The Logical Solution Syracuse Conjecture

By Rolando Zucchini

The first time I faced Syracuse conjecture I thought it was easier to start from any number n and to arrive to ni < n rather than fall down to 1. In this way I could not take into consideration even numbers, because if n is even then n → n/2 < n. So I had only to examine the odd numbers. In addition to the proof proposed by me, I have discovered many properties and peculiarities of this famous conjecture. It hides the magical harmony of odd numbers, and may be a type of law on the expansion of Cosmos based on the power of 2, as prophesied by Plato in some of his writings. So my work takes on a popular and didactic value of this marvelous conjecture. In this paper I have only used arithmetic and elementary number theory, but, in spite of its simple enunciation, Syracuse Conjecture is a difficult topic, therefore this article needs a lot of patience in reading for a well-understanding. I have considered that various applications and examples were needed for better explain my work.

• Language: English
• Publisher: Mnamon
• Release date: Dec 13, 2023
• ISBN: 9788869495885

Rolando Zucchini

Born in Foligno (Umbria – Italy) on June 6, 1947. Degree in Mathematics (1972) at the University of Perugia, with a thesis on non-Euclidean geometry. He taught mathematics in high schools with innovative teaching methods, linking it to its history and philosophy.

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FIRST PART

The conjecture of Syracuse

The conjecture of Syracuse states: If at any natural integer n, non- zero, we apply the algorithm 3n+1 if n is odd, n/2 if n is even, the sequence of the values obtained, precipitates to 1 after a finite number of steps, always in compliance with the final cycle {4; 2; 1}.

Chosen n = 12 (even), applying the algorithm of Collatz we obtain the following sequence: n = 12 → S(12) = {12; 6; 3; 10; 5; 16; 8; 4; 2; 1}

the number 12 falls down to 1 in nine steps.

The sequence (or succession) generated by the number 12 is oscillating (neither increasing nor decreasing) and can be represented with a graph on a Cartesian plane having its origin (0,1). The number 1 can be understood as the minimum level or the sea level. The steps are shown on the horizontal axis and the generating number on the vertical axis. The graph of fluctuation of the sequence S(12) (generated by the number 12) is represented in figure 1. In it the horizontal line parallel to the abscissa axis (steps) is the horizon of the number 12 (indicated with the Greek letter θ). The number 12, during the oscillation of the sequence S(12) by it generated, exceeds its horizon once, exactly in step 5, in correspondence of which reaches a peak 16.

fig1

Fig. 1

Chosen n = 7 (odd), applying the algorithm we obtain the sequence:

n = 7 → S(7) = {7; 22; 11; 34; 17; 52; 26; 13; 40; 20; 10; 5; 16; 8; 4; 2; 1}

the number 7 falls to 1 in 16 steps.

Figure 2 shows the graph regarding the oscillation of the sequence S(7).

fig2

Fig. 2

It is to be noted that if you choose as starting number an odd number, the first element of the sequence is always an even number, since the product of 3 for another odd number still gives an odd number, which becomes even with the addition of 1. This happens for all the successors of an odd number contained in the sequence. An even number, however, may be followed by an odd number or by one or more even numbers.

n = 48 → S(48) = {48; 24; 12; 6; 3; 10; 5; 16; 8; 4; 2; 1}

If you choose an even number power of 2, n = 2p = 2; 4; 8; 16; 32; 64; 128; 256; 512; 1024; 2048;

…, it reaches 1 after a cycle of p applications of the algorithm. For example:

2⁴ = 16 → S(16) = {16; 8; 4; 2; 1} (5 values, 4 steps)

2⁷ = 128 → S(128) = {128; 64; 32; 16; 8; 4; 2; 1} (8 values, 7 steps)

The same observation also applies to the odd numbers, when 3n+1 = 2p, i.e. n = (2p-1)/3. In this case the number of steps needed to get it to fall to 1 is equal to p+1. Bearing in mind that the algorithm is applicable only to natural integers, let's see some examples:

p = 1 → n is not an integer

p = 2 → n = 1 → {1; 4; 2; 1} (4 values, 3 steps) p = 3 → n is not an integer

p = 4 → n = 5 → S(5) = {5; 16; 8; 4; 2; 1} (6 values, 5 steps) p = 5 → n is not an integer

p = 6 → n = 21 → S(21) = {21; 64; 32; 16; 8; 4; 2; 1} (8 values, 7 steps)

p = 7 → n is not an integer

p = 8 → n = 85 → S(85) = {85; 256; 128; 64; 32; 16; 8; 4; 2; 1} (10 values, 9 steps)

p = 9 → n is not an integer

p = 10 → n = 341 → S(341) = {341; 1024; 512; 256; 128; 64; 32; 16; 8; 4; 2; 1} (12 values, 11

steps)

… etc.

