Jelly, 10 bytes

³3,5ḍḄ⁼ʋ#Ṫ

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This uses0 = Number,1 = Buzz,2 = Fizz and3 = FizzBuzz

Jelly, 8 bytes

1g15=ɗ#Ṫ

Try it online!

Thanks toLynn. This uses1 = Number,3 = Fizz,5 = Buzz,15 = FizzBuzz. Included separately as the numbers could encode extra data

How they work

³3,5ḍḄ⁼ʋ#Ṫ - Main link. Takes W=0,1,2,3 on the left, n on the right       ʋ   - Last 4 links as a dyad f(k, W): 3,5ḍ      -   Divisible by 3 or 5? Yields [0,0], [0,1], [1,0], [1,1]     Ḅ     -   From binary; Yields 0, 1, 2, 3      ⁼    -   Equals W?³       #Ṫ - Starting from W, count up k = W, W+1, ..., returning the nth integer such that f(k, W) is true

1g15=ɗ#Ṫ - Main link. Takes W=1,3,5,15 on the left, n on the right     ɗ   - Last 3 links as a dyad f(k, W): g15     -   GCD(k, 15)    =    -   Does that equal W?1     #Ṫ - Count up k = 1, 2, ..., returning the nth integer such that f(k, W) is true

• 5
  \$\begingroup\$1g15=ɗ#Ṫ works for 8, with Number = 1, Fizz = 3, Buzz = 5, FizzBuzz = 15.\$\endgroup\$

  lynn

  – lynn

  2021-12-17 20:06:23 +00:00

  CommentedDec 17, 2021 at 20:06
• \$\begingroup\$@Lynn I've added that as an alternative, as using the specific values might be taken to encode extra data\$\endgroup\$

  caird coinheringaahing

  – caird coinheringaahing♦

  2021-12-17 20:10:13 +00:00

  CommentedDec 17, 2021 at 20:10
• \$\begingroup\$I was going to upvote, but I accidentally downvoted. Now fixed.\$\endgroup\$

  Alan Bagel

  – Alan Bagel

  2021-12-18 11:53:30 +00:00

  CommentedDec 18, 2021 at 11:53

JavaScript (ES6), 44 bytes

Expects(s)(n) with s=0 forNumber, s=1 forFizz, s=2 forBuzz and s=3 forFizzBuzz.

(s,k=0)=>g=n=>n?g(n-=s==!(++k%3)+2*!(k%5)):k

Try it online!

Cheaty version, 35 bytes

Expects 4680 forFizz, 1056 forBuzz, 1 forFizzBuzz and 27030 forNumber.

(s,k=0)=>g=n=>n?g(n-=s>>++k%15&1):k

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Python 2, 47 bytes

f=lambda n,c,k=1:n and-~f(n-(k**4%15==c),c,k+1)

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Take in the category labelc as:

Number -> 1Fizz -> 6Buzz -> 10FizzBuzz -> 0

We fingerprint the category fork usingk**4%15, which produces the corresponding value as listed above. This is wrapped in arecursive function for the n'th number meeting a condition.

Python 2, 54 bytes

lambda n,c:([n*7/8-3*~n/4%2,~-n/4,~-n/2,0][c/2%4]+n)*c

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A short at writing direct formulas for each case, with inputc as one of1,3,5,15.

• \$\begingroup\$For the direct formulas you can combine the Fizz and Buzz cases into one:[n*7/8-3*~n/4%2,0,~-n>>6/c][c/3/-3]\$\endgroup\$

  ovs

  – ovs

  2021-12-18 00:05:33 +00:00

  CommentedDec 18, 2021 at 0:05

Perl 5 -p,52 47 bytes

Saved 5 bytes thanks to @Dom Hastings.

/ /;$_=332312332132330x$';/(.*?$`){$'}/g;$_=pos

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Where 0=FizzBuzz, 1=Buzz, 2=Fizz, 3=Number in the first of the two inputs. Finds the position of the n'th occurrence of what's wanted with a regexp search in a repeated 15 char string encoding number, fizz, buzz and fizzbuzz in their right places.

