Define the odd part of a positive integer $\operatorname{od}(2^i(2j+1))=2j+1$ and define the function $f(n)=\operatorname{od}(n!-2^n)$. For $3\leq n \leq 10$ it holds that:
$$f(2^n)-f(2^n-1)=2^n-2 \tag{1}$$
Does this equality holds for all $n\in\mathbb N_{>2}$?
I don't like to try to prove random conjectures because I'm to bad at it, but I do like finding arithmetical patterns and test them with a computer.
In the table below $n$ and $f(n)$ are listed and the pattern is visible. Of course, there could be a rather easy explanation, only that I haven't found it.
4 1
5 11
6 41
7 307
8 313
9 2831
10 14171
11 155917
12 467771
13 6081067
14 42567517
15 638512859
16 638512873
17 10854718871
18 97692469871
19 1856156927617
20 9280784638121
21 194896477400617
22 2143861251406867
23 49308808782358109
24 147926426347074371
25 3698160658676859367
26 48076088562799171867
27 1298054391195577640609
28 9086380738369043484367
29 263505041412702261046859
30 3952575621190533915703109
31 122529844256906551386796843
32 122529844256906551386796873
33 4043484860477916195764296871
34 68739242628124575327993046871
35 2405873491984360136479756640617
36 21652861427859241228317809765621
37 801155872830791925447758961328117
38 15221961583785046583507420265234367
39 593656501767616816756789390344140609
40 2968282508838084083783946951720703121
41 121699582862361447435141825020548828117
42 2555691240109590396137978325431525390617
43 109894723324712387033933067993555591796859
44 1208841956571836257373263747929111509765617
45 54397888045732631581796868656810017939453109
46 1251151425051850526381327979106630412607421859
47 58804116977436974739922415018011629392548828093
48 176412350932310924219767245054034888177646484371
49 8644205195683235286768595007647709520704677734367
50 216105129892080882169214875191192738017616943359367
51 11021361624496124990629958634750829638898464111328109
52 143277701118449624878189462251760785305680033447265617
53 7593718159277830118544041499343321621201041772705078109
54 205030390300501413200689120482269683772428127863037109359
55 11276671466527577726037901626524832607483547032467041015593
56 78936700265693044082265311385673828252384829227269287109367
57 4499391915144503512689122748983408210385935265954349365234359
58 130482365539190601867984559720518838101192122712676131591796859
59 7698459566812245510211089023510611447970335240047891763916015593
60 115476893502183682653166335352659171719555028600718376458740234359
61 7044090503633204641843146456512209474892856744643820963983154296843
62 218366805612629343897137540151878493721678559083958449883477783203093
63 13757108753595648665519665029568345104465749222289382342659100341796811
64 13757108753595648665519665029568345104465749222289382342659100341796873
65 894212068983717163258778226921942431790273699448809852272841522216796871
66 29508998276462666387539681488424100249079032081810725125003770233154296871
67 1977102884522998647965158659724414716688295149481318583375252605621337890617
68 33610749036890977015407697215315050183701017541182415917379294295562744140621
69 2319141683545477414063131107856738462675370210341586698299171306393829345703117
Yes, it is true.
The first thing is that $$od((2^n)!)=\frac{(2^n)!}{2^{2^n-1}}$$ For example, with $16!$, you get:
• a 2 from the even numbers 2,4,...,16
• another 2 from 4,8,12,16
• another 2 from 8,16
• another 2 from 16.
The gives $8+4+2+1=16-1$ powers of 2.
So when you subtract $2^{2^n}$ from $2^n!$, the od process divides it by $2^{2^n-1}$ leaving $2$.
The second thing is $$od((2^n-1)!)=od(2^n!)=\frac{2^n!}{2^{2^n-1}}\\=\frac{2^n(2^n-1)!}{2^{2^n-1}}\\=\frac{(2^n-1)!}{2^{2^n-1-n}}$$ So when you subtract $2^{2^n-1}$, the od process divides it by $2^{2^n-1-n}$ leaving $2^n$