Fibonacci triangle

Question

Given an integer n, output the following ASCII art to n rows:

1+たす1=わ2
 1+たす2=わ3
 2+たす3=わ5
 3+たす5=わ8
 5+たす8=わ13

Essentially, the first row is 1+1=2 and the nth row (1-indexed) is \$f_n + f_{n+1} = f_{n+2}\$ where \$f\$ is the Fibonacci sequence, padded so the numbers line up with the previous row.

You may instead output it infinitely. You may output a list of lines.

This is code-golf, shortest wins!

Testcase

The output for 20 should be:

1+たす1=わ2
 1+たす2=わ3
 2+たす3=わ5
 3+たす5=わ8
 5+たす8=わ13
 8+たす13=わ21
 13+たす21=わ34
 21+たす34=わ55
 34+たす55=わ89
 55+たす89=わ144
 89+たす144=わ233
 144+たす233=わ377
 233+たす377=わ610
 377+たす610=わ987
 610+たす987=わ1597
 987+たす1597=わ2584
 1597+たす2584=わ4181
 2584+たす4181=わ6765
 4181+たす6765=わ10946
 6765+たす10946=わ17711

• loopy walt: How exactly is this a triangle?
• Greg Martin: Fibonacci swoosh?
• pxeger: Is leading whitespace allowed?
• Darrel Hoffman: @pxeger Allowed? Seems to be required as I read it...
• pxeger: @DarrelHoffman I mean constant extra leading whitespace, like this: gist.github.com/pxeger/768a57edc59ce03a673d1d6a943ff0d7

Answer 1

Python 2, 67 bytes

Outputs the sequence indefinitely.

a=b=l=1
while 1:print'%*d+%d='%(l,a,b)+`a+b`;l-=~len(`b`);a,b=b,a+b

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Answer 2

JavaScript (ES8), 79 bytes

f=(n,A=B=1,p)=>n?''.padEnd(p)+A+`+${B}=${B+=A}
`+f(n-1,B-A,(A+"").length-~p):''

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Commented

f = ( // f is a recursive function taking:
 n, // n = input
 A = B = 1, // A, B = Fibonacci variables
 p // p = padding length, initially undefined
) => //
n ? // if n is not equal to 0:
 ''.padEnd(p) + // append p spaces (none if p is undefined)
 A + // followed by A
 `+${B}=${B += A}\n` + // followed by "+[B]=[B+A]\n" (A is added to B)
 f( // followed by the result of a recursive call:
 n - 1, // decrement n
 B - A, // update A to the previous value of B
 (A + "") // add the length of A coerced to a string + 1
 .length - ~p // to p
 ) // end of recursive call
: // else:
 '' // stop the recursion

Answer 3

C (gcc), (削除) 71 (削除ここまで) 66 bytes

-5 thanks to Sisyphus

i,j;main(k){for(;;)printf("%*d+%d%n=%d\n",i,j=k-j,k+=j,&i,j+k+k);}

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Outputs indefinitely.

• Sisyphus: 66 bytes using %n: Try it online!

Answer 4

Charcoal, 25 bytes

≔E2¦1ηFN«I⌊η+≔⟦⌈ηΣη⟧η⟦⪫η=

Try it online! Link is to verbose version of code. Explanation:

≔E2¦1η

Start with two 1s.

FN«

Repeat n times.

I⌊η

Output the first element of the old pair.

+

Output a +.

≔⟦⌈ηΣη⟧η

Replace the pair with the second element and their sum.

⟦⪫η=

Output the new pair joined with = and move the cursor down.

Answer 5

Pip, 30 bytes

W P[xi::o'+o+:i'=i+o]x.:sX#i+1

Outputs forever. Attempt This Online!

