# Factorials and never ending cycles!

As you may know it, the factorial of a positive integer n is the product of all the positive integers which are equal or smaller to n.

For instance :

6! = 6*5*4*3*2*1 = 720
0! = 1


We will now define a special operation with an irrelevant name like sumFac:

Given a positive integer n, sumFac(n) is the sum of the factorials of the digits.

For instance :

sumFac(132) = 1! + 3! + 2! = 9

Your mission, whether or not you choose to accept it, is to return the sequence (potentially infinite) of the applications of sumFac to an integer given in input.

Example : 132 -> 132, 9, 362880, 81369, 403927, ...

But that's not all! Indeed, the applications of sumFac will eventually create a cycle. You must also return this cycle!

If your language has a built in factorial you can use it. I'm not picky about the return type, you just have to return the sequence of sumFac applications and the cycle in a format understandable by a human.

EDIT : To help you visualize better what should the output look like I copied Leaky Nun's just below:

[132, 9, 362880, 81369, 403927, 367953, 368772, 51128, 40444, 97, 367920, 368649, 404670, 5810, 40442, 75, 5160, 842, 40346, 775, 10200, 6, 720, 5043, 151, 122, 5, 120, 4, 24, 26, 722, 5044, 169, 363601, 1454]


You just need to stop the sequence when the cycle is about to start for the second time!

But this is code-golf so the shortest answer in bytes wins!

• Related – Leaky Nun Apr 25 '17 at 12:47
• Welcome to PPCG! This looks like a nice challenge, BTW. – clismique Apr 25 '17 at 12:49
• @Qwerp-Derp Thank you very much! I tried to be creative ^^ – user68509 Apr 25 '17 at 12:51
• @Zgarb Well it's exactly like the output of Leaky Nun. The sequence of the applications and then it shall end just before the beginning of the second cycle. I'll copy his output in the question so everyone can have a clear understanding. Thanks for pointing that out :) – user68509 Apr 25 '17 at 13:09
• @2501 Hardcoding the value is cheating, but concerning the output formatting you can use any separator you want – user68509 Apr 26 '17 at 8:40

# Jelly, 6 bytes

D!SµÐĿ
ÐĿ  Repeat until the results are no longer unique. Collects all intermediate results.
D           Convert from integer to decimal (list of digits)
!          Factorial (each digit)
S         Sum


Try it online!

I don't see any other way to make it shorter other than to do as told.

## Specs

• Input: 132 (as command-line argument)
• Output: [132, 9, 362880, 81369, 403927, 367953, 368772, 51128, 40444, 97, 367920, 368649, 404670, 5810, 40442, 75, 5160, 842, 40346, 775, 10200, 6, 720, 5043, 151, 122, 5, 120, 4, 24, 26, 722, 5044, 169, 363601, 1454]
• I didn't expect to have an answer that short. Nice :) – user68509 Apr 25 '17 at 12:55
• @Antoine It's Jelly :D It's always shorter than I think it'll be ;) – hyper-neutrino Apr 25 '17 at 12:56
• @HyperNeutrino Somehow Dennis will come with an even shorter answer – Leaky Nun Apr 25 '17 at 14:05
• Somehow, yes. Because Dennis. :P – hyper-neutrino Apr 25 '17 at 14:18
• So... Which character encoding do you use to come up with 6 bytes for those 6 characters? Isn't Jelly supposed to be UTF-8 encoded, meaning that this program is actually 9 bytes? – LordOfThePigs Apr 26 '17 at 11:03

# Python 2, 88 bytes

import math
f=lambda x,l=[]:l*(x in l)or f(sum(math.factorial(int(i))for i inx),l+[x])


Try it online!

• As a python fan I'm glad someone answered with it, nice :) – user68509 Apr 25 '17 at 13:53

# 05AB1E, 12 bytes

[DˆS!O©¯så#®


Try it online!

