Tanay's collatz theory
Creators: name: Gudadhe, Tanay
Publisher: name: Zenodo
Year: 2025

description: TANAY’S COLLATZ THEORY

 

Author: Tanay Gudadhe Date: May 2025 This is original work protected under copyright. Do not reproduce without permission.

 

Contact number-8080395330

 

DEFINITIONS

 

Loop-a Sequence of numbers Repeated

 

Even numbers-Numbers which are divisible by 2

 

Odd numbers-Numbers which are not divisible by 2

 

Collatz function C(n)-the notation that denotes rules of the collatz conjecture

 

Mod-Modular;remainder when divided by a number

 

Log-Logarithm;the base of the log is base of power or exponent . The number is the base^a where a is any number .

 

WHY I THINK IT WORKS:

 

∀ n∈ℕ by following collatz rules always → 1 . it works with every single +ℤ , because of ÷ and that if all digits in a number are even⇔n≡0 (mod 2) and there is 2×even number in units place → we can ÷ it by 2 2× . C(n) is the notation for collatz functions when n is even⇔n≡0 (mod 2) or odd⇔n≡1 (mod 2) and n as any natural number . We will be using mod to define even and odd numbers .

 

Proof

 

It works ∀ n∈ℕ because ÷ 2 →eventually 2^k which is n≡0 (mod 2) ⇔n is even so ÷2 so after ÷2 n→n≡0 (mod 2) until n→1 so it works for all numbers and odd⇔n≡1 (mod 2) →even⇔n≡0 (mod 2) by 3n+1

 

C(n)= 3n+1 If n =odd. C(n)=n/2 If n=even

 

n=odd. C(n)=3n+1. 3n+1=even

 

C(n)=n/2. C(n)=eventually 2^k

 

(2^k)/2 = 2^k

 

until it reaches 2^k=4 =even 4/2=2/2=1

 

(C)n=1 When Followed by collatz rules

 

C(n)=3n+1 if n is odd C(n)= n/2. if n is even

 

n(even)=2k k = + ℤ

 

The 2 rules are:   C(n) = 3n + 1  (if n is odd)    C(n) = n ÷ 2  (if n is even)

 

Even=e Δe=n/2

 

(3n+1)/2 both steps 3n+1=n≡0 (mod 2) and after that n/2 so combining steps = (3n+1)/2

 

Main proof statement

 

Some numbers value increases massively first but n/2 → 2^k ÷2 → 2^k-1 This continues until 2^k=2 2÷2 is 1

 

Now step-by-step:

 

1. 

 

Lets Take any natural number when n∈ℕ . Definition: all ℕ =+ℤ

 

ℕ=1 to ∞

 

2. 

 

n where n≡0 (mod 2) or n=2k ÷2.

 

n ÷ 2 = x. 2 × x = n. n=2×x

 

let n be n=2k which makes an even number ⇔ when n≡0 (mod 2) where n∈ℕ C(n)=n/2

 

3. 

 

If n is n=2k+1or n≡1 (mod 2) in n∈ℕ it is odd , apply 3n+1.

 

3n + 1 always results in an n=2k or n≡0 (mod 2)⇔ when n=even .

 

Odd=3n+1

 

 

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4. 

 

Repeat this process. In every case where any n≡0 (mod 2) , ÷ 2 applies again.

 

Odd→even by 3n+1 Than even apply n/2 so now reapet until it reaches 2^k Even so n/2→even until 2^k=2 than 2÷2=1

 

After each ÷ step: Even numbers are half of its origional value.

 

Eventually leading towards 2^k.

 

3n+1. & n/2=2^k. n=odd in n∈ℕ

 

odd=3n+1. 3n+1⇒even. even=n/2

 

N>n So:

 

n → (after x steps) → 2^k

 

Repeated ÷ by 2 leads to 2^k because 2^k=2^(k) (n) and n/2 n÷2 so 2 in both equations is the same so they always reach 2^k until 2^k=2 which ÷2 is 1

 

Proof

 

This can also be proven by saying 2^k=2^(k) (n) so if we ÷ any number n÷2 = 2×y=n where y is quotient so 2 are common . these equations have a coefficient relation so ÷2 → 2^k . Numbers such as 27 reach numbers like 9000 until its value decreases but it can never be a loop (explained in section 9.) so C(n)=3n+1 when n≡1 (mod 2) C(n)=n/2 when n≡0 (mod 2)

 

Contradiction

 

All paths → shrink because it will reach 2^k lets say 32 even so ÷2 =16 even so ÷2=8 even so ÷2=4 again even so ÷2=2 again even so ÷2=1 this works for all n in n∈ℕ and odd n in n∈ℕ become even when 3n+1 is applied

 

5. 

