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# User:Thomas Scheuerle

I am a software engineer working in development and research for signal-processing algorithms and trajectory computations, used by automotive driver assistance systems.

My interest in number sequences is mostly recreational. Some properties which make number sequences interesting to me are:

- Interesting graph, maybe self-similar or fractal.

- A permutation which has recurrence like formulas.

- sequences which have interesting properties regarding coding-theory, signal-processing or orthogonality.

- sequences which are defined by properties of binary expansions. (personally I regard base-two as more fundamental than all other bases.)

On the other hand there is nothing that appears completely uninteresting to me.

As I am not a professional mathematician, it is always possible that there may be a mathematical error in my opinions or contributions. Of course I hope this case is rare.

I am always open to any ideas and suggestions. Also I am never unhappy if some user uses the "Email this user" link for my account. So don't be shy.

The thing that amazes me most at OEIS: There is a team of people working all time around the clock making this possible. All the editors, administrators and many many more are doing incredible hard work, contributing something with big value for many users around the world, and this for free.

Todolist of minor additions to sequences.User:Thomas_Scheuerle/todolist

## Some random ideas

[[1]]

### Barycentric coordinates on hyper simplexes as a tool to understand the Collatz conjecture

I found that A014682 (The Collatz or 3x+1 function: a(n) = n/2 if n is even, otherwise (3n+1)/2) can be expressed by a sequence which is a permutation (A340615):

A014682(n) = A340615(n - floor(n/5)) for all n - floor(n/5) not congruent to 4 mod 5, else A014682(n) = 3*n/5 - 1. This allows us to compare A340615 and A014682.

Let us imagine a point P whose coordinates are expressed as barycentric coordinates regarding a simplex with infinitely many dimensions, as an infinite sequence 1:2:3:4.. . In short this point P shall have the set of positive integers greater than zero as coordinates. By using the permutation A340615 on the coordinates of P we can define infinitely many points P1 = A340615(P), P2 = A340615(P1), P3 = A340615(P2). All these points have in common that they will lie on the same hyper-ellipsoid, the so called permutation-ellipsoid of P.

If we do the same process and use A014682 as a map instead of the permutation A340615, Q1 = A014682(P), Q2 = A014682(Q1), Q3 = A014682(Q2).. , we will see a different result. All Q1..Qn will lie on individual permutation ellipsoids regarding each other and regarding P. This is because A014682 contains all numbers of the form 6*m - 2 twice.

All these permutation-ellipsoids (of P and Q1..n) are a dilation of the point G = 1:1:1:1.... and also dilations of each other.

More interesting, as A014682 is a union of A340615 and numbers of the form 6*m - 2, we know that the permutation-ellipsoids of Q1 to Qn are smaller than the permutation-ellipsoid of P and that of Q2 is smaller than Q1 and so forth until at infinity a Qn would be reached where not any coordinates are greater than 3. Because of the common centre G and the fact that all ellipsoids share shape and alignment of axis, not any of the ellipsoids of Q1..Qn will overlap when they have different sizes.

My personal opinion is that this is a hint regarding the probable truth of the Collatz conjecture.

Some further details regarding the "size" of the ellipsoids mentioned here:

There exists an hyper-ellipsoid defined by all corner points of our imagined simplex thus that these points all lie on it, the so called Steiner circum-ellipsoid. Based on this can we define the dilation-factor D. If D = 0 then our ellipsoid is contracted to the point G. If D = 1 then our ellipsoid would be on the Steiner circum-ellipsoid, but this will never be the case as we excluded the number zero from the coordinates. we introduce now t for more easy calculation defined by ${\displaystyle D={\sqrt {\frac {t-1}{t+2}}}}$ and t becomes now for a permutation ellipsoid based on a point P = k_1 : k_2 : .. : k_n regarding a simplex with n dimensions this expression: ${\displaystyle t={\frac {k_{1}^{n-1}+k_{2}^{n-1}+..+k_{n}^{n-1}}{\prod _{i=1}^{n-1}k_{i}+\prod _{i=2}^{n}k_{i}+k_{1}\prod _{i=3}^{n}k_{i}+..}}}$.

What does this mean if we try to apply this on A340615 ?