Explanation

[               # infinite loop
Dˆ             # add a copy of current value to the global list (initialized as input)
S            # split current number to digits
!O          # calculate factorial of each and sum
©         # save a copy in register
¯så#     # if the current number is in the global list, exit loop
®    # retrieve the value from the register for the next iteration
# implicitly output the global list

• Short and correct, well played ! – user68509 Apr 25 '17 at 13:54
• Thought I could get rid of the s, was wrong, nice answer. – Magic Octopus Urn Apr 25 '17 at 16:18

# Brachylog, 17 bytes

g:I{tẹḟᵐ+}ᵃ⁾L¬≠Lk


Try it online!

### Explanation

g:I{     }ᵃ⁾         Accumulate I (a variable) times, with [Input] as initial input:
t                  Take the last integer
ẹḟᵐ+              Compute the sum of the factorial of its digits
L        The result of the accumulation is L
L¬≠      Not all elements of L are different
Lk    Output is L minus the last one (which is the start of the loop)

• What does I mean? – Leaky Nun Apr 25 '17 at 13:53
• @LeakyNun It's a parameter to ᵃ⁾. ᵃ³ means "accumulate 3 times". ᵃ⁾ means "accumulate as many times as the last element of the input", which in that case is I. Since I is a completely free variable, it will try values for it from 0 to +inf. – Fatalize Apr 25 '17 at 13:58

# Wolfram Language, 6260 56 bytes

Most@NestWhileList[Tr[IntegerDigits@#!]&,#,UnsameQ,All]&


It is really too bad that Wolfram Language has such abominably long function names. *Sigh*

Explanation:

Most[NestWhileList[Tr[IntegerDigits[#]!]&,#,UnsameQ,All]]&
IntegerDigits[#]                     (*Split input into list of digits*)
!                    (*Factorial each element in the list*)
Tr[                 ]&                  (*Sum the list together*)
NestWhileList[                      ,#,UnsameQ,All]   (*Iterate the function over itself, pushing each to a list, until a repeat is detected*)
Most[                                                   ]& (*Remove the last element in the list*)

• Nice answer. I don't think this can be improved upon. – Kelly Lowder Apr 25 '17 at 21:52
• @KellyLowder Thanks! I was actually able to save two more bytes by mapping the factorial to the list, then summing it with Tr. – Scott Milner Apr 25 '17 at 22:10
• Nice use of NestWhileList[...,All]! – Greg Martin Apr 26 '17 at 0:45

# ClojureScript, 146 109 bytes

#(loop[n[%]](let[f(apply +(for[a(str(last n))](apply *(range 1(int a))))](if(some #{f}n)n(recur(conj n f)))))


Thanks @cliffroot for shaving off a whopping 37 bytes!

This is an anonymous function, to run the function, you have to do this:

(#(...) {arguments})


TIO doesn't have ClojureScript, so here's a link to a ClojureScript REPL.

Here's a link to a Clojure program which prints the last element in the list from 0 to 1000.

Here's the output for 9999:

[9999 1451520 269 363602 1455 265 842 40346 775 10200 6 720 5043 151 122 5 120 4 24 26 722 5044 169 363601 1454]


I have a strong suspicion that all numbers must eventually settle at 1 or the loop [169 363601 1454].

## Ungolfed code:

(defn fact-cycle [n]
(loop [nums [n]]
(let [fact-num
(let [str-n (str (last nums))]
(apply +
(for [a (range (count str-n))]
(apply *
(range 1
(inc (int (nth str-n a))))))))]
(if (some #{fact-num} nums) nums
(recur
(conj nums fact-num))))))


Explanation coming soon!

