While reading about the Goldbach conjecture recently, I thought of Z_{35}.
A few years ago I conjectured that there are 2 integers a and b such that a^{2} = 1 mod n and b^{2} = 1 mod n where a + b = n whenever n is the product of two distinct prime integers greater than 2.
According to the Goldbach Conjecture every even integer greater than 2 can be represented as the sum of two prime integers.
In this respect, Z_{35} is very special because:
35 = 5*7
29^{2} = 1 mod 35
and 29+7 = 36 = 1 mod 35
Furthermore,
29+5 = 34 mod 35
34^{2} = 1 mod 35
Unfortunately, this is not applicable to all products of primes but it is interesting to see where it pops up.
27.9.19
19.3.19
The skewy revisited
A few years ago I boldly introduced a new term called "skewy", which I defined as the smallest square root of (n+1/2)^{2} mod n where n is the product of two distinct odd primes.
The skewy has an interesting relation to the Collatz conjecture and to the prime factorization of n.
Here are a few claims to consider:
Claim #1: 4*((n+1)/2)^{2} = 1 mod n
Claim #2: The row at 4 of the exponentiation table of Z_{n} consists of all squares co-prime with n.
Claim #4: If s is the skewy, then 4*(s^{3} mod n) mod n = s.
Claim #5: 4*(((n+1)/2)^{3} mod n) mod n = (n+1)/2.
The skewy has an interesting relation to the Collatz conjecture and to the prime factorization of n.
Here are a few claims to consider:
Claim #1: 4*((n+1)/2)^{2} = 1 mod n
Claim #2: The row at 4 of the exponentiation table of Z_{n} consists of all squares co-prime with n.
Claim #4: If s is the skewy, then 4*(s^{3} mod n) mod n = s.
Claim #5: 4*(((n+1)/2)^{3} mod n) mod n = (n+1)/2.
Labels:
Collatz Conjecture
,
Mathematics
,
skewy
26.1.19
The Collatz Connection
There is one integer sequence, which is keeping me up at night sometimes if I start thinking about it. It is the sequence of odd integers, which do not have 5 as a Collatz exit point. Why don't they? What is so special about them? But most importantly, how can we predict the next term of this sequence? What is the formula? This sequence is not in OEIS. The first few terms that I found are:
21,75,85,113,151,201
Not found in OEIS |
On a personal note, I just learned today that the mathematician who first thought of the Collatz problem died while attending a Mathematics Conference in the city I was born in.
Labels:
Collatz Conjecture
,
Mathematics
19.1.19
Dreaming of Collatz Exit Points
A few years ago I talked about Collatz exit points. Some Collatz exit points are reached a lot more frequently than others. In fact, it appears that most integers reach only one such point.
Here are the first few Collatz exit points: 5, 21, 85, 341, 1365, 5461, 21845, 87381, 349525, 1398101, 5592405, 22369621, 89478485, 357913941, 1431655765, 5726623061, 22906492245, 91625968981, 366503875925, 1466015503701, 5864062014805, 23456248059221, 93824992236885, 375299968947541
Note: I call them Collatz exit points because for each a_{i} in the set above 3*a_{i} + 1 = 2^{i+1}
As I already mentioned here there are only 3 odd integers below 100, for which 5 is not a Collatz exit. Below 200 there are also only 3, below 300 there are also only 3.
It seems 5 is the most commonly occurring Collatz exit point for all odd integers below 1000.
What about below 10000?
What about below 100000?
...
Here are the first few Collatz exit points: 5, 21, 85, 341, 1365, 5461, 21845, 87381, 349525, 1398101, 5592405, 22369621, 89478485, 357913941, 1431655765, 5726623061, 22906492245, 91625968981, 366503875925, 1466015503701, 5864062014805, 23456248059221, 93824992236885, 375299968947541
Note: I call them Collatz exit points because for each a_{i} in the set above 3*a_{i} + 1 = 2^{i+1}
As I already mentioned here there are only 3 odd integers below 100, for which 5 is not a Collatz exit. Below 200 there are also only 3, below 300 there are also only 3.
It seems 5 is the most commonly occurring Collatz exit point for all odd integers below 1000.
What about below 10000?
What about below 100000?
...
