Output the Iccanobif Sequence

Question

Write a program or named function that will output or return the sequence up to the nth integer in the Iccanobif sequence, documented on OEIS as A014258. Note that only the zeroth element in the sequence (0) will be printed if n is zero.

The sequence is generated by starting like the standard Fibonacci sequence, but after adding the two previous numbers, you flip the result and drop any leading zeros. An interesting fact, to me at least, is that this sequence is not strictly increasing (see the list below). It also seems to be (and probably is) strictly greater than or equal to the Fibonacci sequence.

Your program's input must be an integer.

The first 20 numbers of the sequence are provided here for your viewing pleasure:

0, 1, 1, 2, 3, 5, 8, 31, 93, 421, 415, 638, 3501, 9314, 51821, 53116, 739401, 715297, 8964541, 8389769

Standard loopholes are forbidden.

Shortest program wins.

EDIT: Added a note to clarify that the sequence starts with the zeroth element and should be included if n is zero.

Example IO possibilities:

0    ->    0
1    ->    0 1
6    ->    0 1 1 2 3 5 8
17   ->    [0, 1, 1, 2, 3, 5, 8, 31, 93, 421, 415, 638, 3501, 9314, 51821, 53116, 739401, 715297]

Now that there are several answers, below are my implementations in Python 2 that I worked hard to hide with markup:

Iterative:

#Closest to my initial program. 73 bytes. It should also be noted that this program
 cannot reach a stack overflow. It runs for n=5000 in less than 10 seconds.

i,a,b=input(),0,1
print a
while i:print b;i,a,b=i-1,b,int(str(a+b)[::-1])

Recursive:

#Note that this prints n trailing newlines. 64 bytes.
 Will hit a stack overflow error for large values of n.

def f(n,i=0,j=1):print i,n and f(n-1,j,int(str(i+j)[::-1]))or'';

Answer 1

Pyth, 17 15 14

Pu+Gs_`s>2GQU2

Try it online

Very basic implementation, starts with range(2) and adds a number of elements equal to the input, then slices off the extras pops off the last element.

Thanks @Jakube for pointing out the > reversal thingy.

Explanation

Pu+Gs_`s>2GQU2    : Q = eval(input) (implicit)
P                 : all but the last element
 u         QU2    : reduce Q times starting with [0, 1]
  +G              : add to the previous result
       s>2G       : sum of the last two elements of G
    s_`           : int(reversed(repr(above value)))

Answer 2

Python 2, 58 bytes

a=0;b=1
exec"print a;a,b=b,int(str(a+b)[::-1]);"*-~input()

Uses str to convert rather than backticks because large enough numbers in Python 2 are written with an L at the end. I tried a recursive function, but it turned out longer (61):

f=lambda n,a=0,b=1:-~n*[0]and[a]+f(n-1,b,int(str(a+b)[::-1]))

Answer 3

K, 25 23 bytes

{-1_ x{x,.|$+/-2#x}/!2}

A simple modification of one of the examples at No Stinking Loops.

The phrase .|$ casts a number to a string, reverses it and then evaluates it.

Edit:

Sloppy attention to boundary conditions on my part. More correct now:

  {-1_ x{x,.|$+/-2#x}/!2}'0 1 6 10
(,0
 0 1
 0 1 1 2 3 5 8
 0 1 1 2 3 5 8 31 93 421 415)

Edit 2:

(x+1)# can be replaced with -1_, saving 2 characters. The space is necessary because otherwise _x would be an identifier, when I want the "drop" operator applied to a variable called x.

Answer 4

Julia, 79 bytes

f=n->(t=[0,1];for i=2:n push!(t,t[i]+t[i-1]|>string|>reverse|>int)end;t[1:n+1])

This creates a function that accepts an integer as input and returns an integer array.

