Divinacci Sequence

Question

Divinacci (OEIS)

Perform the Fibonacci sequence but instead of using:

f(n) = f(n-1)+f(n-2)

Use:

f(n) = sum(divisors(f(n-1))) + sum(divisors(f(n-2)))

For an input of n, output the nth term, your program should only have 1 input.

---

First 14 terms (0-indexed, you may 1-index; state which you used):

0 | 0 # Initial | []
1 | 1 # Initial | [1] => 1
2 | 1 # [] + [1] | [1] => 1
3 | 2 # [1] + [1] | [1,2] => 3
4 | 4 # [1] + [1,2] | [1,2,4] => 7
5 | 10 # [1,2] + [1,2,4] | [1,2,5,10] => 18
6 | 25 # [1,2,4] + [1,2,5,10] | [1,5,25] => 31
7 | 49 # [1,2,5,10] + [1,5,25] | [1,7,49] => 57
8 | 88 # [1,5,25] + [1,7,49] | [1, 2, 4, 8, 11, 22, 44, 88] => 180
9 | 237 # [1,7,49] + [180] | [1, 3, 79, 237] => 320
10 | 500 # [180] + [320] | [1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, 500] => 1092
11 | 1412 # [320] + [1092] | [1, 2, 4, 353, 706, 1412] => 2478
12 | 3570 # [1092] + [2478] | [1, 2, 3, 5, 6, 7, 10, 14, 15, 17, 21, 30, 34, 35, 42, 51, 70, 85, 102, 105, 119, 170, 210, 238, 255, 357, 510, 595, 714, 1190, 1785, 3570] => 10368
13 | 12846 # [2478] + [10368] | [1, 2, 3, 6, 2141, 4282, 6423, 12846] => 25704
Etc...

You may choose whether or not to include the leading 0. For those who do: the divisors of 0 are [] for the purpose of this challenge.

It's code-golf lowest byte-count wins...

• Dennis: All natural numbers divide 0, thus its divisor sum is +∞.
• Leaky Nun: @Dennis finally someone who doesn't think that 1 + 2 + 3 + ... = -1/12.
• Magic Octopus Urn: @Dennis We can get rid of the 0 and make this valid though :P. Or you can just submit a Mathematica answer of Infinity if you want.
• Dennis: The Jelly answer would be shorter. :P You can either change the sequence (the answer probably would need tweaking as well) or change its description (start with base values 0, 1, 1).
• Martin Ender: @carusocomputing If it doesn't change the sequence, how can it affect answers?

Answer 1

05AB1E, 9 bytes

XÎFDŠ‚ÑOO

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Explanation

XÎ # initialize stack with 1,0,input
 F # input times do
 D # duplicate
 Š # move down 2 places on the stack
 ‚ # pair the top 2 elements on the stack
 Ñ # compute divisors of each
 OO # sum twice

• Magic Octopus Urn: Tons of swapping going on heh! Interesting.
• Rohan Jhunjhunwala: I like how the last couple bytes are forcefully yelling at the reader.
• Magic Octopus Urn: You won this by 2 minutes lol.

Answer 2

Mathematica, (削除) 45 (削除ここまで) 40 bytes

If[#<3,1,Tr@Divisors@#0[#-i]~Sum~{i,2}]&

Mathematica's divisor related functions Divisors, DivisorSum and DivisorSigma are all undefined for n = 0 (rightly so), so we start from f(1) = f(2) = 1 and don't support input 0.

Defining it as an operator instead of using an unnamed function seems to be two bytes longer:

±1=±2=1
±n_:=Sum[Tr@Divisors@±(n-i),{i,2}]

• CalculatorFeline: *7 bytes longer unless ± is 1 byte in a Mathematica supported encoding.
• Martin Ender: @CalculatorFeline It is. (The default setting for $CharacterEncoding on Windows machines is WindowsANSI, i.e. CP 1252.)
• CalculatorFeline: Good to know. .

Answer 3

Perl 6, 58 bytes

{my&d={sum grep $_%%*,1..$_};(0,1,{d($^a)+d $^b}...*)[$_]}

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Answer 4

Haskell, 55 bytes

f n=sum[a|n>1,k<-f<$>[n-1,n-2],a<-[1..k],mod k a<1]+0^n

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One-indexed.