• Quite long but correct ;) I cannot really help you to golf this sorry ^^ – user68509 Apr 25 '17 at 13:54
• the inner for can be (for[a s](apply *(range 1(-(int a)47)))), can't it? – cliffroot Apr 25 '17 at 14:39
• and this will allow to get rid of the other let #(loop[n[%]](let[f(apply +(for[a(str(last n))](apply *(range 1(-(int a)47)))))](if(some #{f}n)n(recur(conj n f))))) – cliffroot Apr 25 '17 at 14:47
• oh, seems you don't even need (- ... 47) in ClojureScript, just int will suffice – cliffroot Apr 25 '17 at 14:49
• well, (inc(int a)) should do for ClojureScript and (-(int a)47) for Clojure. – cliffroot Apr 25 '17 at 14:59

## JavaScript (ES6), 91 89 bytes

Saved 2 bytes thanks to fəˈnɛtɪk

It turns out to be quite similar to the other JS answer.

f=(n,x=!(a=[n]))=>n?f(n/10|0,x+(F=n=>n?n--*F(n):1)(n%10)):~a.indexOf(x)?a:f(x,!a.push(x))


f=(n,x=!(a=[n]))=>n?f(n/10|0,x+(F=n=>n?n--*F(n):1)(n%10)):~a.indexOf(x)?a:f(x,!a.push(x))

console.log(f(132))

• In your factorial function shouldn't you be able to use n instead of n>1 because 0!=1? – fəˈnɛtɪk Apr 26 '17 at 13:48
• @fəˈnɛtɪk I don't know what I was thinking here. Thank you! – Arnauld Apr 26 '17 at 14:14

# Perl 6, 64 bytes

{my@a;$_,{[+] .comb.map:{[*] 2..$_}}...^{$_∈@a||!@a.push:$_}}


Try it

## Expanded:

{

my @a;             # array of values already seen

$_, # seed sequence with the input { [+] # reduce using &infix:<+> .comb # the digits of$_ (implicit method call)
.map:          # do the following for each
{
[*] 2..$_ # get the factorial of } } ...^ # keep generating values until # (｢^｣ means throw away the last value when done) {$_ ∈ @a        # is it an elem of @a? (｢∈｣ is shorter than ｢(cont)｣)

||               # if it's not

!              # boolean invert so this returns False
@a.push: $_ # add the tested value to @a } }  Every line above that has { starts a new bare block lambda with an implicit parameter of $_.

I used [*] 2..$_ instead of [*] 1..$_ purely as a micro optimization.

# JavaScript, 92 bytes

Thanks @Shaggy for golfing off one byte with includes
Thanks @Neil for golfing off two bytes

Code separated into individual functions 92 bytes

f=(x,a=[])=>a.includes(x)?a:f(k(x),a,a.push(x))
p=y=>y?y*p(y-1):1
k=n=>n?p(n%10)+k(n/10|0):0


Code on one line 92 bytes

f=(x,a=[])=>a.includes(x)?a:f((k=n=>n?(p=y=>y?y*p(y-1):1)(n%10)+k(n/10|0):0)(x),a,a.push(x))


## Explanation

Initially call the function with just a single argument, therefore a=[].

If x exists in the array a return a a.includes(x)?a:...

Otherwise, append x to a and pass the factorial digit sum and a to the function (a.push(x),f(k(x),a))

p=y=>y?y*p(y-1):1
k=n=>n?p(n%10)+k(n/10|0):0


Factorial Digit sum performed so that it will not exceed the maximum recursion limit.

List of all possible endpoints: 1, 2, 145, 169, 871, 872, 1454, 40585, 45361, 45362, 363601

Try it online!

• Gah, I was so close! Save a byte with f=(x,a=[])=>a.includes(x)?a:(a.push(x),f(k(x),a)) – Shaggy Apr 25 '17 at 14:25
• Can you not write f(k(x),a,a.push(x))? Also, I think you can write k=n=>n&& to save another byte. – Neil Apr 25 '17 at 18:10

g#n|elem n g=g|h<-g++[n]=h#sum[product[1..read[d]]|d<-show n]
([]#)


Try it online! Usage: ([]#) 132

Edit: Saved 13 bytes with typs from Ørjan Johansen!

• (1) Test and append n instead of s(same as in ovs's Python answer), then f=([]#). (2) Switch the branches, inline s, and use elem. – Ørjan Johansen Apr 25 '17 at 17:40
• Switch out your ++ for : also. – user68521 Apr 25 '17 at 19:18
• @Canyon It's the wrong order around for that, it would give the final result reversed. You can almost fix it up afterwards, by prepending an extra n: and changing =g to =[], but it seems to be only a tie. – Ørjan Johansen Apr 25 '17 at 20:47

# Pyth, 9 bytes

.us.!MjNT
.us.!MjNTQ  implicit Q

.u          explained below
N      current value
j T     convert to decimal (list of digits)
.!M        factorial of each digit
s           sum


Try it online!