 

Every time a number n where n=2k⇔ n is even appears, ÷2.

 

If n is even⇔2≡0 (mod 2): n ÷ 2

 

Eventually: n ÷ 2 ÷ 2 ÷ ... → 2^k

 

Even n=2k

 

even(appear)=even/2

 

even=(2k)/2

 

n= n≡0 ( mod 2) ⇒ C(n)=n/2

 

6. 

 

Eventually, numbers shrink to 2^k .

 

÷2 repeatedly ensures ∀ n ∃ 2^k ⇒ 2, 4, 8, 16, 32, etc. (2^k).

 

e÷2(x times)=2^k

 

Important:

 

n in n∈ℕ where n≡0 (mod 2)⇔n=2k=½×n≡0 (mod 2)

 

2^(k) (x) = 2^k

 

Thus, even÷ 2^k = 1

 

Sequence of formulas

 

n/2, 3n+1, n/2... or. n/2, n/2, 3n+1... or 3n+1, n/2...

 

we will use log2(n) it is important because it tells us how many times ÷2 is needed for 2^k to reach 1

 

Such as log2(32)=5 because 2⁵=32

 

 

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7. 

 

After reaching 2^k use log2(n) to see how many time ÷2 is needed to fully divide 2^k divisions by 2 continue. ∀ n∈ℕ ∃ 2^k by following the collatz rules

 

2^k ÷ 2 each time.

 

Eventually becomes 2 → then 1.

 

Therefore, every n in n∈ℕ eventually leads to 1.

 

2^k÷2=1

 

This logic works for all n in n∈ℕ {1,2,3,4...∞}

 

Log2(n)=k in 2^k

 

8.Base case

 

The smallest n in n∈ℕ is 1 so C(n)=1 1 is odd⇔n≡1 (mod 2) so we apply 3n+1 which is 4 . 4 is even so ÷2=2 again even so ÷2=1 4 is also 2^k which ÷2=even so 2^k÷2=even so if we keep ÷2 it will reach 1 Use log2(n) to find how many times we need to ÷ to get 1 from 2^k

 

Inductive Hypothesis

 

all n in n∈ℕ can be transformed into 1 by collatz rules and just assume now that it works for k number n=k because 1 reaches 1 by the following rules of the collatz conjecture

 

Induction

 

The conjecture also works for all n=k+1 where n∈ℕ Because, it will reach 2^k so n when ÷2 is even until 2^k is 2 which ÷2 is 1 or use log2(n)=k in 2^k to find how many times ÷2 Is needed to fully ÷ any 2^k and if k is even where k≡0 (mod 2) where k=2n k+1 will be k=2n+1→ odd so 3n+1 which could be written as 3(k+1)+1 is even so ÷2 which leads to 2^k it always leads to 2^k because ÷2 and 2^k have a coefficient relation so they will reach each other so ÷2 will always lead to 2^k and also it can be a series of numbers and to make the process faster use log2(n) to find out how many times 2^k needs to be ÷2 to reach 1

 

n=k,n=k+1,n=k+2,n=k+3,...n=k+x

 

Which always when followed by collatz rules→1 . It always becomes 1 by following collatz rules because 3n+1=even so n/2 this will eventually lead to 2^k which ÷2 is even until 2^k=2 when ÷2 is 1 or use log2(n)

 

Proof that all numbers reach 2^k

 

All numbers eventually reach 2^k because 2^k=2^(k) (x) and n/2=2×x=n So they have 2 in common relation between two terms . So it always reaches 2^k

 

n÷2=2×x=n 2^k=2^(k) (x) Common=2 So, N/2=eventually 2^k

 

 

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Real-world technical note:

 

In extremely rare large numbers, 4 in tens place or other non-units place might delay decrease very slightly, because 4 in tens place when hundred us two will cause 170+x which is not ÷ by 2 two times but still guaranteed by the division process to eventually reach a 2^k.

 

Edge cases

 

Some numbers like 27 grow massively first but they decrease because of 2^k and once they reach a number which has all even numbers in it only 1 odd can also be OK and in units place there is a 2×even number then it can be ÷ 2 2 times so that is its shrinking point of all numbers which grow large

 

9. 

 

No other loops possible:

 

3n + 1 is not equal to n ÷ 2.

 

Expanded:

 

n + n + n + 1 is not equal to n ÷ 2.