Explanation

W P[xi::o'+o+:i'=i+o]x.:sX#i+1
 i is 0, o is 1, x is "", s is " " (implicit)
 [ ] Put the following in a list:
 x The indent: x
 i::o Swap i (the smaller number) with o (the larger number)
 and return the new value of i
 '+ Plus sign
 o+:i Add i to o in-place and return the new value of o
 '= Equals sign
 i+o Add i and o
 P Print the list (concatenating its elements by default)
W Loop while the list is truthy (which is always):
 #i Length of i
 +1 Plus 1
 sX That many spaces
 x.: Concatenate with x in-place

Answer 6

Python 3.8 (pre-release), 66 bytes @emanresu A

a=b=l=1
while l:=len(x:=f"{a:>{l}}+{b}=%d")-3:a,b=b,a+b;print(x%b)

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Old Python 3.8 (pre-release), 67 bytes

a=b=l=1
while l:=len(x:=f"{a:>{l}}+{b}"):a,b=b,a+b;print(x+f'={b}')

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Based on @dingledooper's Python 2 answer.

• emanresu A: You can save a byte with some cursed formatting :)

Answer 7

SNOBOL4 (CSNOBOL4), (削除) 90 87 (削除ここまで) 84 bytes

	N =1
N	N =M + (M =N)
	OUTPUT =P =DUPL(' ',X) M '+' N '=' M + N
	P @X N '='	:(N)
END

Try it online!

Outputs infinitely, but experiences integer overflow after a short while.

Answer 8

Jelly, 22 bytes

+2ÆḞ€DżṖ©;)+=Ɗ6ṁ$®¦F)Y

A full program that accepts an integer and prints the result.

Try it online!

How?

+2ÆḞ€DżṖ©;)+=Ɗ6ṁ$®¦F)Y - Main Link: integer, N
 ) - for each V in [1..N]:
+2 - V add two
 ÆḞ€ - Fibbonacci of each of [1..V+2]
 D - to decimal digits -> [[1],[1],[2],...,Digits(Fib(V+2))]
 call this FibDigits
 Ɗ - last three links as a monad - f(V):
 Ṗ - pop -> [1..V-1]
 © - (copy this to the register)
 )+= - ['+', '=']
 ; - concatenate -> [1,2,3,...,V-1,'+','=']
 call this Fillers
 ż - FibDigits zip with Fillers
 ¦ - sparse application...
 ® - ...to indices: recall from register -> [1..V-1]
 $ - ...action: last two links as a monad:
 6 - space character
 ṁ - mould like (e.g. [[1,4,4],12] -> [[' ',' ',' '], ' ']
 F - flatten
 Y - join with newlines
 - implicit print

Answer 9

Vyxal, 19 bytes

ÞF3l(:n‛+=Y∑꘍,nhL›+

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Outputs infinitely.

ÞF3l # Take the infinite list of Fibonacci numbers, and get overlapping groups of 3
 ( # Looping over that...
 Y∑ # Interleave...
 n # The tuple of three numbers
 ‛+= # With '+='
 : ꘍ # Pad that with the correct amount of spaces, without popping the padding amount
 , # Print that
 + # Add to the padding amount (initially 0)
 nhL› # The length of the first number in the tuple, plus one.

Answer 10

Vyxal j, 30 bytes

ÞFẎ3l⟑‛+=fY∑\$꘍:\=ḟnḢvL=t ̄-h-£

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A big cursed mess of formatting and registering.

Answer 11

Retina 0.8.2, 55 bytes

1=1
{`.*\+
$.&$* 
=
+
\d+.(\d+)
$&=$&$*_1ドル$*_
:`_+
$.&

Try it online! Tries to output indefinitely but runs out of memory after about 50 seconds on TIO. Explanation:

1=1

Initialise the buffer with a hypothetical previous line.

{`

Repeat indefinitely.

.*\+
$.&$* 

Replace up to and including the + with spaces. (This only applies after the first loop.)

=
+

Change the = into a +.

\d+.(\d+)
$&=$&$*_1ドル$*_

Append the sum of the two values in unary.

:`_+
$.&

Convert the unary to decimal and output the result.