• \$\begingroup\$Nice approach. I ended up with a much longer version... You can save a few bytes swapping the order of input too!:Try it online!\$\endgroup\$

  Dom Hastings

  – Dom Hastings

  2021-12-18 08:31:02 +00:00

  CommentedDec 18, 2021 at 8:31
• 1
  \$\begingroup\$Actually, using/ / and$' / `` $` `` works out a bit better:Try it online!\$\endgroup\$

  Dom Hastings

  – Dom Hastings

  2021-12-18 23:03:08 +00:00

  CommentedDec 18, 2021 at 23:03
• 1
  \$\begingroup\$Thx @DomHastings – I need to remember those special vars.\$\endgroup\$

  Kjetil S

  – Kjetil S

  2021-12-19 16:20:14 +00:00

  CommentedDec 19, 2021 at 16:20

Python 2, 51 bytes

lambda d,n,k=0:n and-~f(d,n-(k%3/2*2+k%5/4==d),k+1)

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-2 bytes thanks to AnttiP; becomes -4 using Python 2
-3 bytes thanks to ovs

0 for Number, 1 for Buzz, 2 for Fizz, 3 for FizzBuzz.

• \$\begingroup\$-2 bytes withlambda d,n,k=0:n and f(d,n-(k%3//2*2+k%5//4==d),k+1)or k\$\endgroup\$

  AnttiP

  – AnttiP

  2021-12-17 20:44:51 +00:00

  CommentedDec 17, 2021 at 20:44
• \$\begingroup\$@AnttiP Thanks! Using python 2 gets another 2 bytes as well\$\endgroup\$

  hyperneutrino

  – hyperneutrino♦

  2021-12-17 21:41:04 +00:00

  CommentedDec 17, 2021 at 21:41
• \$\begingroup\$Using Python 2 is cheating, it doesn't officially exist anymore.\$\endgroup\$

  Mark Ransom

  – Mark Ransom

  2021-12-18 03:42:51 +00:00

  CommentedDec 18, 2021 at 3:42
• 3
  \$\begingroup\$@MarkRansom It's not officially supported by Python anymore but as long as a language has an implementation available it's valid.\$\endgroup\$

  hyperneutrino

  – hyperneutrino♦

  2021-12-18 07:04:23 +00:00

  CommentedDec 18, 2021 at 7:04

R,5654 52 bytes

Edit: -4 bytes thanks to pajonk

function(n,i){while(n<-n-!(!T%%3)-i+2*!T%%5)T=T+1;T}

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This really seems to have too many parentheses...
Edit: This is really pajonk's answer, now, after removing all the useless parentheses that I left in the original...

• \$\begingroup\$Here's one pair of parens less.\$\endgroup\$

  pajonk

  – pajonk

  2021-12-18 06:36:35 +00:00

  CommentedDec 18, 2021 at 6:36
• \$\begingroup\$@pajonk - Thanks! I really should try to understand the precedence rules...\$\endgroup\$

  Dominic van Essen

  – Dominic van Essen

  2021-12-18 08:07:47 +00:00

  CommentedDec 18, 2021 at 8:07
• \$\begingroup\$stat.ethz.ch/R-manual/R-devel/library/base/html/Syntax.html is my favourite site when golfing in R ;)\$\endgroup\$

  pajonk

  – pajonk

  2021-12-18 10:04:05 +00:00

  CommentedDec 18, 2021 at 10:04
• \$\begingroup\$And one pair less again.\$\endgroup\$

  pajonk

  – pajonk

  2021-12-18 10:11:18 +00:00

  CommentedDec 18, 2021 at 10:11
• \$\begingroup\$@pajonk - Thanks. I think I ought to transfer this answer to you, after you've fixed all my parentheses...\$\endgroup\$

  Dominic van Essen

  – Dominic van Essen

  2021-12-18 12:42:49 +00:00

  CommentedDec 18, 2021 at 12:42

Python 2, 77 bytes

0 forNumber, 1 forBuzz, 2 forFizz, 3 forFizzBuzz. The formula forNumber is taken fromA229829.

t,n=input()g=n-1print[g/8*15+g%8*12/5+1+g%8/-3,(n+g/4)*3,(n+g/2)*5,n*15][t]

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05AB1E (legacy), 9bytes

µN53SÖ2βQ

First input is the\$number\$; second is a digit\$0\$ forNumber,\$1\$ forFizz,\$2\$ forBuzz, and\$3\$ forFizzBuzz.

Uses the legacy version of 05AB1E, because it'll implicitly output the indexN after a while-loopµ. In the new 05AB1E version, an explicit trailing}N should be added to accomplish that.

Try it online orverify all test cases.

Explanation:

µ          # Loop until the counter_variable is equal to the first (implicit) input N         #  Push the loop-index  53S      #  Push [5,3]     Ö     #  Check if the loop-index is divisible by either 5 or 3           #  (resulting in [0,0], [1,0], [0,1], or [1,1])      2β   #  Convert it from a base-2 list to an integer           #  ([0,0],[1,0],[0,1],[1,1] will become 0,1,2,3 respectively)        Q  #  And check if it's equal to the second (implicit) input           #  (if this is truthy: implicitly increase the counter_variable by 1)           # (after the loop, the loop-index `N` is output implicitly)

05AB1E (legacy), 6bytes

µN15¿Q

First input is the\$number\$; second is an integer\$1\$ forNumber,\$3\$ forFizz,\$5\$ forBuzz, and\$15\$ forFizzBuzz.

