# Recover the power from the prime power

It seems that many people would like to have this, so it's now a sequel to this challenge!

Definition: a prime power is a natural number that can be expressed in the form pn where p is a prime and n is a natural number.

Task: Given a prime power pn > 1, return the power n.

Testcases:

input output
9     2
16    4
343   3
2687  1
59049 10


Scoring: This is . Shortest answer in bytes wins.

• Note: This challenge might be trivial in some golfing languages, but it's not so trivial for some mainstream languages, as well as the language of June 2018, QBasic. Commented Jun 20, 2018 at 23:55
• Can we output True instead of 1? Alternatively, float instead of ints?
– Jo King
Commented Jun 21, 2018 at 0:33
• @JoKing yes, yes. Commented Jun 21, 2018 at 0:34
• @EriktheOutgolfer Challenge accepted :D Commented Jun 22, 2018 at 4:46

# Python 2, 37 bytes

f=lambda n,i=2:i/n or(n%i<1)+f(n,i+1)


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Counts factors. Apparently I wrote the same golf in 2015.

Narrowly beats out the non-recursive

Python 2, 38 bytes

lambda n:sum(n%i<1for i in range(1,n))


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# 05AB1E, 2 bytes

Òg


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# Python 3, 49 bytes

f=lambda n,x=2:n%x and f(n,x+1)or n/x<2or-~f(n/x)


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Outputs True instead of 1 (as allowed by OP). Recursive function that repeatedly finds the lowest factor and then calls the function again with the next lowest power until it reaches 1. This is an extension of my answer to the previous question.

# Stax, $\require{cancel}\xcancel 4 3$ bytes

|f%


Run and debug it

Length of prime factorization.

• Ahh.. you're breaking the crossed out 4 is still regular 4 ;( meme. ;p (It was getting old anyway though.. So well done I guess) Commented Jun 21, 2018 at 7:47
• $\text{Yay, MathJax abuse!}$ But remember to put the cross before the actual bytecount otherwiae the leaderboard snippet may not be able to recognize it. Commented Jun 22, 2018 at 8:53

# Bash + GNU utilities, 22

• 2 bytes saved thanks to @H.PWiz and @Cowsquack
factor|tr -cd \ |wc -c


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• Does factor|sed s/\ //|wc -w work? Commented Jun 21, 2018 at 10:32
• What about factor|tr -cd \ |wc -c? Commented Jun 21, 2018 at 11:10
• Turns out awk is indeed shorter, factor|awk $0=NF-1 Commented Mar 31, 2021 at 9:40 # R 22 bytes Power n is the number of multiples of p in p^n when p is prime: sum(!(b<-scan())%%2:b)  Try it online! ## Nibbles 5 bytes ,|,~^%@  This is 9 nibbles each of which is encoded in a half byte in the binary form. I think this is the shortest solution that doesn't use built in factoring. Translation: , length | filter , 0..input ~ not \x-> ^ pow (so that 0 which would have been 0 from the mod isn't) % mod @ input implicit  (x) implicit  (x)  It works by just counting the number of numbers the input divides evenly You could run it passing in a list of numbers to process in stdin or as a command line arg. Nibbles isn't on TIO.run yet... # dc, 50 41 bytes 1si[dli1+dsi%0<X]dsXx[dli/dli<Y]sYdli<Yzp  Try it online! Takes input from the top of the stack (in TIO, put the input in the header to load it onto the stack before execution). Outputs to stdout. # Explanation Registers used: i: the current trial divisor, while X is running. Later, the divisor we've found. X: the macro dli1+dsi%0<X, which has the effect "increment i, then check the modulus with the value on the stack (which will be the original input). If it's not zero, repeat". Y: the macro dli/dli<Y, which has the effect "Add to the stack a copy of the current top of the stack, divided by i. Repeat until i is reached." Full program: 1si Initialize i [dli1+dsi%0<X]dsXx Define and run X [dli/dli<Y]sY Define Y dli<Y Run Y, but only if needed (if the input wasn't just i) z The stack is i^n, i^(n-1), ... ,i, so print the stack depth  • I found a much better solution! Editing now... Commented Jun 21, 2018 at 1:13 # Pyth, 2 Count prime factors: lP  # face, 86 bytes (%d@)$*,c'$,io>Av"[""mN*c?*m1*mp*m%*s1"$pN1p:~+p1p%%Np?%~:=/NNp+?1?-%N1?%=p%'$i?w1'%>  Hooray, longer than Java! Try it online! I am particularly fond of the trick of using the return value of sscanf. Normally the return value would be discarded, but here it will always be 1, because we're always reading a single number as input. We can take advantage of this by assigning its return value to the variable 1, saving the 2 bytes that would otherwise be required to assign 1 to 1 explicitly. (%d@) \$*,c'$,io> ( setup - assign$ to "%d", * to a number, o to stdout )
Av"[""mN*    ( set " to input and allocate space for N for int conversion )
c?*          ( calloc ?, starting it at zero - this will be the output )
m1*          ( allocate variable "1", which gets the value 1 eventually )
mp*m%*       ( p is the prime, % will be used to store N mod p )

