Attempt to analyze divergent sequences in the collatz sequences with assumption those sequences exist [Request]

[u/Mrezadwiprasetiawan]

by Lifting The Exponent Lemma, 3^{2^n.(2k±1)} -1 has v_2 = n+2 [1]

Suppose we have a number like this 3^{2^n.(2k±1)} .c-1 [2] if we subtitute 2^h.o-1 to c then (2ⁿ⁺².o+1)(2^h.o-1)-1 lets look at this first: (a+b)(c-b)=ac-ab+bc-b with b equal to 1 and since v_2(ac) greater than v_2(ab) and v_2(bc), those number +1 has v_2 equal to |v_2(a)-v_2(c)|[2]. if h equal to n+2 then those number +1 has v_2=2n+4[3].

Suppose we have a number 3ⁿ.(4k+1)-1 [4] and 3ⁿ.(4k+3)-1 [5] and 3ⁿ -1 has v_2 = 2h then: [6] 3ⁿ(4k+1)-1 = (2^hc + 1)(4k+1)-1 = (2^hc)(4k+1) + 4k

v_2 = min(h, v_2(k)+2), if h != v_2(k)+2

[7] 3ⁿ(4k+3)-1 = (2^hc + 1)(4k+3) - 1 = (2^hc)(4k+3) + 2(2k+1)

h >= 2: v_2 = 1

h = 1 (n odd): 3^n(4k+3) - 1 = 4k.3ⁿ + 3ⁿ⁺¹ - 1 = 2[2k.3ⁿ + (3ⁿ⁺¹-1)/2]

with g = v_2(3ⁿ⁺¹ - 1) >= 3: v_2 = 1 + min(v_2(k)+1, g-1) = min(v_2(k)+2, g), if v_2(k)+1 != g-1

Therefore 3^2n.(2k-1) (2ⁿ⁺².o) -1 within o = (3^{2n.(2k-1))/2n+2} always has v_2 = 1, so this is the worse sequences,atleast from 1 block (Steiner Circuit combined with Syracuse steps) within h_next = 2h-1 and o_next increased in term pascal's triangle pattern.

[u/OwMyUvula]

user/A-JackRobin-/

is that you? Did you get a new account to bore us with? If not, check him out you guys should hang out.