This answer uses .u ("Cumulative fixed-point. Apply until a result that has occurred before is found. Return all intermediate results.")

# Pyth, 30 bytes

W!hxYQQ=+YQKQ=Q0WK=+Q.!%KT=/KT


Try It Here

• Pyth has more useful functions than one imagines. See my answer as reference. – Leaky Nun Apr 26 '17 at 7:48

# Python 3, 110 bytes

g=lambda x:x<1or x*g(x-1)
def f(n):
a=[];b=n
while b not in a:a+=[b];yield b;b=sum(g(int(x))for x in str(b))


Try it online!

# R, 120 Bytes

o=scan()
repeat {
q=sum(factorial(as.double(el(strsplit(as.character(o[length(o)]), "")))))
if(q%in%o)break
o=c(o,q)
}
o

• you can do o=scan(), use el() instead of [[1]], and gamma(n+1)=factorial(n) which I believe saves a byte, and I think as.numeric is the same as as.double for integers, which also saves a byte, and you can use toString instead of as.character. – Giuseppe Apr 25 '17 at 15:26
• @Giuseppe Thanks for the input, updated. – Neil Apr 25 '17 at 16:07

# Java 9 JSHell, 213 bytes

n->{Set<Integer>s=new HashSet<>();
return IntStream.iterate(n,i->(""+i).chars()
.map(x->x<50?1:IntStream.rangeClosed(2,x-48)
.reduce(1,(a,b)->a*b)).sum()).boxed()


Try it online!

Note: This solution relies on the string representation of a number having code points in the range 48-57. Works for ASCII, UTF-8, Latin-1, all ISO-8859-* character sets, most code pages. Does not work for EBCDIC. I don't think anyone will deduct points for that. :)

Ungolfed:

Function<Integer, List<Integer>> f =        // function from Integer to List of Integer
n -> {
Set<Integer> s = new HashSet<>();       // memo of values we've seen
return IntStream.iterate(n,             // iterate over n, f(n), f(f(n)), etc.
i -> (""+i).chars()                     // the sumFac function; for all chars
.map(x -> x < 50? 1 :               // give 1 for 0! or 1!
IntStream.rangeClosed(2, x-48)      // else produce range 2..d
.reduce(1,(a,b)->a*b))              // reduction to get the factorial
.sum())                             // and sum up the factorii!

// now we have a stream of ints
// from applying sumFac repeatedly
.boxed()                            // box them into Integers (thanks, Java)
.takeWhile(x->s.add(x))             // and take them while not in the memo
.collect(Collectors.toList());      // collect them into a list
}


Notes:

• The return value of Set::add is very helpful here; returns true iff the item was not in the set
• I was being sarcastic when I said "Thanks, Java"
• factorii isn't really a word; I just made that up
• I admit this is original ! Nice job :) – user68509 Apr 25 '17 at 22:05
• @Antoine I admit, Java isn't the best language to golf in, but it's hardly the craziest thing I've done lately. :) codegolf.stackexchange.com/a/117644/794 – David Conrad Apr 25 '17 at 22:18

# Pyth, 22 11 bytes

.usm.!sd+Nk


Try it online!

Lots of credit to Leaky Nun's answer, which introduced me to .u, and helped save a massive 11 bytes of this program.

