## 26.3.17

### Other Collatz-like Algorithms

I wrote about the Collatz conjecture in relation to finite fields and I came across a few interesting sequences but I didn't discuss the possibility of other Collatz-like algorithms that always reach 1.

Once again, the Collatz problem is defined as follows:

Collatz Algorithm: Given any positive integer n, let a0 = n. For i > 0, let:

ai = ai-1/2 if ai-1 is even
ai = 3*ai-1 + 1 if ai-1 is odd

Conjecture: Given any positive integer, the above recurrence relation returns 1.

This conjecture seems to hold although it hasn't been proven yet.

While looking at a different problem related to generating powers of 2 backwards, I stumbled upon a Collatz-like algorithm, which I generalize below for all positive integers.

Collatz-like Algorithm #1: Given any positive integer n, let a0 = (n+1)/2 if n is odd or a0 = n/2 if n is even. For i > 0, let:

ai = ai-1/2 if ai-1 is even
ai = (ai-1 - 1)/2 +  a0 if ai-1 is odd

Conjecture: Given any positive integer n, the above recurrence relation returns 1, and all integers it reaches before it reaches 1 are strictly less than n

In fact, I claim that:

Claim: If n is even then the Collatz-like Algorithm #1 generates all powers of 2 mod (n-1) in reverse order and if n is odd then it generates all powers of 2 mod n in reverse order.

Example: let n = 36 so a0 = 18
Then:
a1 = 18/2 = 9
a2 = ((9-1)/2) + 18 = 22
a3= 22/2 = 11
a4 = ((11-1)/2) + 18 = 23
a5 = ((23-1)/2) + 18 = 29
a6 = ((29-1)/2) + 18 = 32
a7 = 32/2 = 16
a8 = 16//2 = 8
a9 = 8/2 = 4
a10 = 4/2 = 2
a11 = 2/2 = 1

## 16.3.17

### Ghetto Superstar

Last year I visited the world's oldest ghetto, namely the Venetian Ghetto Nuovo. It is the first place in the world where the word ghetto was used to describe a poor segregated community.

The Venetian Ghetto is a part of Venice, Italy where gondolas don't go. That's what I was told when I approached a gondolier asking for directions.

When both Apple maps and Google maps failed to provide a coherent path to my desired destination as well, I had to resort to ancient technology: ask a local who is not in the transportation industry.

A store clerk from a local souvenir shop kindly provided me with the following map, and I somehow managed to find my way.

Venice is by far the most surreal place that I had the opportunity to visit.
 An emergency vehicle
Many parts of the city are only accessible by boat, and only emergency service boats are allowed to speed.

According to my slightly cryptic but nevertheless useful map, I had to go to the Grand Canal first and take a waterbus in the opposite direction from where I was initially headed to until reaching the edge of the city.

 A waterbus stop near San Marco Square

The Grand Canal runs through the entire length of the Veneto region, which is composed of 117 small islands connected by bridges. The Venetian Ghetto is one of these islands.

 Rialto Bridge, Venice

I got off near (what looked like) the edge of the city and I went down a small street, then up another small street, and then I somehow ended up on one of the bridges connecting the Venetian Ghetto with the rest of Venice.

At the other side of the bridge there is a tunnel that leads to the main square. The Venetian ghetto was once home to thousands of Sephardic and Ashkenazi Jews, mostly refugees from other parts of Europe where they were unwanted at one point or another.

 Venetian Ghetto Nuovo

Although there was a Jewish presence in Venice for centuries prior to the establishment of the Venetian Ghetto, they were not segregated until Venice saw an influx of Jews from Spain after the Alhambra Decree and from other parts of Europe with similar sentiments at the time.

For many of the Sephardic Jews fleeing Spain, Venice was merely a pit stop before reaching the Ottoman Empire.

The Venetian Republic began rounding up and confining Jews to a small island on the edge of the Veneto region on the 29th of March 1516. The island was home to ancient foundries.

There are several synagogues in the ghetto tucked away in common areas of unassuming buildings, and some of them are still in use today. All of them honoured the Venetian Republic in some way with their decor in addition to honouring Ashkenazi, Sephardic, and Italian Jewish tradition.

 Sephardic Synagogue in the Venetian Ghetto: red and gold were the official colours of the Venetian Republic

The word ghetto as we know it today comes from an old Italian word for foundry pronounced as jetto (geto with a soft g as in jet). Apparently, the etymology of the word is a controversial topic but the most plausible theory is that Ashkenazi Jews had trouble pronouncing the soft g so they transformed it into a hard g effectively coining the word ghetto.