s1"$pN ( scan " into N with$ as format; also assigns 1 to 1 )

1p:~         ( begin loop, starting p at 1 )
+p1p       ( increment p )
%%Np       ( set % to N mod p )
?%~          ( repeat if the result is nonzero, so that we reach the factor )

:=           ( another loop to repeatedly divide N by p )
/NNp       ( divide N by p in-place )
+?1?       ( increment the counter )
-%N1       ( reuse % as a temp variable to store N-1 )
?%=          ( repeat while N-1 is not 0 -- i.e. break when N = 1 )

p%'$i? ( sprintf ? into ', reusing the input format string ) w1'%> ( write to stdout )  # Attache and Wolfram Language (Mathematica) polyglot, 10 bytes PrimeOmega  Simply a builtin for computing the number of prime factors N has. ## Explanation Since N = pk, Ω(N) = Ω(pk) = k, the desired result. # Whitespace, 141 bytes [S S S N _Push_0][S N S _Duplicate_0][T N T T _Read_STDIN_as_number][T T T _Retrieve][S S S T N _Push_1][N S S N _Create_Label_LOOP_1][S S S T N _Push_1][T S S S _Add][S N S _Duplicate][S T S S T S N _Copy_2nd_input][S N T _Swap_top_two][T S T T _Modulo][N T S S N _If_0_Jump_to_Label_BREAK_1][N S N N _Jump_to_Label_LOOP_1][N S S S N _Create_Label_BREAK_1][S S S N _Push_0][S T S S T S N _Copy_2nd_input][N S S T N _Create_Label_LOOP_2][S N S _Duplicate_input][S S S T N _Push_1][T S S T _Subtract][N T S S S N _If_0_Jump_to_Label_BREAK_2][S N T _Swap_top_two][S S S T N _Push_1][T S S S _Add][S N T _Swap_top_two][S T S S T S N Copy_2nd_factor][T S T S _Integer_divide][N S N T N _Jump_to_Label_LOOP_2][N S S S S N _Create_Label_BREAK_2][S N N _Discard_top][T N S T _Print_as_number]  Letters S (space), T (tab), and N (new-line) added as highlighting only. [..._some_action] added as explanation only. Try it online (with raw spaces, tabs and new-lines only). Explanation in pseudo-code: Integer n = STDIN as input Integer f = 1 Start LOOP_1: f = f + 1 if(n modulo-f == 0) Call function BREAK_1 Go to next iteration of LOOP_1 function BREAK_1: Integer r = 0 Start LOOP_2: if(n == 1) Call function BREAK_2 r = r + 1 n = n integer-divided by f Go to next iteration of LOOP_2 function BREAK_2: Print r as number to STDOUT Program stops with an error: Exit not defined  Example run: input = 9 Command Explanation Stack Heap STDIN STDOUT STDERR SSSN Push 0 [0] SNS Duplicate top (0) [0,0] TNTT Read STDIN as number [0] {0:9} 9 TTT