Ungolfed + explanation:

function f(n)
    # Start with the usual Fibonacci stuff
    t = [0,1]

    # Looooooooooooooop
    for i = 2:n
        # Compute the Iccanobif number by piping results
        iccanobif = t[i] + t[i-1] |> string |> reverse |> int

        # Jam it into t
        push!(t, iccanobif)
    end

    # Return the first n + 1
    t[1:n+1]
end

Examples:

julia> f(1)
2-element Array{Int64,1}:
 0
 1

julia> f(17)
18-element Array{Int64,1}:
      0
      1
      1
      2
      3
      5
      8
     31
     93
    421
    415
    638
   3501
   9314
  51821
  53116
 739401
 715297

Answer 5

Haskell, 64 49 bytes

a!b=a:b!(read$reverse$show$a+b)
q n=0:take n(1!1)

Usage example: q 15 -> [0,1,1,2,3,5,8,31,93,421,415,638,3501,9314,51821,53116]

How it works: ! recursively builds an infinite list of iccanobif numbers starting with it's first argument (the second argument must be the next iccanobif number). q takes the first n numbers from the iccanobif list starting with 1, 1 and prepends a 0.

Answer 6

T-SQL, 149

Very straight forward inline table function that uses a recursive CTE query. As it is using INTs this will top out at 37. Adding in CASTs for bigints will allow it to go further to 63

create function i(@ int)returns table return with r as(select 0I,1N union all select n,reverse(i+n)+0from r)select 0n union all select top(@)n from r

It is used as follows

select * from i(0)
n
-----------
0

(1 row(s) affected)

select * from i(1)
n
-----------
0
1

(2 row(s) affected)

select * from i(6)
n
-----------
0
1
1
2
3
5
8

(7 row(s) affected)

select * from i(17)
n
-----------
0
1
1
2
3
5
8
31
93
415
421
638
3501
9314
51821
53116
715297
739401

(18 row(s) affected)

Answer 7

PHP, 114, 109 bytes

function f($n){if($n==0)return 0;$f=[0,1];for($i=2;$i<=$n;++$i){$f[$i]=strrev($f[$i-1]+$f[$i-2]);}return $f;}

Nothing fancy, just an average fibonacci algorithm with the string reverse magic.

Ungolfed:

function f($n)
{
    if($n == 0) return 0;
    $f = [0, 1];
    for ($i=2; $i<=$n; ++$i){
        $f[$i] = strrev($f[$i-1] + $f[$i-2]);
    }
    return $f;
}

Answer 8

CJam, 18 bytes

U1{_2$+sW%i}ri*;]p

How it works

U1                      e# First two numbers in the series
  {        }ri*         e# Run the loop input numbers times
   _2$+                 e# Get sum of last two numbers in the series
       sW%i             e# Convert to string, inverse and convert back to a number
                ;       e# Remove the last number to get only first n + 1 numbers.
                 ]p     e# Wrap all numbers in an array and print the array

Try it online here

Answer 9

Java - 126 124

I haven't seen Java around this site in awhile...

void f(int b){for(int c=0,d=1,g;b-->=0;d=Integer.valueOf(new StringBuilder(c+(c=d)+"").reverse()+""))System.out.println(c);}

f(5) prints 0 1 1 2 3 5 8 31 93 421 415 638

Answer 10

SWI-Prolog, 141 131 121 bytes

a(X,R):-X>1,A is X-1,a(A,B),reverse(B,[K,L|_]),W is K+L,name(W,Z),reverse(Z,Y),name(E,Y),nth0(X,R,E,B);X=1,R=[0,1];R=[0].

Running a(17,X). outputs:

[0, 1, 1, 2, 3, 5, 8, 31, 93, 421, 415, 638, 3501, 9314, 51821, 53116, 739401, 715297] 

Takes about 10 seconds to output the result of a(10000,X). on my computer.

Edit: The 121 bytes version above is a one predicate definition = one liner. The old 131 bytes version is the following (must be runned as p(17,X)):

a(0,[0]).
a(1,[1,0]).
a(X,[E|B]):-A is X-1,a(A,B),B=[K,L|_],W is K+L,name(W,Z),reverse(Z,Y),name(E,Y).
p(X,Y):-a(X,Z),reverse(Z,Y).