Life was not easy in the shanty part of town. Just like in the rest of the world at the time, Jewish people in the Venetian ghetto were only allowed to do the jobs that nobody else liked doing.

They were also not allowed to leave the ghetto at night and they had to wear identifiers that they were Jewish. Living quarters were cramped: large stories inside the buildings were separated into smaller ones to accommodate more people. Entire families were crammed inside single small rooms with low ceilings and sometimes no windows.

 The view above the main plaza across from Museo Ebraico di Venezia where I learned about the history of the Venetian Ghetto

Although people who spent time in the ghetto endured many hardships, some of them and many of their descendants thrived and prospered in different parts of the world.

Meanwhile in Spain 500 years after a bill was passed to expel all Jews from the Iberian peninsula, a new bill was introduced to grant citizenship to the descendants of those who were expelled. According to a DNA test I did, it is possible that I'm one such descendant.

## 23.2.17

### Generating Powers of 2 Again

While looking into proving the algorithm I described here, I stumbled upon yet another algorithm for generating powers of 2.

This new algorithm is based on the following recurrence relation:

a1 = (n+1)/2        given an odd n
ai = ai-1/2              if ai-1 is even
ai = (ai-1-1)/2 + a1 if ai-1 is odd

I claim that the recurrence relation above generates all powers of (n+1)/2 mod n in consecutive order and all powers of 2 mod n in reverse order.

This follows from the first part of the algorithm that I previously described, namely the claim below:

Claim: Given an odd integer n and k = (n+1)/2, then all consecutive multiples of k mod n, namely 1*k mod n, 2*k mod n, 3*k mod n,..., (n-2)*k, (n-1)*k, are equal to k, 1,  k + 1, 2,  k + 2,  3, k + 3, .... , n-1,  k - 1 respectively.

Proof:
Since n is an odd integer then Zn is a commutative ring closed under multiplication and addition with an identity element and so the following algebraic expressions hold.

Since k = (n+1)/2, then 2*k mod n = 2(n+1)/2 mod n = n+1 mod n and since n mod n = 0 then
2*k mod n = 1 mod n.

Similarly for all even multiples x of k: x*k = x/2.

On the other hand, 3*k mod n = ((n+1)/2  + (n+1)/2  + (n+1)/2) mod n
= (2(n+1)/2 + (n+1)/2) mod n
= 2(n+1)/2 mod n + (n+1)/2 mod n
= 1 mod n + (n+1)/2 mod n
= (1 +  (n+1)/2) mod n.

Similarly, for all odd multiples y of k: y*k = (n+1)/2 + (y - 1)/2
.'.

The index of each term in the sequence k, 1,  k + 1, 2,  k + 2,  3, k + 3, .... , n-1,  k - 1 , which is essentially the ((n+1)/2)th row of the multiplication table of Zn, reveals a few ways of finding powers of 2 mod n.

My previous algorithm for generating consecutive powers of 2 mod n in reverse order required the explicit generation of this particular row but the recurrence relation for the new algorithm that I described in the beginning of this post does not.

I implemented the new algorithm in Javascript below for reference.

``````/* -------------------------------------------------
This content is released under the GNU License
http://www.gnu.org/copyleft/gpl.html
Author: Marina Ibrishimova
Version: 1.0
Purpose: Find all powers of 2 and (n+1)/2 mod n
---------------------------------------------------- */
//generate powers of 2 from highest to lowest exponent
function powers_of_two_backwards_again(n)
{
var k = (n+1)/2;
var halfie = k;
var powers = new Array();
do{
powers.push(k);
if (k%2 == 0)
{k = k/2;}
else
{k = ((k-1)/2)+halfie;}
}while(k != 1)
return powers;
}
``````

And embedded bellow is a demo.

This algorithm is reminiscent of the algorithm behind the Collatz conjecture.

## 15.2.17

### I Can Do Proofs

In my last entry I made a typo when describing a sequence using two different formulas.

The first formula is 2i + (2i-1 - 3) if i is even or 2i + (2i-1 + 3) if i is odd for all i > 2
The second formula is 3*(2i-1 - 1) if i is even or 3*(2i-1 + 1) if i is odd for all i > 2

I claim that these two formulas are equivalent.

It may look counterintuitive that if i is even then 2i + (2i-1 - 3) = 3*(2i-1 - 1) and that if i is odd then 2i + (2i-1 + 3) = 3*(2i-1 + 1) but I can actually prove it.

Below I provide a proof using mathematical induction to show off my mad skillz.