Here we stop, but it’s interesting to note that for some natural numbers, either odd or even, which constitute a partition of N (v. Fig. 3), it’s possible to establish a priori the number of values contained in their respective sequences obtained by applying to them the algorithm of Collatz, and then establish the exact number of steps needed to reach 1.

fig3

Fig. 3

Precisely, indicating the set of even numbers E and O the set of odd numbers:

If n ∈ E : n = 2p → the corresponding sequence S(n) contains p+1 values and it reaches 1 in p steps

.

If n ∈ O : n = (2p–1)/3 = (2²p–1)/3 = (4p-1)/3 → the corresponding sequence S(n) contains p+2 values and it reaches 1 in p+1 steps.

Things get complicated if the choice is a number that does not satisfy the above conditions. For it is not possible to establish the number of steps needed to get to 1. If we choose, for example, the number 25 we will have the following sequence:

25 → 76 → 38 → 19 → 58 → 29 → 88 → 44 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

It consists of 24 values, 23 steps. We observe that the sequence of values does not contain two same numbers, but they are all different from each other. This happens for all sequences of all the numbers. In short, the sequences obtained through the algorithm of Collatz precipitate to 1 taking different values between them. This is obvious. In fact, if in a succession two numbers were the same, from them onwards we would have the same sequence, giving rise to a vicious circle, and this is impossible (clearly with the exclusion of the number 1 that generates the final cycle). It should be noted also that, when a succession is built, many more are known. With reference to the previous example, the sequence of the number 25 allows us to know the sequences of numbers:

19 → 58 → 29 → 88 → 44 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

88 → 44 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

And so on for all the others. The sequence of number 25 allows us to know the sequences of twenty natural numbers, odd or even, (excluding the values of final cycle: 4, 2, 1 and 25). If a number generates a sequence of 400 values, from it will know the sequences of 396 numbers in it.

When calculating the values of sequences we can also observe that each of them connects to a previous one, when one of its values becomes equal to that of another already computed. Figure 4 shows the diagram of such a consideration in reference to the consecutive numbers 5, 6, 7.

fig4

Fig. 4

The sequence of number 7, from value 20, continues in that of number 6, which in turn, from value 10, continues in that of 5. If we would continue the computation of successive sequences we would realize that the sequence of number 8 is already present in the sequence of number 5, while that of number 9, from the value 14, continues in that of number 7, which in turn and so on.

Again with reference to the number 25, however this reasoning is valid for any other number, another important observation we can make is that when the value of the sequence falls below 25, in this case 19, its succession from here on becomes perfectly equal to that of this number. In this way the sequence of number 25 is linked to the succession of number 19. In short, the endless sequences generated by the algorithm give rise to a kind of interlacement, in which each of the sequences is linked to one of the previous ones. In figure 5 is shown the interlacement produced by the first seven natural numbers.

fig5

Fig. 5

The numbers 2 and 3 snap to sequence of number 1, those of number 4, number 5 and number 6, to that of number 3, while the sequence generated by number 7 attaches to the one generated by number 5. We could build an intertwining of hundreds of consecutive numbers.

Then, to successions Collatz we can associate a specific procedure.

If we take 33 as generating number, applying the algorithm, we obtain the sequence:

33 → 100 → 50 → 25 < 33

25 → 76 → 38 → 19 < 25

19 → 58 → 29 → 88 → 44 → 22 → 11 < 19

11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 < 11

10 → 5 < 10

5 → 16 → 8 → 4 < 5

4 → 2 → 1 final cycle. (v. Fig. 6)

fig6

Fig. 6

If we take 34 as generating number, applying the algorithm, we obtain the sequence:

34 → 17 < 34

17 → 52 → 26 → 13 < 17

13 → 40 → 20 → 10 < 13

10 → 5 < 10

5 → 16 → 8 → 4 < 5

4 → 2 → 1 final cycle. (v. Fig. 7)

fig7

Fig. 7

If we take 49 as generating number, applying the algorithm, we obtain the sequence:

49 → 148 → 74 → 37 < 49

37 → 112 → 56 → 28 < 37

28 → 14 < 28

14 → 7 < 14

7 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 < 7

5 → 16 → 8 → 4 < 5

4 → 2 → 1 final cycle.