Explanation:

.usm.!sd+NkQ | ending Q is implicitly added
| Implicit: Q = eval(input())
.u         Q | Repeat the function with initial value Q until a previous value is found. Return all intermediate values
s          | Summation
m.!sd     | For each character 'd' in the string, convert to integer and take the factorial
+Nk  | Convert function argument to string

• Pyth has more useful functions than one imagines. See my answer as reference. – Leaky Nun Apr 26 '17 at 7:48
• @LeakyNun I've rewritten my answer to use .u. I guess I'll need to take a look through the character reference again to see if there are any other useful functions there. – K Zhang Apr 27 '17 at 0:11
• You can use N to convert to string instead of +Nk. – Leaky Nun Apr 27 '17 at 2:10
• @LeakyNun Where the N would be obsolete then, and there comes a 9-byte solution... – Erik the Outgolfer Nov 4 '17 at 22:26

# Husk, 6 bytes

U¡oṁΠd


Try it online!

# Axiom, 231 bytes

l(a:NNI):List NNI==(r:List NNI:=[];repeat(r:=cons(a rem 10,r);a:=a quo 10;a=0=>break);r)
g(a:NNI):NNI==reduce(+,[factorial(x) for x in l(a)])
h(a:NNI):List NNI==(r:=[a];repeat(a:=g(a);member?(a,r)=>break;r:=cons(a,r));reverse(r))


not golfed functions and some test

-- convert one NNI in its list of digits
listify(a:NNI):List NNI==
r:List NNI:=[]
repeat
r:=cons(a rem 10,r)
a:=     a quo 10
a=0=>break
r

-- g(1234)=1!+2!+3!+4!
SumfactorialDigits(a:NNI):NNI==reduce(+,[factorial(x) for x in listify(a)])

ListGenerateFromSumFactorialDigits(a:NNI):List NNI==
r:=[a]
repeat
a:=SumfactorialDigits(a)
member?(a,r)=>break
r:=cons(a,r)
reverse(r)

(9) -> h 132
(9)
[132, 9, 362880, 81369, 403927, 367953, 368772, 51128, 40444, 97, 367920,
368649, 404670, 5810, 40442, 75, 5160, 842, 40346, 775, 10200, 6, 720,
5043, 151, 122, 5, 120, 4, 24, 26, 722, 5044, 169, 363601, 1454]


# Java 7, 220 bytes

String c(int n){String r=n+",",c;for(;!r.matches("^"+(c=(n=d(n))+",")+".*|.*,"+c+".*");r+=c);return r;}int d(int n){int s=0;for(String i:(n+"").split(""))s+=f(new Long(i));return s;}long f(long x){return x<2?1:x*f(x-1);}


Explanation:

String c(int n){                            // Method with integer parameter and String return-type
String r=n+",",                           //  Result-String (which starts with the input integer + a comma
c;                                 //  Temp String
for(;!r.matches(                          //  Loop as long as the result-String doesn't match the following regex:
"^"+(c=(n=d(n))+",")+".*|.*,"+c+".*");  //    "^i,.*|.*,i,.*" where i is the current integer
//   n=d(n) calculates the next integer in line
//   c=(n=d(n))+"," sets the temp String to this integer + a comma
r+=c                                    //   And append the result-String with this temp String
);                                        //  End of loop
return r;                                 //  Return the result-String
}                                           // End of method

int d(int n){                               // Separate method (1) with integer parameter and integer return-type
int s=0;                                  //  Sum
for(String i:(n+"").split(""))            //  Loop over the digits of n
s+=f(new Long(i));                      //   And add the factorial of these digits to the sum
//  End of loop (implicit / single-line body)
return s;                                 //  Return the sum
}                                           // End of separate method (1)

long f(long x){                             // Separate method (2) with long parameter and long return-type (calculates the factorial)
// (NOTE: 2x long and the new Long(i) is shorter than 2x int and new Integer(i), hence long instead of int)
return x<2?                               //  If x is 1:
1                                     //   return 1
:                                       //  Else:
x*f(x-1);                             //   return x multiplied by the recursive-call of x-1
}                                           // End of method (2)


Test code:

Try it here.

class M{
String c(int n){String r=n+",",c;for(;!r.matches("^"+(c=(n=d(n))+",")+".*|.*,"+c+".*");r+=c);return r;}int d(int n){int s=0;for(String i:(n+"").split(""))s+=f(new Long(i));return s;}long f(long x){return x<2?1:x*f(x-1);}

public static void main(String[] a){
System.out.println(new M().c(132));
}
}


Output:

