3n + 1 cycles for n = 2^57,885,161-1

Any estimate as to the run time (on a typical pc say)?

I was wondering about the 'loop count' for this hailstone sequence, and whether it'd be feasible to run such a program to test such a thing.

[science_man_88]

Originally Posted by Unregistered

Any estimate as to the run time (on a typical pc say)?

I was wondering about the 'loop count' for this hailstone sequence, and whether it'd be feasible to run such a program to test such a thing.

2^57885161-1
3*2^57885161-2
3*2^57885160-1
9*2^57885160-2
9*2^57885159-1
27*2^57885159-2


is the start by the looks of it. as for the run time I'm not sure since I did it in my head.

[henryzz]

Originally Posted by science_man_88

2^57885161-1
3*2^57885161-2
3*2^57885160-1
9*2^57885160-2
9*2^57885159-1
27*2^57885159-2


is the start by the looks of it. as for the run time I'm not sure since I did it in my head.

Eventually I think it will get to 3^57885161-1.

[firejuggler]

( 3^57885161-1)/2, at least?

[henryzz]

Originally Posted by firejuggler

( 3^57885161-1)/2, at least?

Yes you can divide that step by 2.
I don't think we can do more without actual calculation.

[Batalov]

You could do another 50 (or a few thousand, if you want) steps, like this:

? n=lift(Mod(3,2^100)^57885161-1)
%5 = 526113967643187600285488863586
? for(i=1,50,if(n%2,n=(3*n+1)/2;print(n" =(3n+1)/2"),n/=2;print(n" =n/2")))
263056983821593800142744431793 =n/2
394585475732390700214116647690 =(3n+1)/2
197292737866195350107058323845 =n/2
295939106799293025160587485768 =(3n+1)/2
147969553399646512580293742884 =n/2
73984776699823256290146871442 =n/2
36992388349911628145073435721 =n/2
55488582524867442217610153582 =(3n+1)/2
27744291262433721108805076791 =n/2
41616436893650581663207615187 =(3n+1)/2
62424655340475872494811422781 =(3n+1)/2
93636983010713808742217134172 =(3n+1)/2
46818491505356904371108567086 =n/2
23409245752678452185554283543 =n/2
35113868629017678278331425315 =(3n+1)/2
52670802943526517417497137973 =(3n+1)/2
79006204415289776126245706960 =(3n+1)/2
39503102207644888063122853480 =n/2
19751551103822444031561426740 =n/2
9875775551911222015780713370 =n/2
4937887775955611007890356685 =n/2
7406831663933416511835535028 =(3n+1)/2
3703415831966708255917767514 =n/2
1851707915983354127958883757 =n/2
2777561873975031191938325636 =(3n+1)/2
1388780936987515595969162818 =n/2
694390468493757797984581409 =n/2
1041585702740636696976872114 =(3n+1)/2
520792851370318348488436057 =n/2
781189277055477522732654086 =(3n+1)/2
390594638527738761366327043 =n/2
585891957791608142049490565 =(3n+1)/2
878837936687412213074235848 =(3n+1)/2
439418968343706106537117924 =n/2
219709484171853053268558962 =n/2
109854742085926526634279481 =n/2
164782113128889789951419222 =(3n+1)/2
82391056564444894975709611 =n/2
123586584846667342463564417 =(3n+1)/2
185379877270001013695346626 =(3n+1)/2
92689938635000506847673313 =n/2
139034907952500760271509970 =(3n+1)/2
69517453976250380135754985 =n/2
104276180964375570203632478 =(3n+1)/2
52138090482187785101816239 =n/2
78207135723281677652724359 =(3n+1)/2
117310703584922516479086539 =(3n+1)/2
175966055377383774718629809 =(3n+1)/2
263949083066075662077944714 =(3n+1)/2
131974541533037831038972357 =n/2
...

...but that would not really lead anywhere useful.

[ATH]

Originally Posted by henryzz

Eventually I think it will get to 3^57885161-1.

Yeah after 2*57885161 = 115770322 iterations we get to 3^57885161-1:

ite1: 3^1 * 2^57885161 - 2
ite2: 3^1 * 2^57885160 - 1
ite3: 3^2 * 2^57885160 - 2
ite4: 3^2 * 2^57885159 - 1
ite5: 3^3 * 2^57885159 - 2
ite6: 3^3 * 2^57885158 - 1
.
.
.

[ATH]

Starting from 3^57885161-1 (27,618,241 digits) after 1,000,000 iterations the number is down to 27,576,443 digits.