Claim: For all i > 2, if i is even then 2i + (2i-1 - 3) = 3*(2i-1 - 1) and if i is odd then 2i + (2i-1 + 3) = 3*(2i-1 + 1).

Proof by weak mathematical induction:

Base Case: i = 3

23 + (23-1 + 3) = 8 + (4 + 3) = 15
3*(23-1 + 1) = 3*(4+1) = 15            Clearly, base case holds

Inductive Step:

Suppose that 2i + (2i-1 + 3) = 3*(2i-1 + 1) is true for an odd i, must show that the claim is also true for the next odd i, namely i + 2.

Since 2i + (2i-1 + 3) = 3*(2i-1 + 1) then 2i  3*(2i-1 + 1) - (2i-1 + 3)

Replacing i with i+2 yields

2i + 2 = 3*(2i +2 - 1 +  1) - ( 2i + 2 - 1 + 3)
2i + 2 = 3*2i+1  + 3 -  2i+1  - 3
2i + 2 = 3*2i+1  + 0 -  2i+1
2i + 2 = 2*2i+1   = 2i + 2

Similarly, suppose that 2i + (2i-1 - 3) = 3*(2i-1 - 1) is true for an even i, must show that the claim is also true for the next even i, namely i + 2

Replacing i with i+2 yields

2i + 2 = 3*(2i +2 - 1 -  1) - ( 2i + 2 - 1 - 3)
2i + 2 = 3*2i+1  - 3 -  2i+1  + 3
2i + 2 = 3*2i+1  - 0 -  2i+1
2i + 2 = 2*2i+1   = 2i + 2
.'.

## 11.2.17

### A Shrewd Sequence

In previous entries I categorized integers n between 2i and 2i+1 where the order of 2 mod n has a common formula for each i.

For example:
1. I proved that if n is a Mersenne number (an integer of the form 2i - 1), then the order of 2 mod n is equal to i.

2. Then I conjectured that if n is of the form 2+ 1 then the order of 2 mod n is equal to 2*i.

3. I also conjectured that if n is of the form:

2i + (2(i-1) - 3) if i is even
or 2i + (2i-1 + 3) if i is odd

or in other words of the form (here's the proof that these two formulas are equal)

3*(2i-1 - 1) if i is even
3*(2i-1 + 1) if i is odd

then the order of 2 mod n is i + (i - 2)

Another example that I have not previously published on this blog is also based on just a conjecture that may turn out to be false.

Conjecture: For each i > 3 there exists at least one integer n such that 2< n < 2i+1 and the order of 2 mod n is equal to (i-1)*(i-2)

Below is a sequence generated using the first occurrence of such integers between 2i and 2i+1 for each i > 3

21, 35, 75, 231, 301, 731, 1241, 2079, 7513, 8337, 16485, 39173, 66591, 131241, 371365, 539973, 1125441, 2153525,...

Below is a slow calculator for finding the order of 2 mod n for any odd integer n:

## 19.1.17

### Wiener Werkstaette Kunstlerinnen

Another city I visited in 2016 with the intention of checking out a modern art exhibit was Vienna, Austria.

I went to see "Das bessere Haelfte: Judische Kunstlerinen bis 1938", which celebrates the contribution of Jewish women to the development of Viennese modernism.

The exhibit blends together the works of well established Viennese artists with the works of the unfairly forgotten ones.

Contrary to popular belief at the time, these artists came from different backgrounds, and were not just "daughters of high dignitaries or products of wealthy households" as Adolf Loos called the women of Wiener Werkstaette in 1927.

One of the artist featured in the exhibit is Friedl Dicker, who was born in Vienna in 1898 into a poor Jewish family. In 1915 Dicker joined the textile department of the School of Art and Crafts in Vienna.
In 1919 she went to Weimar Bauhaus to study. She returned to Vienna four years later where she established a furniture atelier, which received awards at several exhibitions.

The photo below depicts a reproduction based on a collage by Friedl Dicker entitled "The Bourgeoisie Becomes Fascist" originally made in 1932.

Everyone can be an artist but not every artist can be a visionary. Friedl Dicker was a visionary.

In the above piece she managed to capture the spirit of the problem, which 10 years later nearly destroyed Europe, like no other artist before or after her.

Modern art played a central role in World War II. The Nazis felt threatened by it to such an extent that they organized a propaganda exhibit where popular modernist works were derided. Psychologists such as Carl Jung,  who enjoyed a blossoming career during the Nazi regime, and whose pseudo-scientific ramblings still corrode the popular Western culture today, saw modern art as a symbol of "corrosive character".  According to letters written by Jung during the Nazi regime in Germany, he was a Nazi sympathizer, yet he denied being a Nazi sympathizer after Nazi Germany lost WWII.

