Tanay's collatz theory (improved)

TANAY’S COLLATZ THEORY Author: Tanay Gudadhe Date: May 2025 This is original work protected under copyright. Do not reproduce without permission. ABSTRACT This theory proves the collatz conjecture to be true by power of 2 , mods , logs , Contradiction , Induction and more . This theory shows every number when followed by collatz rules -: 3n+1 when n is odd and n/2 when n is even reach 1. 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 the 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 +ℤ , because of ÷ and that if every digit in a number n are even⇔n≡0 (mod 2) and there is 2×even number in units place → we can n/2 2 times . 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. Let's 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+1 or 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 --- 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 repeat 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 original 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 → decrease 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 decrease 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 how they might occur 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 --- 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 it 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 works for all n=k+1 where n∈ℕ Because, it will reach 2^k so n/2 → n≡0 (mod 2) until 2^k is 2 so 2/2 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) ⇔ k=2n so k+1 will be n≡1 (mod 2) ⇔ k=2n+1 so 3n+1 which could be written as 3(k+1)+1 is N≡0 (mod 2) so ÷2 which→2^k it always→2^k because ÷2 and 2^k have a coefficient relation so they will reach each other so ÷2 will always→2^k which ÷2 until k in 2^k is 2 than ÷2 is 1 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 is 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 --- Real-world technical note: In extremely rare large numbers, 4 in tens place or other non-units place might 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 let's 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 original 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 Loops existence by mods Loops also don't exist because 3n+1 makes N and N/2 makes x so N>x so a loop can never exist and complex Loops also can't exist because when 3n+1 where n≡1 (mod 2) which makes N even so N≡0 (mod 2) so N/2 this can never make a loop because remainders are different so it can never be same numbers every time 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 let's 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 specific 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 th