Answer 12

Python, ^{(削除) 117 (削除ここまで) (削除) 107 (削除ここまで) (削除) 103 (削除ここまで) (削除) 99 (削除ここまで)} 88 bytes

p="1=1"
while p:=" "*(a:=1+p.find("+"))+eval("f'{%s=}'"%p[a:].replace("=","+")):print(p)

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Outputs infinitely.

Can probably be much improved.

• Fmbalbuena: How do you use ":=" in a lambda?
• pxeger: @Fmbalbuena What do you mean by "how"? It's the same as anywhere else

Answer 13

05AB1E, (削除) 23 (削除ここまで) 22 bytes

ÌLÅfü3ε„+=.ιJ ̄Oú,yнg>ˆ

-1 byte after being inspired by @JonathanAllan's Jelly answer

Outputs the first \$n\$ lines.

Try it online.

Or a minor 22 bytes alternative:

ÌLÅfDü3„+=δ.ιJs€g>.\ú»

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And the infinite sequence would be 22 bytes as well:

∞Åfü3vy„+=.ιJ ̄Oú,yнg>ˆ

Try it online.

Explanation:

Ì # Increase the (implicit) input-integer by 2
 L # Pop and push a list in the range [1,input+2]
 Åf # Get the 0-based n'th Fibonacci value for each of these
 ü3 # Pop and push all overlapping triplets of this list
 ε # Foreach over the triplets:
 # (implicitly push the current triplet)
 „+=.ι # Intersperse it with "+" and "=" delimiters
 J # Join this list together to a string
 ̄ # Push the global_array (empty by default)
 O # Sum it together
 ú # Pad the string with that many leading spaces
 , # Pop and output this line with trailing newline
 y # Push the pair again
 н # Pop and push its first item
 g # Pop and push its length
 > # Increase it by 1
 ˆ # Pop and add it to the global_array

• emanresu A: You might be able to save bytes on the infinite version by porting my Vyxal answer and keeping the padding amount on the stack?
• Kevin Cruijssen: @emanresuA I'm afraid not. It would be 22 bytes as well. 05AB1E remembers the last value if it needs an argument and the stack is now empty, so I will have to push an explicit 0 at the start (otherwise it will use the infinite overlapping triplets list for the padding). In addition, I need an explicit Duplicate within the loop. That, combined with the 1 byte longer Fibonacci builtin compared to your Vyxal answer, is an alternative infinite 22-byter.

Answer 14

R, 76 bytes

l=b=1
repeat{cat(sprintf("%*d+%d=%d
",l,T,b,s<-T+b));l=l+nchar(b)+1;T=b;b=s}

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Outputs infinitely.

---

If leading whitespace is allowed:

R, (削除) 74 (削除ここまで) 69 bytes

b=1
repeat cat(sprintf("%*d+%d=%d
",F<-F+nchar(+T)+1,T,b,b<-T+(T=b)))

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Answer 15

Retina, 54 bytes

K`1+1=2
"$+"+`(\d+)=(\d+)$
$&¶$.%`* 1ドル+2ドル=$.(*_2ドル*
A`$

Try it online! No test suite because of the way the program uses history. Explanation:

K`1+1=2

Replace the input with the first line of the output.

"$+"+`

Repeat n times.

(\d+)=(\d+)$

Match fn=fn+1 from the previous line.

$&¶$.%`* 1ドル+2ドル=$.(*_2ドル*

Append fn+fn+1= plus their sum, with the correct amount of indent.

A`$

Since the program started with one line and appended n lines, it now has one line too many, so delete the last line.

Answer 16

Bash, ^{(削除) 139 (削除ここまで) . . . (削除) 104 (削除ここまで)} 103 bytes

_()($[a=b=1];for n in `seq 1ドル`;{ printf "%${s}s";echo -n $a+$b;$[c=a+b,s+=~${#a},a=b,b=c];echo "=$c";})

Try it online!