 

Thus, a cycle cannot exist except the cycle 1.

 

If ∃ a loop like 33 it would have to be 3n+1 = even n/2=33 which cannot because 3n+1≠ n/2

 

¬ other loops exist than 1 because if we apply that 3n+1≠n/2 but it is only one because [3n+1]+[n/2]=(3n+1)/2 which when n=1 becomes (3×1+1)/2=4/2=2 which ÷2=1 but it cannot exist for all n in n∈ℕ because when n=1 3n+1⇒3+1=2^k→k=2 and any other ×3 +1 ≠2^k→k=2 .

 

¬ complex cycles also exist because any n in n∈ℕ when ÷2→value becomes less and × by 3 and + 1→value greater . therefore a cycle ∃ ! when 3×1 where n=1 therefore, it only happens for 1

 

Other loops cannot exist because every n in n∈ℕ cannot be same if 3n+1 makes the value of n more and ÷2 makes the value of n ½ of n so they cannot exist unless n=1⇒3n=3

 

Also loops can't exist because any time we use 3n+1 than n/2 no matter how many times we do it it cannot be same because it reaches 2^k which ÷2 until 2^k→k=1 2÷2=1

 

Contradiction of cycles not existing

 

∃ ! 1 loop because lets Take any n in n∈ℕ like n=18 lets just say for 18 ∃ a loop than ÷2→ value smaller lets say 18÷2=9 and 9=x for no confusion than 3x+1(because n is x) than new n so x<new n so they can never be same after n/2→ ½ of n so this also proves that numbers such as 43,34,67 dont have complex or normal loops

 

Loops than 1 also can't exist because all numbers when 3n+1 it's value becomes bigger than the origional value . And ÷2 makes the value half or ½ of n . Means after 3n+1 we have to do ½n which makes n smaller so thus the values of n never natch so this rules out normal loops now complex loops , complex loops can never occur because all number reach 2^k which ÷2 is 2^(k-1) so use log2(n) to know how many times to divide by 2 ro reach one this rules out complex loops

 

Formula for numbers which hit the same 2^k

 

I the author have made these formulas to show in collatz which numbers hit a specific 2^k . So we will see a formula for which numbers hit specific 2^k which is

 

Term (n)=n1+(n2-n1)n

 

which says numbers which hit a specific power of two (denoted as term (n) ) is the first number which hits it + (second-first number) × number which says that what numbers reach a specific 2^k lets say 2^k=16 than 2⁵=9+(13-9)n=9+4n this is the formula for 2^k=2⁵ . Thus this formula shows which numbers hit a spesific 2^k (in this case , 16) .

 

My decreasing function in collatz

 

So my decreasing function is T(n)→1=[∑δ=N]|[∑σ=n]|N>n so N-n=1 where δ is the sum of all the answers of 3n+1 for n and σ is the sum of all the answers of n/2 for n

 

PROOFS

 

 

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MAIN PROOF

 

All odd numbers⇒even by 3n+1 this happens every time because 3 is odd n×3=odd+1=even

 

1st PROOF:

 

Every n in n∈ℕ when n=2k which is even can be ÷2 because 2×k=x where x is even so we transpose 2 it becomes k=x÷2 where x is even means x=2k so every even number can be ÷2.

 

Even numbers are built from 2 × 2^k:

 

Even = e. Even = e ÷ 2. 2 × 2^k = even numbers

 

Even=e. e=e/2

 

 

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2nd PROOF:

 

All even numbers are multiples of 2, so they can be expressed as:

 

Even = 2 × k Even ÷ 2^k

 

Eventually reaching 1.

 

Multiples of 2 = 2×x

 

3rd PROOF:

 

All even numbers are divisible by 2 which creates 2^k:

 

Even = e. Even = e ÷ 2. 2 × 2 × 2 × ... × 2 = 2^k

 

Proof for 1,2,3 profes

 

Numbers such as 27 value grow very much before decreasing but it's value always decreases because n/2 which leads to 2^k which ÷2 repeatedly because 2^k÷2=even always because they are made by 2×2...×2 and 2×x=2k (even) becomes 1

 

Proof It works for all number same

 

It doesn't behave differently for any number because every number when 3n+1 is 2k (even) and than n/2 which always eventually 2^k which repeatedly ÷2 =1

 

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

 