Labels:
Collatz Conjecture
,
Mathematics
12.1.19
Non-Complimentary Primes
On the other side of complimentary primes are the non-complimentary primes, which are primes p for which the order of 2 mod p is odd.
The first few primes are 2,7,23,31,47,71,73,79,89,103,127,151 and more can be generated here using the following Mathematica line
Select[Prime[Range[800]], OddQ[MultiplicativeOrder[2, #/(2^IntegerExponent[ #, 2])]]&]
In OEIS these primes are known as primes p that do not divide 2^x+1 for any x>=1. What's interesting to me about these primes is that their exponentiation table follows a very different pattern than that of complimentary primes, and it all seems to be dictated by the order of 2 mod p.
The first few primes are 2,7,23,31,47,71,73,79,89,103,127,151 and more can be generated here using the following Mathematica line
Select[Prime[Range[800]], OddQ[MultiplicativeOrder[2, #/(2^IntegerExponent[ #, 2])]]&]
In OEIS these primes are known as primes p that do not divide 2^x+1 for any x>=1. What's interesting to me about these primes is that their exponentiation table follows a very different pattern than that of complimentary primes, and it all seems to be dictated by the order of 2 mod p.
Labels:
Mathematics
5.1.19
Complimentary Primes
Definition: Complimentary primes are prime integers p such that 2^{k/2} mod p = p-1 where k is the order of 2 mod p.
Example: Let p = 43. It can be verified that k = 14 since 2^{14} mod 43 = 1 and 14 is the smallest such integer so 14 is the order of 2 mod 43. Since 2^{7} mod 43 = 42 then 43 is a complimentary prime.
The first few complimentary primes are 5,11,13,17,19,29,37,41,43,53,59,61,67,83,97,101.
This is a sequence documented in oeis.org (more or less) as the sequence of primes p where the order of 2 mod p is even. What isn't documented there and what I find special about complimentary primes is the following conjecture that I made:
Conjecture: If p is a complimentary prime and k is the order of 2 mod p then
Example: Let p = 43. Since k is the order of 2 and k = 14, then k/2 = 7
2^{0} mod 43 = 1 and 2^{7} mod 43 = 42,
2^{1} mod 43 = 2 and 2^{8} mod 43 = 41,
2^{2} mod 43 = 4 and 2^{9} mod 43 = 39
2^{3} mod 43 = 8 and 2^{10} mod 43 = 35,
2^{4} mod 43 = 16 and 2^{11} mod 43 = 27
2^{5} mod 43 = 32 and 2^{12} mod 43 = 11,
2^{6} mod 43 = 21 and 2^{13} mod 43 = 22
2^{7} mod 43 = 42 and 2^{14} mod 43 = 1
You can use the iFrame below to find powers of 2 modulo an odd integer.
Example: Let p = 43. It can be verified that k = 14 since 2^{14} mod 43 = 1 and 14 is the smallest such integer so 14 is the order of 2 mod 43. Since 2^{7} mod 43 = 42 then 43 is a complimentary prime.
Verified result |
The first few complimentary primes are 5,11,13,17,19,29,37,41,43,53,59,61,67,83,97,101.
This is a sequence documented in oeis.org (more or less) as the sequence of primes p where the order of 2 mod p is even. What isn't documented there and what I find special about complimentary primes is the following conjecture that I made:
Conjecture: If p is a complimentary prime and k is the order of 2 mod p then
(2^{i} (mod p)) + (2^{(k/2)+i}(mod p)) = p for all i greater or equal to 0
Example: Let p = 43. Since k is the order of 2 and k = 14, then k/2 = 7
2^{0} mod 43 = 1 and 2^{7} mod 43 = 42,
2^{1} mod 43 = 2 and 2^{8} mod 43 = 41,
2^{2} mod 43 = 4 and 2^{9} mod 43 = 39
2^{3} mod 43 = 8 and 2^{10} mod 43 = 35,
2^{4} mod 43 = 16 and 2^{11} mod 43 = 27
2^{5} mod 43 = 32 and 2^{12} mod 43 = 11,
2^{6} mod 43 = 21 and 2^{13} mod 43 = 22
2^{7} mod 43 = 42 and 2^{14} mod 43 = 1
Labels:
Mathematics
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