Meanwhile, Friedl Dicker had the option to flee Nazi Europe shortly before she was sent to a concentration camp but she chose to stay with those who couldn't go anywhere and tried to lift their spirits with art workshops where they drew flowers and landscapes to escape their gruesome reality. In 1944 she was gassed in Auschwitz along with some of her students.

## 3.1.17

### Multiple Positions

Previously, I generated a sequence using the following recurrence relation and I made a few conjectures about it.

2i + (2(i-1) - 3) if i is even
or 2i + (2i-1 + 3) if i is odd

I also wrote an algorithm based on this recurrence relation to generate the first few such integers up to some limit. The first 20 terms of the sequence with initial term i = 3 are:

Obviously, each term in this sequence is divisible by 1 or more primes.

Claim: If a term in this sequence at position b is divisible by a prime p then the term at position
b + (p-1)  is divisible by p.

This claim holds for all other recurrence relations of the form

2i + (2(i-1) - k) if i is even
or 2i + (2i-1 + k) if i is odd

where k is any odd integer

It is easy to see it because it is a result of the fundamental theorem of algebra in that every integer can be represented as a product of primes.

For example, the first 20 terms of the sequence generated with the recurrence relation

2i + (2(i-1) - 7) if i is even
or 2i + (2i-1 + 7) if i is odd

are:

The first term 17 is a prime at position 0, and the term at position 0 + (17 - 1) = 16 is divisible by 17.
In fact 1572857/17 = 92521.

Similarly, 55 = 11* 5 is at position 1, so the term in the sequence at position 1 + (11 - 1) = 11 is divisible by 11 and the term at 1 + (5 - 1) = 5 is divisible by 5.

I'm willing to bet that the 90th term in the sequence is divisible by 89 and the 201st term is divisible by 199.

377 = 13*29 is at position 4 in the sequence and the term in the sequence that's at position 4+(13-1) = 16 is 1572857, which is divisible by 13. In fact, the term at position 4+(29-1) = 32 is 103079215097 and it is divisible by 29 among other integers.

And so on.

One of the claims I made for the sequence where k = 3 is that every single element in it is divisible by 3. It is also easy to see why this is so using the claim above since the first 2 terms of the sequence at position 0 and 1 respectively are divisible by 3. Namely, if the initial i = 3 then clearly the first term of the sequence at position 0 is 23 + (22 - 3) = 15, which is divisible by 3 so the term at position 0+(3-1) = 2 will also be divisible by 3. Therefore, every second term will be divisible by 3. To calculate the term at position 1, I plug in i = 4 so 24+ (23 - 3) = 21, which is also divisible by 3 so every third element of the sequence will also be divisible by 3.

Similarly, if k is equal to any other odd multiple of 3, then the sequence generated using the relation

2i + (2(i-1) - k) if i is even
or 2i + (2i-1 + k) if i is odd

is composed of terms such that every term is divisible by 3.

I also made another claim related to the sequence when k = 3. I claimed that for each term n where

n = 2i + (2(i-1) - 3) if i is even
or n = 2i + (2i-1 + 3) if i is odd

the order of 2 mod n is i+(i-2)

I haven't been able to see why this is the case yet but it holds for the first 50 terms of the sequence, and beyond as far as I can see.

Below is a graphic I made of different sequences generated for different k with initial i = 4.

The interesting thing about this is that only the sequence for k = 1 exists in the OEIS.org database of sequences but it is generated using a different relation, and the sequence itself is an amalgamation of sequences concerning powers of 3.

Neither of the other sequences I presented here have found their way into the OEIS.org database of integer sequences.

## 31.12.16

### 2016 Sequences

I came up with 7 different sequences in 2016 and I thought I'd put them all in one place as some of them are slightly related to one another.

