# Primes ’n’ Digits

This has no practical purpose but it could be fun to golf.

# Challenge

Given a number n,

1. Count the amount of each digit in n and add 1 to each count
2. Take the prime factorization of n
3. Count the amount of each digit in the prime factorization of n, without including duplicate primes
4. Create a new list by multiplying together the respective elements of the lists from steps 1 and 3
5. Return the sum of that list

For example, 121 has two 1s and a 2, so you would get the following list from step 1:

0 1 2 3 4 5 6 7 8 9
1 3 2 1 1 1 1 1 1 1


The prime factorization of 121 is 112, which gives the following list for step 3:

0 1 2 3 4 5 6 7 8 9
0 2 0 0 0 0 0 0 0 0


Note how we did not count the exponent. These multiply together to get:

0 1 2 3 4 5 6 7 8 9
0 6 0 0 0 0 0 0 0 0


And the sum of this list is 6.

# Test cases

1 -> 0
2 -> 2
3 -> 2
4 -> 1
5 -> 2
10 -> 2
13 -> 4
121 -> 6


# Notes

• Standard loopholes are forbidden.
• Input and output can be in any reasonable format.
• You should leave ones (or zeros for step 3) in the list for digits that did not appear in the number.
• This is , so the shortest solution in bytes wins.
• Does 667 (=23*29) make for two 2s, one 3, and one 9 in step 3? Dec 23 '17 at 18:10
• @JonathanAllan Yes. Dec 23 '17 at 18:19
• @wizzwizz4 232792560 -> [2,1,4,2,1,2,2,2,1,2] (step 1); 2*2*2*2*3*3*5*7*14*17*19 (step 2); so [0,5,1,2,0,1,0,2,0,1] (step 3); then [0,5,4,4,0,2,0,4,0,2] (Step 4); and hence should output 21. Dec 23 '17 at 18:26
• @JonathanAllan It would be nice if I could count. :-/ Dec 23 '17 at 18:36

# Jelly, 16 bytes

ṾċÐ€ØD
ÆfQÇ×Ç‘$S  Try it online! Developed independently from and not exactly the same as the other Jelly solution. Explanation I'm gong to use 242 as an example input. ṾċÐ€ØD Helper link Ṿ Uneval. In this case, turns it's argument into a string. 242Ṿ → ['2','4','2']. [2,11] → ['2', ',', '1', '1']. The ',' won't end up doing anything. ØD Digits: ['0','1',...,'9'] ċÐ€ Count the occurrence of €ach digit in the result of Ṿ ÆfQÇ×Ç‘$S  Main link. Argument 242
Æf         Prime factors that multiply to 242 → [2,11,11]
Q        Unique elements → [2,11]
Ç       Apply helper link to this list → [0,2,1,0,0,0,0,0,0,0]
Ç‘$Apply helper link to 242 then add 1 to each element → [1,1,3,1,2,1,1,1,1,1] × Multiply the two lists element-wise → [0,2,3,0,0,0,0,0,0,0] S Sum of the product → 5  # Jelly, 18 17 bytes -1 byte thanks to caird coinheringaahing & H.PWiz (avoid pairing the two vectors) DF‘ċÐ€⁵ ÆfQÇæ.Ç‘$


A monadic link taking a positive integer and returning a non-negative integer.

Try it online!

### How?

DF‘ċÐ€⁵ - Link 1, digitalCount: number(s)    e.g. [13,17]
D       - to decimal list (vectorises)            [[1,3],[1,7]]
F      - flatten                                 [1,3,1,7]
‘     - increment (vectorises)                  [2,4,2,8]
⁵ - literal ten                             10
Ð€  - map across              (implicit range [1,2,3,4,5,6,7,8,9,10])
ċ    - count                                   [0,2,0,1,0,0,0,1,0,0]

ÆfQÇæ.Ç‘$- Main link: positive integer, n e.g. 11999$ - last two links as a monad:
‘  -   increment (vectorises)              [1,3,1,1,1,1,1,1,1,4]
Æf        - prime factorisation                   [13,13,71]
Q       - deduplicate                           [13,17]
æ.    - dot product                           8

• 17 bytes Dec 23 '17 at 19:34
• Or use dot product Dec 23 '17 at 19:45

# APL (Dyalog), 43 41 bytes

⎕CY'dfns'
+/×/+/¨⎕D∘.=⍕¨(⎕D,r)(∪3pco r←⎕)


Try it online!

How?

r←⎕ - input into r

3pco - prime factors

∪ - unique

⎕D,r - r prepended with 0-9

⍕¨ - format the factors and the prepended range

⎕D∘.= - cartesian comparison with every element of the string 0123456789

+/¨ - sum each row of the two tables formed

×/ - multiply the two vectors left

+/ - sum the last vector formed

# Pip, 44 bytes

Y_N_.aM,tT++o>aTa%o{a/:olPBo}\$+y*Y_N JUQlM,t


Takes input from command-line argument. Try it online!

# Python 2, 136 127 bytes

lambda a:sum(''.join(u(a)).count(i)*-~a.count(i)for i in range(10))
u=lambda a:[jfor j in range(2,a)if a%j<1>len(u(j))]


Try it online!

# Credits

• 127 bytes Dec 23 '17 at 20:42
• @Mr.Xcoder Updated, thanks for showing me the use of -~ I was always a bit confused on that. And I need to start remembering the <1 thing. Thanks for the help.
– Neil
Dec 23 '17 at 20:45
• You can take a look through this for -~ and related stuff. Dec 23 '17 at 20:48