A340615 contains all positive integers > 0 exactly once. Let us now ignore the infinity of n here, then we could define A = 1^(n-1) + 2^(n-1) + 3^(n-1) +...+ n^(n-1) and B = Stir(r+(n-1),r),r = {1..n} ( Stir() means Stirling numbers of the first kind) and t = A/B.

And for A014682 ?

A014682 contains all numbers already included in A340615 plus duplicates of the form 6*m - 2. It can be shown that with each added duplicate to a permutation, t will become closer to 1. For example with only 3 coordinates (1^2+2^2+3^2)/(1*2+2*3+3*1) > (1^2+3^2+3^2)/(1*3+3*3+3*1). If t becomes closer to 1, the dilation-factor D will become closer to zero.

### partial conjugate of Collatz mapping

let a'(m) be the inverse permutation of A342842, such that a'(A342842(n)) = n. Let p = A006370^k(6*(A342842(n) + 1) - 2) and choose k such that p is of the form m*6 + 4, then a'((p + 2)/6 - 1) < n.

### Playing with geometric series and A014682

More details here in A343684.

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{A014682(n)}}}=1+{\frac {2}{7}}}$

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{A014682^{2}(n)}}}=1+{\frac {6}{7}}+{\frac {2}{511}}}$

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{A014682^{3}(n)}}}=1+{\frac {8}{7}}+{\frac {162}{511}}+{\frac {2}{134217727}}}$

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{A014682^{4}(n)}}}=1+{\frac {12}{7}}+{\frac {548}{511}}+{\frac {17538}{134217727}}+{\frac {2}{2^{(3^{4})}-1}}}$

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{A014682^{5}(n)}}}=1+{\frac {14}{7}}+{\frac {890}{511}}+{\frac {33704228}{134217727}}+{\frac {2^{61}+2^{55}+2^{41}+2^{10}+1}{2^{(3^{4})}-1}}+{\frac {2}{2^{(3^{5})}-1}}}$

${\displaystyle \sum _{n=1}^{\infty }{\frac {2^{A014682^{k}(n)\cdot A014682^{k+1}(n)}}{2^{(A014682^{k}(n)+A014682^{k+1}(n))^{2}}}}<\sum _{n=1}^{\infty }{\frac {2^{A014682^{k}(n)\cdot A014682^{k+j}(n)}}{2^{(A014682^{k}(n)+A014682^{k+j}(n))^{2}}}}<{\frac {1}{4}}\cdot \sum _{n=1}^{\infty }{\frac {1}{2^{A014682^{k+j}(n)^{2}}}}}$ , if j > 1 and A014682 is free of non-trivial cycles.

This evaluates to a hypergeometric series, this would allow to research into this deeper by computation with algorithms based on hypergeometric functions, so even results for complex k and j may then be explored.

### Constants from A014682

A014682^k(j*2^k+m) < A014682^k((j+n)*2^k+m) is true for all n > 0 and m in the range {1..2^k}.

${\displaystyle \sum _{n=1}^{2^{k}}A014682^{k}(n)+j\cdot 2^{2\cdot k}=\sum _{n=1}^{2^{k}}A014682^{k}(j\cdot (2^{k})+n)}$ for all j > 0.

This leads me to the idea to calculate a constant from A014682:

2.35369801 > ${\displaystyle 1+\sum _{k=1}^{\infty }{\frac {2^{k}}{\sum _{n=1}^{2^{k}}A014682^{k}(n)}}\gtrapprox 2.353697995\approx {\frac {\pi ^{2}}{2}}\cdot {\frac {207}{434}}}$ This is a very rough empirical approximation.

This constant is somehow a measure how fast the iteration of A014682 on any number will converge to 1. If it would be smaller than ${\displaystyle {\frac {\pi ^{2}}{6}}}$, the iteration of A014682 would not converge to 1 for any number.

If A014682 would converge to 1 for all numbers in the first iteration, we would have an infinite sum of ones. It appears that if this series would not converge, A014682 would converge to 1 for all numbers and all permutations of A014682. As this series is convergent only some permutations of A014682 may converge to 1 for all numbers under iteration.

### The distribution of numbers < 3 in A014682^k(1...2^k)

We know that all numbers in A014682^k(1...2^k) are of the form 3^m-n and we know that n is > 0 and < 3^m.