132,9,362880,81369,403927,367953,368772,51128,40444,97,367920,368649,404670,5810,40442,75,5160,842,40346,775,10200,6,720,5043,151,122,5,120,4,24,26,722,5044,169,363601,1454,


# GolfScript, 44 bytes

~]{..)\;10base{,1\{)*}/}%{+}*.@\+@@?)!}do);

~                                             eval input
]                                            initialization
{                                   }do     do...
..)\;                                          initialization
10base                                    get digits
{,1\{)*}/}%                         factorial each
{+}*                     sum
);   drop last element
  tostring


Try it online!

The factorial part is from here.

## C, 161 bytes

f(a){return a?a*f(a-1):1;}
a(l){return l?f(l%10)+a(l/10):0;}
c,t,o;r(i){for(t=o=i;t=a(t),o=a(a(o)),c=t^o;);for(t=i;t^c;printf("%d ",t),c=c|t^o?c:o,t=a(t),o=a(o));}


See it work online.

# TI-BASIC, 857964 60 bytes

:Prompt L₁                             //Get input as 1 length list, 4 bytes
:Lbl C                                //create marker for looping, see below, 3 bytes
:int(10fPart(Xseq(10^(~A-1),A,0,log(X //split input into list of digits, 20 bytes
:sum(Ans!→X                           //factorial and sum the list, write to new input, 6 bytes
:If prod(L₁-X                         //Test to see if new element is repeated, see below, 7 bytes
:Then                                 //Part of If statement, 2 bytes
:augment(L₁,{X→L₁                     //Push new input to List 1, 10 bytes
:Goto C                               //Loop back to beginning, 3 bytes
:Else                                 //Part of If statement, 2 bytes


Since this is running on a graphing calculator, there is limited RAM. Try testing it with numbers that loop quickly, like 169.

More Explanation:

:int(10fPart(Xseq(10^(~A-1),A,0,log(X
seq(10^(~A-1),A,0,log(X //Get a list of powers of 10 for each digit (i.e. 1, 0.1, 0.01, etc.)
X                        //Multiply by input
fPart(                         //Remove everything but the decimal
10                               //Multiply by 10 (move one digit in front of the decimal
:int(                                 //Truncate to an integer


If prod(L₁-X works by subtracting the new element from the old list, then multiplying all the elements of the list together. If the element was already in the list, the product will be 0, a falsey value. Otherwise, the product will be a positive integer, a truthy value.

# J, 40 31 bytes

Edit: 9 bytes saved using the improvements by FrownyFrog. Thanks!

f=.$:@,~]@.e.~[:+/@:!10#.inv{:  Original code: f=.[($:@,)@.([:-.e.~)[:+/!@("."0&":@{:)


In this case I decided to count the bytes for the verb definition, since otherwise it doesn't work in the interpreter.

Explanation:


(        {:) - Takes the last element of the array
":@    - converts it to string
"."0&       - converts each character back to integer
!@             - finds the factorials
+/               - sums them
[:                 - cap (there are 2 derived verbs above, we need 3 for a fork)
(    e.~)                   - check if the result is present in the list
-.                       - negates the above check
[:                         - cap
@.                            - Agenda conjunction, needed for the recursion
($:@,) - if the result is not in the list, add it to the list and recur [ - if the result is in the list, display it and stop  Try it online! • ([:-.e.~) -> (1-e.~) – FrownyFrog Nov 4 '17 at 17:34 • 31 bytes – FrownyFrog Nov 4 '17 at 18:13 • @FrownyFrog Thanks, your code is much better! I tried 10#.inv early while experimenting, but then I wrote it 10&#.inv and it was longer, so I rejected it. Thanks for all your suggestions! I have much to learn :) – Galen Ivanov Nov 4 '17 at 18:46 • @FrownyFrog Reversing the cases for the agenda is so good, I regret I didn't see it :) – Galen Ivanov Nov 4 '17 at 18:53 • [:+/!@"."0@":@{: is the same length, so there's no improvement with 10#.inv. Just had to drop the (). – FrownyFrog Nov 4 '17 at 18:54 # Tcl, 143 bytes proc P {n L\ ""} {proc F n {expr$n?($n)*\[F$n-1]:1}
while {[set n [expr [join [lmap d [split $n ""] {F$d}] +]]] ni $L} {lappend L$n}
set L}


Try it online!

