[seqfan] Re: This sequence proves the Collatz Conjecture

[Ali Sada]

Hi everyone,
It turns out the array (H(i,j)) is already in the OEIS (A257852.) That made the analysis shorter. And I have made additional changes and the text now is 1 page. 
Also, I have added the sequence to the OEIS. It's A371949. However, I was told it could be deleted. 

Best,
Ali  



Let's call 3x+1 the I operation, and let's call applying m +(3m+1) * (1, 4,16..,4^(m-1)) on each even number, Odd * 2^m, the A operation.The IA operation is a combination of the two, in this order. If we can provethat by applying the IA operation several times on any odd number weunavoidably reach A002450, then the Collatz is proven. 

A096773 is combination of A002450 and A072197. And since termsof A072197 reach A00245 in a single IA operation (in other words reach a powerof 2 in two Collatz iterations), that means that if applying the IA operation multipletimes on any odd number leads unavoidably to A096773, then the Collatz isproven. Let’s call A096773 A1, and A079319 A2. 

To prove that let’s start with A257852, which is an arraythat contains all odd numbers. Let’s call it H(i,j). 

Applying the IA operation once on any H(i,j):

IA(H(i,j)) = 3 * 2^j * H(i,j)+ A2(j) = 3 * H(i,j) *2^j + 2^j+ (4 ^ j-1)/3

Let d1(x)​denote the number of A1​terms that H(i,j) jumps following one IA operation. Let d2(x)​ represent the differencebetween the last number resulting from the jump and the nearest preceding A1​ term that is less than or equalto IA(H(i,j)).

Assume H(i,j) falls within the interval 2^k1 and 2^(k1+1). Withinthis range, the number yielding the highest result after one IA operation isthe largest number from sequence A096773 that fits within this interval. Thisnumber is referred to as the “peak” or p(x,x+1). 

p(k1,k1+1) = (2^(k1+2)-1) /3 if k1 is even, and (5*2^k1-1)/3 if k1 is odd.

When the IA operation is applied once to p(k1,k+1), theoutcome is (2^(6+4*(Int(k1/2)))-1)/3​,regardless of k1 being odd or even. After one IA iteration on the peak we get:  

d1(p(k1,k1+1)) = 4 if k1 = 0 

d1(p(k1,k1+1)) = 3 if k1 = 1

d1(p(k1,k1+1)) = k1+5 if k1 is an odd number >1

d1(p(k1,k1+1)) = k1+3 if k1 is an even number > 0 

And for any peak, d2(p(k1,k1+1)) equals zero.   

When we apply the IA on H(i,j), which is any other number inthe range, that’s not a member of A1 already, d1(H(i,j)) is less than d1(p(k1,k1+1)),however, d2(H(I,j) is > zero. Which means that d1(H(i’,j’) will graduallyincrease with multiple IA iteration until it reaches k1+3, meaning H(i’,j’)itself became a peak.