1. The first sequence is the sequence generated by powers of 2 mod n when n is the product of 2 distinct safe primes. The sequence was published in the OEIS.org with sequence number is A269453. Here are the first few terms of the sequence:

12, 20, 30, 44, 33, 92, 110, 116, 69, 174, 164, 230, 212, 246, 290, 318, 332, 356, 410, 253, 452, 249, 530, 534, 524, 638, 678, 692, 716, 830, 393, 902, 764, 890, 932, 956, 1038, 1166, 1130, 537, 1004, 573, 1334, 1124, 1310, 1172, 1398, 717, 753, 1436, 1730, 913, 1886, 1686, 1790

2. Then came the the sequence of the products of two distinct safe primes, which is now published in the OEIS.org under sequence number A269452

 24, 40, 60, 88, 132, 184, 220, 232, 276, 348, 328, 460, 424, 492, 580, 636, 664, 712, 820, 1012, 904, 996, 1060, 1068, 1048, 1276, 1356, 1384, 1432, 1660, 1572, 1804, 1528, 1780, 1864, 1912, 2076, 2332, 2260, 2148, 2008, 2292, 2668, 2248, 2620, 2344, 2796, 2868, 3012, 2872, 3460, 3652, 3772, 3372 3. Looking at these 2 sequences made me discover a third sequence, which I dedicated to my great grandmother Yona because at the time I thought it was the most important result I'd come up with. This sequence consists of safe primes not congruent to -1 mod 8 and it is now published with a sequence number A269454. 5, 11, 59, 83, 107, 179, 227, 347, 467, 563, 587, 1019, 1187, 1283, 1307, 1523, 1619, 1907, 2027, 2099, 2459, 2579, 2819, 2963, 3203, 3467, 3779, 3803, 3947, 4139, 4259, 4283, 4547, 4787, 5099, 5387, 5483, 5507, 5939, 6659, 6779, 6827, 6899, 7187, 7523 The next 3 sequences I came up with involve the Collatz conjecture, which is an open problem in Mathematics. 2 of these sequence were published and the last one was denied and so it remains unpublished because it seemed "artificial" 4. The first sequence of the Collatz trio is the sequence of Collatz primes(A276260), and it only has 9 terms. If a 10th term exists, it would have to be really really large. 5, 13, 17, 53, 61, 107, 251, 283, 1367 5. The second sequence is the sequence of Collatz products, and it has many more terms. It is published under the sequence number A276290 . 25, 35, 55, 65, 77, 85, 95, 115, 133, 143, 145, 155, 161, 185, 203, 205, 209, 215, 217, 235, 253, 259, 265, 287, 295, 305, 329, 341, 355, 365, 371, 391, 395, 403, 407, 415, 427, 437, 445 6. The third sequence of the Collatz trio was never published and here I explain it in some detail.  The first few terms are: 5,27,21,107,85,427,341,1707,1365,6827, 5461
7. The last sequence that I came up with in 2016 is the sequence I described here. I haven't attempted to publish this sequence anywhere yet. The first few terms are:

15, 21, 51, 93, 195, 381, 771, 1533, 3075, 6141

2016 has been the year of discovery for me. I discovered many different results, some of which I even managed to prove. Also in 2016 I learned not to be intimidated by sharing my results with the public.

## 30.12.16

### Just another sequence

In my previous post I talked about the order of 2 mod n where n is an integer such that either n = 2^i + (2^(i-1) - 3) when i is even or n = 2^i + (2^(i-1) + 3) when i is odd.

I conjectured that the order of 2 mod n for such n is i + (i - 2) regardless of whether i is even or odd.

The recurrence relation for generating all such n up to a limit is :

2i + (2(i-1) - 3) if i is even
or 2i + (2i-1 + 3) if i is odd

For the fun of it, I wrote a simple script to generate the first few such integers up to some limit.

``````/* -------------------------------------------------
This content is released under the GNU License
http://www.gnu.org/copyleft/gpl.html
Author: Marina Ibrishimova
Version: 1.0
Purpose: Generate the 3sequence

---------------------------------------------------- */
function gen_seq(i, limit, a)
{
var j = i - 1;
var n = Math.pow(2,j);
var m = Math.pow(2,i);
if (i < limit)
{
if (i % 2 != 0)
{
n = n + 3;
}
else
{
n = n - 3;
}
a.push(m+n);
i = i + 1;
gen_seq(i, limit, a)
}
return a;
}``````

Here's the output for limit = 12:

15, 21, 51, 93, 195, 381, 771, 1533, 3075, 6141

Obviously, there is another way to generate this sequence than the one I use in my script. Each term in the sequence can also be generated by multiplying 3 and 2^i - 1 if i is odd or 3 and 2^i + 1 if i is even.

3*5 = 15 where 5 = 2^2 + 1
3*7 = 21 where 7 = 2^3 - 1
3*17 = 51 where 17 = 2^4 + 1
3*31 = 93 where 31 = 2^5 - 1

and so on.

## 28.12.16

### Between 2 i's

Previously I talked about the order of 2 mod n when n is either equal to 2^i + 1 or 2^i - 1

How about the order of 2 mod n for other integers n, which are between 2^i and 2^(i+1)?