sorry mistake

2^t+1= 0 (mod n)
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Old 2011-01-26, 09:15   #7
allasc
 
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All elements of this sequence can be represented as
n=(2^t+1) / (2^p + x)
x is an integer

Quote:
Originally Posted by allasc View Post
Properties of the numbers in this sequence

.....
4) If m = 0 (mod p), then n = (2^t +1) / (2^p +1), ie x = 1
5) If the m-even number and (p = m-2), then n = (2^t +1) / (2^p+2^p-2^(m/2) +1) ie x = 2^p-2^(m/2)+1
6) If the m+3=2*p, then n = (2^t +1) / (2^p+2^(p-1) +3); ie x = 2^(p-1)+3

still believe that there is a universal rule x = F (m, p),
I will seek...

Last fiddled with by allasc on 2011-01-26 at 09:22
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Old 2011-01-26, 09:45   #8
cmd
 
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allask ... allrun ... allarm ... ( we must be serious sometimes )

Last fiddled with by cmd on 2011-01-26 at 09:58 Reason: |_|
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Old 2011-01-26, 10:25   #9
allasc
 
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Quote:
Originally Posted by allasc View Post
All elements of this sequence can be represented as
n=(2^t+1) / (2^p + x)
x is an integer



6) If the m+3=2*p, then n = (2^t +1) / (2^p+2^(p-1) +3); ie x = 2^(p-1)+3

still believe that there is a universal rule x = F (m, p),
I will seek...
sorry mistake
6) If the (m+3=2*p) and m=0(mod 3), then n = (2^t +1) / (2^p+2^(p-1) +3); ie x = 2^(p-1)+3
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Old 2011-01-26, 14:31   #10
allasc
 
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Quote:
Originally Posted by allasc View Post
sorry mistake
6) If the (m+3=2*p) and m=0(mod 3), then n = (2^t +1) / (2^p+2^(p-1) +3); ie x = 2^(p-1)+3
sorry mistake
6) If the (m+3=2*p) and t=0(mod 3); (t=m+p), then n = (2^t +1) / (2^p+2^(p-1) +3); ie x = 2^(p-1)+3

Last fiddled with by allasc on 2011-01-26 at 14:35
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Old 2011-01-26, 18:43   #11
tichy
 
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I think this is a better way to read it.
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