Reduced Collatz function

The Collatz problem could also be restated as asking whether the iterated reduced Collatz function $\scriptstyle R(2n-1),\,n\,\geq \,1,\,$ where $\scriptstyle R(k)\,=\,{\frac {3k+1}{2^{r}}},\,$ with $\scriptstyle r\,$ as large as possible, always reaches an even power of 2 (a power of 4), in which case the hailstone height then decreases exponentially down to 1.

Reduced Collatz function

From $(2n-1)\cdot 2^{\alpha }$		To $R((2n-1)\cdot 2^{\alpha })$		Power (even) of 2
$1\cdot 2^{\alpha }$	→	$1\cdot 2^{\beta }$	$(1\cdot 3+1=4=2^{2}\cdot 1)\,$
$3\cdot 2^{\alpha }$	→	$5\cdot 2^{\beta }$	$(3\cdot 3+1=10=2^{1}\cdot 5)\,$
$5\cdot 2^{\alpha }$	→	$1\cdot 2^{\beta }$	$(3\cdot 5+1=16=2^{4}\cdot 1)\,$	*
$7\cdot 2^{\alpha }$	→	$11\cdot 2^{\beta }$	$(3\cdot 7+1=22=2^{1}\cdot 11)\,$
$9\cdot 2^{\alpha }$	→	$7\cdot 2^{\beta }$	$(3\cdot 9+1=28=2^{2}\cdot 7)\,$
$11\cdot 2^{\alpha }$	→	$17\cdot 2^{\beta }$	$(3\cdot 11+1=34=2^{1}\cdot 17)\,$
$13\cdot 2^{\alpha }$	→	$5\cdot 2^{\beta }$	$(3\cdot 13+1=40=2^{3}\cdot 5)\,$
$15\cdot 2^{\alpha }$	→	$23\cdot 2^{\beta }$	$(3\cdot 15+1=46=2^{1}\cdot 23)\,$
$17\cdot 2^{\alpha }$	→	$13\cdot 2^{\beta }$	$(3\cdot 17+1=52=2^{2}\cdot 13)\,$
$19\cdot 2^{\alpha }$	→	$29\cdot 2^{\beta }$	$(3\cdot 19+1=58=2^{1}\cdot 29)\,$
$21\cdot 2^{\alpha }$	→	$1\cdot 2^{\beta }$	$(3\cdot 21+1=64=2^{6}\cdot 1)\,$	*
$23\cdot 2^{\alpha }$	→	$35\cdot 2^{\beta }$	$(3\cdot 23+1=70=2^{1}\cdot 35)\,$
$25\cdot 2^{\alpha }$	→	$19\cdot 2^{\beta }$	$(3\cdot 25+1=76=2^{2}\cdot 19)\,$
$27\cdot 2^{\alpha }$	→	$41\cdot 2^{\beta }$	$(3\cdot 27+1=82=2^{1}\cdot 41)\,$
$29\cdot 2^{\alpha }$	→	$11\cdot 2^{\beta }$	$(3\cdot 29+1=88=2^{3}\cdot 11)\,$
$31\cdot 2^{\alpha }$	→	$47\cdot 2^{\beta }$	$(3\cdot 31+1=94=2^{1}\cdot 47)\,$
$33\cdot 2^{\alpha }$	→	$25\cdot 2^{\beta }$	$(3\cdot 33+1=100=2^{2}\cdot 25)\,$
$35\cdot 2^{\alpha }$	→	$53\cdot 2^{\beta }$	$(3\cdot 35+1=106=2^{1}\cdot 53)\,$
$37\cdot 2^{\alpha }$	→	$7\cdot 2^{\beta }$	$(3\cdot 37+1=112=2^{4}\cdot 7)\,$
$39\cdot 2^{\alpha }$	→	$59\cdot 2^{\beta }$	$(3\cdot 39+1=118=2^{1}\cdot 59)\,$
$41\cdot 2^{\alpha }$	→	$31\cdot 2^{\beta }$	$(3\cdot 41+1=124=2^{2}\cdot 31)\,$
$43\cdot 2^{\alpha }$	→	$65\cdot 2^{\beta }$	$(3\cdot 43+1=130=2^{1}\cdot 65)\,$
$45\cdot 2^{\alpha }$	→	$17\cdot 2^{\beta }$	$(3\cdot 45+1=136=2^{3}\cdot 17)\,$
$47\cdot 2^{\alpha }$	→	$71\cdot 2^{\beta }$	$(3\cdot 47+1=142=2^{1}\cdot 71)\,$
$49\cdot 2^{\alpha }$	→	$37\cdot 2^{\beta }$	$(3\cdot 49+1=148=2^{2}\cdot 37)\,$
$51\cdot 2^{\alpha }$	→	$77\cdot 2^{\beta }$	$(3\cdot 51+1=154=2^{1}\cdot 77)\,$
$53\cdot 2^{\alpha }$	→	$5\cdot 2^{\beta }$	$(3\cdot 53+1=160=2^{5}\cdot 5)\,$
$55\cdot 2^{\alpha }$	→	$83\cdot 2^{\beta }$	$(3\cdot 55+1=166=2^{1}\cdot 83)\,$
$57\cdot 2^{\alpha }$	→	$43\cdot 2^{\beta }$	$(3\cdot 57+1=172=2^{2}\cdot 43)\,$
$59\cdot 2^{\alpha }$	→	$89\cdot 2^{\beta }$	$(3\cdot 59+1=178=2^{1}\cdot 89)\,$
$61\cdot 2^{\alpha }$	→	$23\cdot 2^{\beta }$	$(3\cdot 61+1=184=2^{3}\cdot 23)\,$
$63\cdot 2^{\alpha }$	→	$95\cdot 2^{\beta }$	$(3\cdot 63+1=190=2^{1}\cdot 95)\,$

A075677 Reduced Collatz function R applied to the odd integers: a(n) = R(2n-1), where R(k) = (3k+1)/2^r, with r as large as possible.

{1, 5, 1, 11, 7, 17, 5, 23, 13, 29, 1, 35, 19, 41, 11, 47, 25, 53, 7, 59, 31, 65, 17, 71, 37, 77, 5, 83, 43, 89, 23, 95, 49, 101, 13, 107, 55, 113, 29, 119, 61, 125, 1, 131, 67, ...}