Answer 11

><> (Fish) 592 254 Bytes

Not super golfed (42/43 blanks that do nothing and a total of 30 redirection tokens) , but was an interesting exercise getting it working in the first place.

10!/{:}0=?v/{1-}}:{+:0}!/a,:1%-:0=?!v~{:1!/$:@0=?!v$~}}:&{{&*\
/-$/    ;n/\oo", "n:    \       }+1{/     \$-1$*a /|.!20}}01@/
* :{:}(?v:{!":}-1!/$:@0=?!v$~{:}1!/$:@0=?!v$~}}}:&{{{&*:1%-*&{{&+}}{1+}02.
b .1 +bb   \      \$-1$*a /       \$-1$,a /
\*9{~~~{/

You can test it here, providing the desired length in the initial stack.

EDIT: More than halved byte count

Answer 12

Pip, 13 bytes

I'm pretty sure all the features used in this program were present in Pip before this question was asked.

LaSio:+RVi+oi

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

Explanation

               a is 1st cmdline arg; i is 0; o is 1 (implicit)
La             Loop (a) times:
       RVi+o   Reverse of i+o
      +        Unary + treats its operand as a number, thus removing leading 0's
    o:         Assign the result to o...
  Si           ... before swapping i and o
            i  After the loop, output i

The two variables' values evolve like so:

Iter   o   i (output)
   0   1   0
   1   0   1
   2   1   1
   3   1   2
   4   2   3
   5   3   5
   6   5   8
   7   8  31
   8  31  93
   9  93 421
  10 421 415

Answer 13

JavaScript (ES2015), 81 73 bytes

(a,b=0,c=1)=>{for(;a-->-1;c=[...(b+(b=+c)+"")].reverse().join``)alert(b)}

Running this function (named f) with 6:

f(6);// alerts: 0, 1, 1, 2, 3, 5, 8

Answer 14

Excel VBA, 279 bytes

n = InputBox("n")
For i = 0 To n
If i < 2 Then
Cells(i + 1, 1) = i
ElseIf i > 6 Then
x = Cells(i, 1) + Cells(i - 1, 1)
l = Len(x)
v = CStr(x)
For j = 1 To l
r = r + Right(v, 1)
v = Left(v, l - j)
Next j
Cells(i + 1, 1) = r
r = ""
Else
Cells(i + 1, 1) = Cells(i, 1) + Cells(i - 1, 1)
End If
Next i

Running the macro will prompt the user to enter a value for n.

The results will then be printed row by row in column A:

Answer 15

Pushy, 18 bytes (non-competing)

Z1{:2d+vFs@KjkvF;_

Try it online!

It's not the most elegant of programs, but it works.

Z1     \ Push 0 and 1 to begin the sequence
{:     \ Input times do:
 2d+   \   Add the last two terms
 vF    \   Send to second stack
 s     \   Split into digits
 @Kjk  \   Reverse and join into one number
 vF;    \   Send back to first stack
_      \ At the end of the program, print the whole stack.

Answer 16

Jelly, 9 bytes (non-competing)

0+ṚV$¥Ð¡1

Try it online!

Answer 17

R, 134 bytes

i=function(n){s=c(0,1);for(i in 3:n){s[i]=as.numeric(paste0(rev(strsplit(as.character(s[i-2]+s[i-1]),'')[[1]]),collapse=''))};cat(s)}

Example:

> i(10)
0 1 1 2 3 5 8 31 93 421

Would love to see if somebody had a better R alternative than to take your number, make it a string, reverse it and turn it back into a number again.

Answer 18

Groovy, 70 bytes

{r={"$it".reverse() as int};f={n->n<3?1:r(f(n-1))+r(f(n-2))};r(f(it))}

---

{
    r={"$it".reverse() as int};       // Reverse digits, costly using string.
    f={n->n<3?1:r(f(n-1))+r(f(n-2))}; // Recursive Iccanobbif implementation.
    r(f(it))                          // Reverse final output.
}