Recamán's triangular fibonacci

Question

This challenge is based upon three sequences, below are their formulae:

• Recamán's sequence:

a₁ = 0; for n > 0, a_n = a_n-1 - n

if positive and not already in the sequence, otherwise

a_n = a_n-1 + n

• Fibonacci sequence:

a_n = (Φⁿ – (–Φ)^–n) / √5

where Φ denotes the golden ratio, and

a₁ = 0

• Triangular numbers sequence:

a₁ = 0; a_n = n * (n+1) / 2

Your task is to take specified nth terms of each sequence and perform certain basic mathematical operations on them, namely addition, subtraction, multiplication, division, and the support of parentheses, nothing more is required.

Input is always in infix notation, and output formats are unrestricted.

Your code must be able to handle all real numbers from -10,000,000 to +10,000,000

Standard loopholes apply, and built-in functions for these sequences are not allowed.

Examples:

(These examples use an example output format of 1R = 1st term of R, 3430T = 3430th term of T, and so on, however you are free to use whichever output format you desire.)

Examples:

• (1R + 1F) * 3

= ( 0 + 0 ) * 3

= 0

• (1R * 1F) - 3

= ( 0 * 0 ) - 3

= - 3

• (1R + 2R) + (57R - 35R) / (51R - 41R) ) + (31F + 32F) - (54T)

= { [ (0 + 1) + (30 - 113) ] / (33 - 38) } + (832040 + 1346269) - (1431)

= { [ (0 + 1) + (-83) ] / (-5) } + (2178309) - (1431)

= [ (1 - 83 ) / (-5) ] + (2178309) - (1431)

= (-82 / -5) + (2178309) - (1431)

= (16.4) + (2178309) - (1431)

= 2176894.4

Answer 1

JavaScript (ES6), 164 163 148 bytes

Input format: F(n), R(n), T(n)

s=>eval(s.replace(/[FRT]/g,c=>`(F=n=>${c<'R'?'n>1?F(n-2)+F(n-1):n':c>'R'?'n*++n/2':'(N=>{for(l=[1];--n;)l[x=y+=++N>y|l[y-N]?N:-N]=1})(x=y=0)|x'})`))

Example

This may take a few seconds to complete.

let f =

s=>eval(s.replace(/[FRT]/g,c=>`(F=n=>${c<'R'?'n>1?F(n-2)+F(n-1):n':c>'R'?'n*++n/2':'(N=>{for(l=[1];--n;)l[x=y+=++N>y|l[y-N]?N:-N]=1})(x=y=0)|x'})`))

console.log(f("((R(1) + R(2)) + (R(57) - R(35))) / (R(51) - R(41)) + (F(30) + F(31)) - (T(53))"));

Answer 2

Perl, 201 199 169 + 1 = 170 bytes

Run with the -n flag.

Whitespace is provided for readability and is not part of the code.

sub R{
  $_=pop;
  %m=(@_=0,1);
  $m{push@_,$m{$a=$_[-1]-@_}||$a<0?$a+2*@_:$a}=1 for 2..$_;
  pop
}
sub F{
  @_=(0,1,@_);
  push@_,$_[-1]+$_[-2]for 3..pop;
  pop
}
sub T{
  ($_=pop)*$_++/2
}
say eval

Pretty straightforward. R calculates the nth Recaman number, F calculates the nth Fibonacci number, and T calculates the nth Triangular number. The script accepts input of an expression with terms such as R(##), T(##), and F(##), and evaluates it.

I'm sure I can reduce the byte-count for R, but I'll have to spend more time thinking about it.

Answer 3

Ruby, 135 132 129 bytes

->s{R,F,T=->x{*y=r=0;(x-1).times{|i|r<<y;y=r-[z=y+~i]!=r||z<0?y-~i:z};y},->x{a,b=1,0;(x-1).times{a=b+b=a};b},->x{x*~-x/2};eval s}

Input format: string using the sequence names and subscript. Example:

( (1R + 2R) + (57R - 35R) / (51R - 41R) ) + (31F + 32F) - (54T)

becomes

"( (R[1] + R[2]) + (R[57] - R[35]) / (R[51] - R[41]) ) + (F[31] + F[32]) - (T[54])"

Sorry for adding the missing bracket to the example, it wouldn't work otherwise. :-)

Explanation

The evalis using the definition of R, T, and F functions.

Fibonacci: my own entry in the Fibonacci contest: https://codegolf.stackexchange.com/a/103329/18535 changed to start from 0 instead of 1.

Recamán: inspired by Martin Ender's answer: https://codegolf.stackexchange.com/a/37640/18535 then golfed away a couple of bytes.

Triangular: starting from 0, instead of n*(n+1)/2 I could use n*(n-1)/2