Answer 5

Python 2, 76 bytes

f=lambda n:sum(a for k in[1,2][:n]for a in range(1,3**n-8)if f(n-k)%a<1)or n

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Ridiculously slow.

Answer 6

Jelly, (削除) 10 (削除ここまで) 9 bytes

ð,ÆDẎSð¡1

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Thanks to Dennis for -1.

• Dennis: 9 bytes
• Erik the Outgolfer: @Dennis The 0 was implicit?
• Dennis: When you take the number of iterations from STDIN, you get a niladic chain, and 0 is the implicit argument of niladic chains.
• Erik the Outgolfer: @Dennis So ¡ and others will just try to take an argument from everywhere, even with a Ɠ? That's quite unexpected...
• Dennis: Unless specified explicitly, ¡ et al. take the last command-line argument and, if there aren't any, reads a line from STDIN.

Answer 7

Python 3, (削除) 88 (削除ここまで) (削除) 83 (削除ここまで) 81 bytes

f=lambda n:+(n<3)or g(f(n-1))+g(f(n-2))
g=lambda n,i=1:n>=i and(n%i<1)*i+g(n,i+1)

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Excludes the 0

Answer 8

MATL, (削除) 16 (削除ここまで) 15 bytes

Oliq:",yZ\s]+]&

This solution uses 0-based indexing.

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Explanation

O % Push the number literal 0 to the stack
l % Push the number literal 1 to the stack
i % Explicitly grab the input (n)
q % Subtract 1
: % Create the array [1...(n - 1)]
" % For each element in this array...
 , % Do the following twice
 y % Copy the stack element that is 1-deep
 Z\ % Compute the divisors
 s % Sum the divisors
 ] % End of do-twice loop
 + % Add these two numbers together
] % End of for loop
& % Display the top stack element

Answer 9

Haskell, (削除) 64 (削除ここまで) 60 bytes

d n=sum[a|a<-[1..n],mod n a<1]
f=0:scanl((.d).(+).d)1f
(!!)f

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Answer 10

PHP, 97 bytes

for($f=[0,$y=1];++$i<$argn;$x=$y,$y=$r)for($f[]=$r=$v=$n=$x+$y;--$v;)$n%$v?:$r+=$v;echo$f[$argn];

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PHP, 101 bytes

for($f=$d=[0,1];$i<$argn;$d[]=$r)for($f[]=$r=$v=$n=$d[$i]+$d[++$i];--$v;)$n%$v?:$r+=$v;echo$f[$argn];

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Answer 11

Pari/GP, 39 bytes

f(n)=if(n<3,1,sum(i=1,2,sigma(f(n-i))))

Based on Martin Ender's Mathematica answer.

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Answer 12

R, 81 bytes

f=function(n,a=1,b=1,d=numbers::divisors)`if`(n-1,f(n-1,b,sum(d(a))+sum(d(b))),a)

1-indexed, and excludes the 0 at the start of the sequence. That zero gave me a lot of trouble to implement, because the builtin numbers::divisors doesnt handle it well.

The rest is a modified version of the standard recursive function that implements the fibonacci sequence.

> f(1)
[1] 1
> f(2)
[1] 1
> f(3)
[1] 2
> f(5)
[1] 10
> f(13)
[1] 12846

Answer 13

Vyxal, 9 bytes

1{~"vKf∑...

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Prints the sequence forever.

1 # Push 1
 { # Forever...
 ~" # Grab top two elements, without popping
 vK # Get divisors of each
 f∑ # Deep sum
 ... # Print that without popping

Answer 14

Julia, ^{(削除) 47 (削除ここまで)} 44 bytes

~n=(n.%(r=1:n).<1)'r
!N=N<3||~!(N-1)+~!(N-2)

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-3 bytes thanks to Czylabson Asa. (I forgot how a'*b worked)