# APL (Dyalog Extended), 1716 15 bytes

Saved 1 byte thanks to @Adám and another later

f←(#+.(!⍎)⍕)⌂traj


Try it online!

traj is a function in the dfns workspace that runs a function until a cycle is detected. Here, that function is #+.(!⍎)⍕. ⍕ turns the argument into a string. Each character/digit (¨) is executed/turned into a number (⍎). ! finds the factorial of each of these digits. Then + adds up all these factorials. # is used because #⍎foo just executes foo, so it's #f.⍎ is practically the same as {f/⍎¨⍵} here.

# APL (Dyalog Unicode), 25 20 bytes

Saved 5 bytes thanks to @Adám!

1↓{⍵,⍨+/!⍎¨⍕⊃⍵}⍣(⊃∊)


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Output is reversed.

This is very similar to the code above, but uses the power operator (⍣) to run the dfn until a cycle is reached, then drops the first element (1↓) because it runs 1 extra time. {⍵,⍨+/!⍎¨⍕⊃⍵} applies the function in the question to the first element of the argument (⊃⍵), then prepends it to the argument (⍵,⍨). Each time it is called, the list grows by one. The train ⊃∊ is applied with the next result on the left side and the previous result on the right side and tells ⍣ when to stop. ∊ is membership and tells which of the next result's elements are in the previous results (if one of them is repeated, there is a cycle). Then it picks the first of these booleans (⊃) because we need a single value and the rest have been checked already in previous iterations.

# PowerShell Core, 87 bytes

param($r)for(;$r-notin$l){$l+=,$r$r=("$r"|% t*y|%{1.."$_"-ne0-join'*'})-join'+'|iex}$l  Try it online! Thanks to mazzy for saving a lot of bytes! ## Explained (old implementation, but close enough) param($r)for(;$r-notin$l                       # Checks that the last computed value isn't an existing one
){$l+=,$r              # Adds it to the list

$r=("$r"|                                      # Converts r as a string
% t*y|                                   # Converts it to a char array
%{1.."$_"-ne0-join'*'} # Builds the factorial calculation strings and handles 0!. e.g. 1*2*3*4 )-join'+'|iex} # Joins them with "+" and computes it. e.g. 1*2*3*4+1*2*3$l                                             # Returns all the values in the list

• Thanks. I think you can save some bytes if you use an array instead of hashtable – mazzy Dec 23 '20 at 9:53
• I tried but arrays have fixed size and I couldn't find any short way of instantiating one. i.e. [System.Collections.ArrayList]::New() – Julian Dec 23 '20 at 20:17
• Another 117 bytes option that raises a lot of error but returns a correct result – Julian Dec 24 '20 at 1:50
• – mazzy Dec 24 '20 at 7:43

# Perl 5-MList::Util=product,sum -n, 49 bytes

say&&($_=sum map{product 1..$_}/./g)until$k{$_}++


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• You can save 2 bytes by changing && to ,, ) to a space, and by removing (. Try it online! – mik Mar 15 at 13:21

# Wolfram Language (Mathematica), 72 bytes

Most@NestWhileList[Total[IntegerDigits@#!]&,#,DuplicateFreeQ@*List,All]&


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Keeps applying sumFac to it's own output until it sees a duplicate value, then returns a list of all values excluding that final duplicate.

# Stax, 10 bytes

│¼µ\Ä▬Éε╩X


Run and debug it

# Perl 5-n, 65 bytes

say,(map{$f=1;map{$f*=$_}2..$_;$.+=$f}/./g),$_=$.until$.=$r{$_}++  Try it online! Uses map as regular loops (ignoring return value), and $. as a temporary variable (to save space before until).