We also know for each k the distribution of the exponents m:

k = 1: m = {0,1}; k = 2: m = {0,1,1,2}; k = 3: m = {0,1,1,2,1,2,2,3}; k = 4: m = {0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4};... (Original A343422 was recycled after some time.User:Thomas_Scheuerle/Archived_A343422).

We know that at least for each exponent 0 and 1 we will get a number < 3 in the sequence, thus we know a lower limit for the overall count here:

Minimum count of numbers < 3 for each k in A014682^k(1...2^k): 1+k.

In the previous example we considered only numbers < 3 generated by 3^1. For other exponents there are some pattern to be found. The count of numbers < 3 generated by the terms of 3^2 are for k = 1..6: 0,0,1,3,5,7. The full sequence will be found in OEIS as A268292(k+4). Thus if we consider 3^1 and 3^2 together our minimum count of numbers < 3 for each k in A014682^k(1...2^k) will become: 1+k+A268292(k+4).

If we consider 3^3, for k = {1..20} we get this yet unknown sequence: 0,0,0,0,1,3,5,8,12,17,24,32,40,49,60,73,88,105,124,145.

If we consider 3^4, for k = {1..20} we get this yet unknown sequence: 0,0,0,0,0,0,1,3,6,11,17,24,34,48,65,85,109,138,172,212.

If we consider all exponents together we get another unknown sequence. Count of numbers < 3 for each k in A014682^k(1...2^k) for k = {1..20}: 2,3,5,8,12,17,23,31,43,61,85,116,155,205,273,364,484,643,854,1136. (grows slightly faster than ceiling((2*n^3 + 3*n^2 + 37*n)/66)).

The more interesting question is: Is there a minimum for the amount of numbers < 3 which appear in an uninterrupted sequence from n = 1 to n = ?

Here a computation result for k = 1..25. It shows the (count of uninterrupted consecutive numbers A014682^k(1...y<=2^k) < 3)-k over k :

Collatz conjecture would be known as true if it could be proven that this curve will always leave the negative domain for some greater value of k.

### Calculating an upper limit on number of 3x+1 steps in Collatz conjecture

${\displaystyle Upperlimit=A014682^{\lceil {log_{2}(n)}\rceil }(n+2^{\lceil {log_{2}(n)}\rceil })-A014682^{\lceil {log_{2}(n)}\rceil }(n)}$

If this Upperlimit is summed over an interval n = {1..2^k}, then this sum will become 2^2k.

### A system of involutions which are residue-class-wise affine at the same time

A general algorithm to generate self inverse permutations of integers > 0 :

a(n) = n for n < s, for all n not in a(1..n-1) a(n) = M(n) else a(n) = M'(n). M'(M(n)) = n.

If we define the function M(n) to be of the form k*n - j*(k-1) and s = j, we will generate involutions which are residue-class-wise affine. Some examples:

k = 2; j = 0; a(1...10): 2,  1,  6,  8, 10,  3, 14,  4, 18,  5, 22, 24, 26,  7, 30, 32, 34,  9, 38, 40  A073675
k = 2; j = 1; a(1...10): 1,  3,  2,  7,  9, 11,  4, 15,  5, 19,  6, 23, 25, 27,  8, 31, 33, 35, 10, 39  A118966
k = 3; j = 0; a(1...10): 3,  6,  1, 12, 15, 18, 21, 24, 27, 30, 33,  4, 39, 42,  5, 48, 51,  6, 57, 60
k = 3; j = 1; a(1...10): 1,  4,  7,  2, 13, 16,  3, 22, 25, 28, 31, 34,  5, 40, 43,  6, 49, 52, 55, 58
k = 3; j = 2; a(1...10): 1,  2,  5,  8,  3, 14, 17,  4, 23, 26, 29, 32, 35,  6, 41, 44,  7, 50, 53, 56


It seems for j only values up to k-1 are necessary to get a interesting system.

Let B be a permutation such that it is a finite product of some of these involutions with some k's and j's. Let C be any arbitrary permutation chosen of the residue-class-wise affine group.

Can we always find a B such that C(n+r) = p+B(n+q) for some constants p,q,r ? Is this also true if C is any permutation which is also a linear recurrences with constant coefficients ?