Port of@Lynn's Jelly comment (I now also notice my original program above uses a similar approach as@cairdCoinheringaahing's 10-bytes Jelly program, even though we came up with it independently. Not too surprising, since it's a pretty straight-forward approach.)

Try it online orverify all test cases.

Explanation:

µ       # Loop until the counter_variable is equal to the first (implicit) input N      #  Push the loop-index  15¿   #  Get the GCD (Greatest Common Divisor) of this index and 15     Q  #  And check if it's equal to the second (implicit) input        #  (if this is truthy: implicitly increase the counter_variable by 1)        # (after the loop, the loop-index `N` is output implicitly)

Retina, 60 bytes

L$`^.$'*$&¶$'*$(NNFNBFNNFBNFNNZL$`(.)+¶(?<-1>.*?\1)+$.>%`

Try it online! Link includes test cases. Takes input as the letterF,B,Z orN followed by the numbern. Explanation:

L$`^.

Match the letter. At this point,$' refers to the numbern following it.

$'*$&¶$'*$(NNFNBFNNFBNFNNZ

Repeat the lettern times, then on the next line repeat the Fizz Buzz sequence15n times.

L$`(.)+¶(?<-1>.*?\1)+

Matchn copies of the letter on the first line, and use those to find thenth match of the letter on the second line.

$.>%`

Output the offset (.`) of the end (>) of the match relative to the start of the second line (%).

Desmos,112 99 bytes

Input into the function\$f(n,k)\$, where\$n\$ is the number, and\$k\$ is the word.

Number = 0, Fizz = 3, Buzz = 5, FizzBuzz = 15

h(n)=floor(n)a=mod(n-1,8)f(n,k)=\{k=0:15h(n/8-1/8)+2a+h(2a/5+1)-h(a/3+2/3),k=15:kn,kn+kh(kn/15)\}

Try It On Desmos!

Try It On Desmos! - Prettified

Uses formula inOEIS A229829:

a(n) = 15*floor((n-1)/8) +2*f(n) +floor((2*f(n)+5)/5) -floor((f(n)+2)/3), where f(n) = (n-1) mod 8.

99.99% sure this could be golfed. Maybe I could shorten the OEIS formula somehow.

Husk, 8 bytes

!¥⁰m⌋15N

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Input as1=number,3=Fizz,5=Buzz,15=FizzBuzz. Same approach asLynn's comment to caird coinheringaahing's answer.

Alternative versions with successively less pre-processing squeezed into the values chosen as input:

12 bytes, with input as[0,0]=number,[1,0]=Fizz,[0,1]=Buzz &[1,1]=FizzBuzz.

!fo≡⁰§e¦3¦5N

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TheHuskcongr command -≡ - checks whether two lists have the same distribution of truthy/falsy elements: in this case, divisibility (¦) by3 and5.

or14 bytes, finally with input just as0=number,1=Fizz,2=Buzz &3=FizzBuzz.

!¥mȯḋm¬§e%5%3N

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Treats list of not-modulo (¬ &%) 3 or 5 as binary digits (ḋ).

Vyxal, 8 bytes

15ġ⁰=)ȯt

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Port of the other answers.

Charcoal, 22 bytes

Ｉ⊕§⌕Ａ×”{⊞‴¡*ＲX⭆eＲ”ＮＳ⊖θ

Try it online! Link is to verbose version of code. Takes the number as the first input and one of the lettersF,B,Z orN as the second input. Explanation:

      ...       Compressed string `NNFNBFNNFBNFNNZ`     ×          Repeated by         Ｎ      First input as a number   ⌕Ａ           Find all indices of          Ｓ     Second input  §             Indexed by            θ   First input           ⊖    Decremented ⊕              IncrementedＩ               Cast to string                Implicitly print

Python 2, 26 bytes

lambda(a,b),c:c/8*b+a[c%8]

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The value (a,b) should be a list/tuple of 2:

[[-1,1,2,4,7,8,11,13],15] for Number[[-3,3,6,9,12,18,21,24],30] for Fizz[[-5,5,10,20,25,35,40,50],60] for Buzz[[0,15,30,45,60,75,90,105],120] for FizzBuzz

Wonder if this violatesthis loophole as it's just a list. If it does, I will delete this answer.

• code-golf
• number
• sequence
• fizzbuzz