• emanresu A: Is all that whitespace necessary?
• sinvec: @emanresuA not sure :)
• emanresu A: Removed

Answer 17

Haskell, 89 bytes

l!s@[a,b,c]|[x,y,z]<-show<$>s=(l++x++'+':y++'=':z):(' '<$' ':x++l)![b,c,b+c]
f=""![1,1,2]

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f is a stream of lines.
l!s formats current line and computes next triplet s and leading whitespace l

Answer 18

CJam, ^{(削除) 49 (削除ここまで) (削除) 47 (削除ここまで)} 45 bytes

liX:Y;{[ST*X`_,T)+:T;'+Y:K'=XY+:YN]oK:X;(_}g;

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Answer 19

C (gcc), ^{(削除) 102 (削除ここまで) (削除) 76 (削除ここまで)} 72 bytes

b;s;f(a){for(a=b=1;s=printf("%*d+%d",s,a,b);a=b-a)printf("=%d\n",b+=a);}

Try it online!

Saved a whopping 26 bytes thanks to ceilingcat!!!
Saved 3 bytes thanks to m90!!!

Outputs forever!

• m90: A further improvement
• Noodle9: @m90 Nice one - thanks! :D

Answer 20

Julia 1.0, ^{(削除) 66 (削除ここまで)} 65 bytes

f(a=1,b=1,n=0)=print(" "^n,"$a+$b=$(a+b)
")f(b,a+b,n-~ndigits(a))

Try it online!

prints infinitely. To avoid Int64 overflow, replace a=1 with a=big(1) (+5 bytes)

Answer 21

Python 3, (削除) 146 (削除ここまで) 141 bytes

print("1+1=2")
a=[1,1,2]
w=2
while 1:print(" "*w+str(a[1])+"+"+str(a[2])+"="+str(a[1]+a[2]));del a[0];w+=len(str(a[0]))+1;a.append(a[0]+a[1])

Try it online!

Answer 22

Japt -mR, (削除) 23 (削除ここまで) 22 bytes

Includes a trailing space on each line of output.

"+= "¬ËiMg°UÃvÈùT±XÊì

Try it

"+= "¬ËiMg°UÃvÈùT±XÊì :Implicit map of each U in the range [0,input)
"+= "¬ : Split "+= " to an array of characters
 Ë : Map
 i : Prepend
 °U : Prefix increment U
 Mg : Get the Uth Fibonacci number, 0-indexed
 Ã : End map
 v : Modify the first element
 È : By passing it through the following function as X
 ù : Left pad with spaces to length
 T± : Increment T (initially 0) by
 XÊ : Length of X
 Ã : End modification
 ¬ : Join
 :Implicit output joined with newlines

Answer 23

Scala, (削除) 228 (削除ここまで) (削除) 226 (削除ここまで) 225 bytes

Saved 3 bytes thanks to the comment of @ceilingcat

---

Golfed version. Try it online!

object Main extends App{def f(n:Int)={def g(a:Int,b:Int,n:Int):List[(Int,Int,Int)]=n match{case 0=>Nil case _=>(a,b,a+b)::g(b,a+b,n-1)};g(1,1,n)};f(20).zipWithIndex.map{case((a,b,c),i)=>(" "*i)+s"$a+$b=$c"}.foreach(println)}

Ungolfed version. Try it online!

object Main extends App {
 
 def formatList(list: List[(Int, Int, Int)]): List[String] = {
 list.zipWithIndex.map { 
 case ((a, b, c), i) => (" " * (2 * i)) + s"$a + $b = $c"
 }
 }
 
 def fibonacciSeq(n: Int): List[(Int, Int, Int)] = {
 def fibHelper(a: Int, b: Int, n: Int): List[(Int, Int, Int)] = n match {
 case 0 => Nil
 case _ => (a, b, a + b) :: fibHelper(b, a + b, n - 1)
 }
 fibHelper(1, 1, n)
 }
 
 val formatted = formatList(fibonacciSeq(20))
 formatted.foreach(println)
}

• Shaggy: It looks like 20 is hardcoded into this; you must be able to take the value as input, or output infinitely.