Collatz conjecture rules are 3n+1→even after that n/2 if we combine⇒(3n+1)/2 and 3n+1 always will be divisible by 2 because when an odd number is ×3 it is odd because it is × odd number and +1 makes n even so always it is ÷ by 2 so this formula has a meaning the upper when solved always is an even number so after solving it becomes a number n so now 3n+1=n So (3n+1)/2=n/2 simplifed where n in n∈ℕ can be both n=2k and n=2k+ especially n=2k+1 so again applys 3n+1 this process repeats so it seems like a infinite cycle but it cannot be possible because this process gets 2^k as an answer eventually which ÷2=1 thus is how every number which is in n∈ℕ by following these rules reaches 1 and by following these rules it always reaches 2^k because 2^k=n(even)/2 Because n(even)/2=Q(x) Q is for quotient So Q=2^k Eventually as 2 is common in both ÷2 and 2^k and let's assume a number x when ÷2 it becomes half of its origional value and then 3n+1=n then again n is half of its origional value so eventually 2^k=1/2×n(even) so it becomes 2^k

 

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

 

Graph explanation

 

Let's assume a graph which contains odd even 1 3n+1 n/2 2^k and a equation 2x+y substitute any value to make number 2 to infinite to show natural number set use 2x+y to form any number such as 2 , 3, 4 , 5 ... n if it's 2^k it will go to 1 and if 3n+1=even and n/2=odd or even it will always reach 1 for all 2x+y because of that point on the graph 2^k so when it reaches that point ÷2 will always be even until it reaches 2 so 2÷2 is 1 so this is how all positive integers reach 1 by following the collatz rules

 

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

 

Algebraic explanation

 

2x+y=n So, n/2 when even so n/2= eventually reaching 2^k so every number can be made by 2x+y and it canalso be made by 3x-y so 2x+y= all n and 3x-y=all n so thus following and substituting values of x and y as anything so one equation that has any constant(coefficient) and any values of x and y can make up all n so we can denote n as one equation 2x+y so 3n+1=3(2x+y)+1=n(even) So after that even so n/2 and as n=2x+y it becomes n/2=(2x+y)/2 Or 1/2×(2x+y) which is odd or even

 

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

 

Deeper proof of numbers reaching 2^k

 

All numbers reach 2^k before getting to 1 because 2^k÷2 always is even so until it reaches 2 I will always be even so 2÷2=1 because numbers when n/2 they become half of their origional value and 2^k also reach when even numbers are half of their origional value so eventually every number reach 2^k to become 1 So if we ÷ even numbers by 2 and relations of two terms n/2 and 2^k have 2 in common so dividing by 2 leads to 2^k

 

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

 

Algebraic notation of 2^k and relation in variables of n/2

 

2^k can be expressed as a rational number 2^k/1 and expansion of 2^k is 2×2...×2(x times ) 2^1 2^2 2^3 ... 2^k so expansion of 2^k is 2^1 ,2^2,2^3,...,2^k and n/2 is n÷2=Q and n division is n=divisor(x)×quotient(x)+remainder(x) or n=D(x)×Q(x)+R(x) where D is always 2 so n=2×Q(y)+R(z) so they have a coefficient relation 2 that tells us 2^k÷2=even and n/2= eventually 2^k . The odd numbers after 3n+1 even than n/2 becomes odd or even mostly odd so after some steps the odd number always is half of the new even number generated from 3n+1

 

Σn∈ℕ=∞ this shows that this works for all natural numbers till infinity 1 to ∞

 

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

 

This logic works ∀ n in n∈ℕ or +ℤ or {1 , 2 , 3 , 4 , 5 ,... , ∞} or it works for ∑ℕ₁ or ℕ₁∩+ℤ we can also say that all n in collatz reach 1 by following collatz rules and also not only because that n÷2 and 2^k have a coefficient relation but also ÷2 means factor of that n which ×2 is n so we can say in (n÷2)=new n where new n=(n÷2)×2 and 2^k is 2^(k) (f) or simply 2×2...×2 they have ×2 in common so each will reach another if followed by an operation so ÷2 will lead to 2^k.

 

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Conclusion

 

All natural numbers reach 1 by following collatz rules because of 2^k and ÷2 . ÷2 of even numbers lead to 2^k which ÷ by answer of log2(n) is 1 so they always reach 1 . No other cycles exist because 3n+1≠n2 .

 

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Originality statement

 

I tanay gudadhe confirm this title TANAYS COLLATZ THEORY is my own origional work . The ideas , equations , symbols , logic and proofs are all my work .

 

This statement is originally the work of Tanay gudadhe on April 2025 theory created and development since than all math symbols equations and logic are mine only formatting has been done by ai (chatgpt)