Retrieve [9] {0:9} SSSTN Push 1 [9,1] {0:9} NSSN Create Label_LOOP_1 [9,1] {0:9} SSSTN Push 1 [9,1,1] {0:9} TSSS Add top two (1+1) [9,2] {0:9} SNS Duplicate top (2) [9,2,2] {0:9} STSSTSN Copy 2nd from top [9,2,2,9] {0:9} SNT Swap top two [9,2,9,2] {0:9} TSTT Modulo top two (9%2) [9,2,1] {0:9} NTSSN If 0: Jump to Label_BREAK_1 [9,2] {0:9} NSNN Jump to Label_LOOP_1 [9,2] {0:9} SSSTN Push 1 [9,2,1] {0:9} TSSS Add top two (2+1) [9,3] {0:9} SNS Duplicate top (3) [9,3,3] {0:9} STSSTSN Copy 2nd [9,3,3,9] {0:9} SNT Swap top two [9,3,9,3] {0:9} TSTT Modulo top two (9%3) [9,3,0] {0:9} NTSSN If 0: Jump to Label_BREAK_1 [9,3] {0:9} NSSSN Create Label_BREAK_1 [9,3] {0:9} SSSN Push 0 [9,3,0] {0:9} STSSTSN Copy 2nd from top [9,3,0,9] {0:9} NSSTN Create Label_LOOP_2 [9,3,0,9] {0:9} SNS Duplicate top (9) [9,3,0,9,9] {0:9} SSSTN Push 1 [9,3,0,9,9,1] {0:9} TSST Subtract top two (9-1) [9,3,0,9,8] {0:9} NTSSSN If 0: Jump to Label_BREAK_2 [9,3,0,9] {0:9} SNT Swap top two [9,3,9,0] {0:9} SSSTN Push 1 [9,3,9,0,1] {0:9} TSSS Add top two (0+1) [9,3,9,1] {0:9} SNT Swap top two [9,3,1,9] {0:9} STSSTSN Copy 2nd from top [9,3,1,9,3] {0:9} TSTS Integer-divide top two (9/3) [9,3,1,3] {0:9} NSNTN Jump to Label_LOOP_2 [9,3,1,3] {0:9} SNS Duplicate top (3) [9,3,1,3,3] {0:9} SSSTN Push 1 [9,3,1,3,3,1] {0:9} TSST Subtract top two (3-1) [9,3,1,3,2] {0:9} NTSSSN If 0: Jump to Label_BREAK_2 [9,3,1,3] {0:9} SNT Swap top two [9,3,3,1] {0:9} SSSTN Push 1 [9,3,3,1,1] {0:9} TSSS Add top two (1+1) [9,3,3,2] {0:9} SNT Swap top two [9,3,2,3] {0:9} STSSTSN Copy 2nd from top [9,3,2,3,3] {0:9} TSTS Integer-divide top two (3/3) [9,3,2,1] {0:9} NSNTN Jump to Label_LOOP_2 [9,3,2,1] {0:9} SNS Duplicate top (1) [9,3,2,1,1] {0:9} SSSTN Push 1 [9,3,2,1,1,1] {0:9} TSST Subtract top two (1-1) [9,3,2,1,0] {0:9} NTSSSN If 0: Jump to Label_BREAK_2 [9,3,2,1] {0:9} NSSSSN Create Label_BREAK_2 [9,3,2,1] {0:9} SNN Discard top [9,3,2] {0:9} TNST Print as integer [9,3] {0:9} 2 error  Program stops with an error: No exit found. # Brachylog, 2 bytes ḋl  Try it online! ### Explanation ḋ Prime decomposition l Length  # APL (Dyalog Extended), 8 2 bytes ≢⍭  Try it online! ⍭ finds factors, ≢ counts how many of them there are. # Add++, 42 bytes D,f,@,bUbU$^=
L,dVfbUG$XGRzGGXzÞ{f}bUbU0$:


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I don't even know where to begin explaining this mess.