• Czylabson Asa: gives true for N=1,2
• MarcMush: @CzylabsonAsa true == 1, so I think it's ok
• Czylabson Asa: oh, really. just a short note, your ~ can be ~m=(r=1:m;r'*(m.%r.<1)), not shorter, but a little bit different.
• MarcMush: thanks, I knew I could do better but I forgot about how finicky a'*b was to get a number instead of a 1x1 matrix
• Czylabson Asa: w/similar method i got 6th (20220615) at code.golf/abundant-numbers :-)

Answer 15

Husk, 11 bytes

!¡ȯΣṁḊ↑_2l·2

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Explanation

!¡ȯΣṁḊ↑_2l·2
 l·2 array [0,1]
 ¡ iterate on the array, expanding it:
 ȯ using the following 4 functions:
 ↑_2 get last 2 elements
 ṁḊ get divisors of both
 Σ sum them
! Get element at index n

Answer 16

Factor + math.primes.factors math.unicode, 53 bytes

[ 0 1 rot [ tuck [ divisors Σ ] bi@ + ] times drop ]

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0-indexed.

Answer 17

C (gcc), (削除) 93 (削除ここまで) (削除) 87 (削除ここまで) 86 bytes

f(n){return n<2?n:s(f(n-1))+s(f(n-2));}s(n,c,i){for(c=i=0;n/++i;)c+=n%i?0:i;return c;}

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A simple recursive solution that finds divisors by brute force.

Thanks to @ceilingcat for -6 bytes on the for loop and -1 on some superfluous parens.

Answer 18

Burlesque, 23 bytes

0 1 2{fc++jfc++.+}C~j!!

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0 1 # Initial 
2 # Keep top 2 stack elements
{
 fc++ # Factors, Sum
 j # Swap stack
 fc++ # Factors, Sum
 .+ # Sum
}C~ # Continue forever
j!! # Get input element

Answer 19

05AB1E, 7 bytes

1λè‚Ñ ̃O

Outputs the 1-based \$n^{th}\$ value.

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With some small modifications, we could output:

• 7 bytes: The first 1-based \$n\$ values by replacing è with £: try it online.
• 5 bytes: The infinite 1-based sequence by removing the 1 and è: try it online.
• Or with 0-based sequence by adding Ý after the 1:
  • 8 bytes: 0-based \$n^{th}\$: try it online.
  • 8 bytes: 0-based first \$n\$ values: try it online.
  • 7 bytes: 0-based infinite sequence: try it online.

So with default sequence rules, this could have been 5 bytes.

Explanation:

 λ # Start a recursive environment,
 è # to output the a(input)'th value
 # (which is output implicitly at the end)
1 # Starting at a(0)=1
 # And where every following a(n) is calculated as:
 ‚ # Pair the (implicit) previous two terms together: [a(n-2),a(n-1)]
 # (a(n-2) is unknown in the first iteration, so will default to 0)
 Ñ # Get the divisors of both inner values
 ̃ # Flatten this list of lists of integers
 O # Sum them together

Explanation of the 5-bytes infinite 1-based sequence:

λ # Start a recursive environment,
 # to output the infinite sequence
 # (which is output implicitly at the end)
 # Implicitly starting at a(0)=1
 # And where every following a(n) is calculated as:
 ‚Ñ ̃O # See above

Answer 20

Scala, 123 bytes

Golfed version. Attempt it online!

def f(n:Int):Int=if(n<3)1 else{def g(n:Int,i:Int=1):Int=if(n>=i)if(n%i<1)i+g(n,i+1)else g(n,i+1)else 0;g(f(n-1))+g(f(n-2))}

Ungolfed version. Attempt it online!

object Main {
 def main(args: Array[String]): Unit = {
 val sequence = (1 to 10).map(f).toList
 println(sequence)
 }
 def f(n: Int): Int = {
 if (n < 3) 1
 else g(f(n - 1)) + g(f(n - 2))
 }
 def g(n: Int, i: Int = 1): Int = {
 if (n >= i) {
 if (n % i < 1) i + g(n, i + 1)
 else g(n, i + 1)
 } else 0
 }
}

Answer 21

Japt, 11 bytes

Starts with 0 and outputs the nth term, zero-indexed. Replace g with h to output the first n terms instead.

@Zo2 câ x}g

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