## Explained

D,f,@,bUbU$^= D,f,@, ; a helper function f that given a list [number, [x, y]] bUbU$^  ;   returns whether x ^ y
= ;   equals number

L,dVfbUG$XGRzGGXzÞ{f}bUbU0$:
L,                            ; a lambda that
dV                          ; places its input into the register
fbU                       ; and gets the prime factor of the input. This is guaranteed to be a single item because the input is a prime raised to a power.
G$X ; push a list of input copies of that power GRz ; and zip that with the range [1...input] GGX ; also, push input copies of the input z ; and zip that with our big list. I'm calling it a big list because it is what it is. Þ{f} ; filter that list based on the results of the helper function f bUbU0$:  ; get the power out of the many nested lists returned.

• 34 bytes. You can replace f with g, per this tip to save 2 bytes, use the full flatten command BF instead of two unpacks and use some stack manipulation to replace 0\$: Commented Jan 5, 2022 at 13:18

# QBasic, 51 41 bytes

INPUT n
FOR i=2TO n
f=f-(n/i=n\i)
NEXT
?f


-10 bytes by copying the approach from Darren Smith's Nibbles answer: For a prime power input, the desired output equals the number of integers between 1 (exclusive) and the input (inclusive) that evenly divide the input.

INPUT number
FOR testFactor = 2 TO number
' number is divisible by testFactor if their float division equals
' their int division
isDivisible = (number / testFactor = number \ testFactor)
' Truthy is -1 in QBasic, so we subtract rather than add to the tally
numFactors = numFactors - isDivisible
NEXT testFactor
PRINT numFactors


# HBL, 7 bytes

(or possibly 6.5 depending on how this meta question shakes out)

+(*'?(*%.(02.


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### Explanation

+(*'?(*%.(
(0    Inclusive range
2    from 2
.   to the argument
(*        Map over each value x in that list:
%.       Argument mod x
(*            Map over each value in that list:
'?           Logical negation
The result is a list containing 1 for each number that divides
the argument, 0 otherwise
+              Take its sum


# Risky, 3 bytes

!\?___


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A basically built-in solution for now, working on a non-trivial version (might not be possible given Risky's heavy investment in specific operators, and generally awful control flow).

!      Count
\     Prime factors
?    Input


# Nekomata, 1 byte

ƒ


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ƒ factorizes the input number, and returns a list of unique prime factors and a list of exponents.

Only the top of the stack will be printed, which is the list of exponents. Since the input is a prime power, this will be a singleton list.

# Java 8, 59 bytes

A lambda from int to int.

x->{int f=1,c=0;while(x%++f>0);for(;x>1;c++)x/=f;return c;}


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# J, 4 bytes

#@q:


q: gives the list of prime factors, # gives the length of the list.

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# R, 37 bytes

length(numbers::primeFactors(scan()))


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• sum(x|1) is nearly always shorter than length(x) Commented Jun 21, 2018 at 15:56

# MATL, 3 bytes

Yfz


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### Explanation:

     % Implicit input: 59049
Yf   % Factorize input [3, 3, 3, 3, 3, 3, 3, 3, 3, 3]
z  % Number of non-zero elements: 10
% Implicit output


# Jelly, 3 2 bytes

Æḍ


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# Vyxall, 1 byte

ǐ


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-1 byte thx to @lyxal

Length of the prime factors, the flags are cheaty+awesome

• Try it Online! for 1 byte Commented May 22, 2021 at 11:44

# MS Excel, 33 bytes

An anonymous worksheet function that takes input from cell A1 and outputs to the calling cell

-SUM(-(MOD(A1,SEQUENCE(A1))<1))-1


# Thunno 2L, 1 byte

f


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Push the prime factors of the input then take the Length of the list.

# Python 2, 62 bytes

def f(n,p=2,i=0):
while n%p:p+=1
while n>p**i:i+=1
return i


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Nothing fancy here.

• You can save three bytes by making it a full program: Try it online! Commented Jun 21, 2018 at 0:23

# Japt, 3 bytes

k l


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## Explanation:

k l
k     Get the prime factors of the input
l   Return the length


# Actually, 2 bytes

ol


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