# Decompose a number into triangles

Given an integer n, decompose it into a sum of maximal triangular numbers (where Tm represents the mth triangular number, or the sum of the integers from 1 to m) as follows:

• while n > 0,

• find the largest possible triangular number Tm such that Tm ≤ n.

• append m to the triangular-decomposition representation of n.

• subtract Tm from n.

For example, an input of 44 would yield an output of 8311, because:

• 1+2+3+4+5+6+7+8 = 36 < 44, but 1+2+3+4+5+6+7+8+9 = 45 > 44.

• the first digit is 8; subtract 36 from 44 to get 8 left over.
• 1+2+3 = 6 < 8, but 1+2+3+4 = 10 > 8.

• the second digit is 3; subtract 6 from 8 to get 2 left over.
• 1 < 2, but 1+2 = 3 > 2.

• the third and fourth digits must be 1 and 1.

Use the digits 1 through 9 to represent the first 9 triangular numbers, then use the letters a through z (can be capitalized or lowercase) to represent the 10th through 35th triangular number. You will never be given an input that will necessitate the use of a larger "digit".

The bounds on the input are 1 ≤ n < 666, and it will always be an integer.

All possible inputs and outputs, and some selected test cases (listed as input, then output):

1 1
2 11
3 2
4 21
5 211
6 3
100 d32
230 k5211
435 t
665 z731


An output of for an input of -1/12 is not required. :)

• But does an input of need to have an output of ∞? Oct 26, 2017 at 9:22

## JavaScript (ES6), 52 bytes

f=(n,t=0)=>t<n?f(n-++t,t):t.toString(36)+(n?f(n):'')


### How?

Rather than explicitly computing Ti = 1 + 2 + 3 + … + i, we start with t = 0 and iteratively subtract t + 1 from n while t < n, incrementing t at each iteration. When the condition is not fulfilled anymore, a total of Tt has been subtracted from n and the output is updated accordingly. We repeat the process until n = 0.

Below is a summary of all operations for n = 100.

 n  |  t | t < n | output
----+----+-------+--------
100 |  0 | yes   | ""
99 |  1 | yes   | ""
97 |  2 | yes   | ""
94 |  3 | yes   | ""
90 |  4 | yes   | ""
85 |  5 | yes   | ""
79 |  6 | yes   | ""
72 |  7 | yes   | ""
64 |  8 | yes   | ""
55 |  9 | yes   | ""
45 | 10 | yes   | ""
34 | 11 | yes   | ""
22 | 12 | yes   | ""
9 | 13 | no    | "d"
----+----+-------+--------
9 |  0 | yes   | "d"
8 |  1 | yes   | "d"
6 |  2 | yes   | "d"
3 |  3 | no    | "d3"
----+----+-------+--------
3 |  0 | yes   | "d3"
2 |  1 | yes   | "d3"
0 |  2 | no    | "d32"


### Test cases

f=(n,t=0)=>t<n?f(n-++t,t):t.toString(36)+(n?f(n):'')

console.log(f(1))   // 1
console.log(f(2))   // 11
console.log(f(3))   // 2
console.log(f(4))   // 21
console.log(f(5))   // 211
console.log(f(6))   // 3
console.log(f(100)) // d32
console.log(f(230)) // k5211
console.log(f(435)) // t
console.log(f(665)) // z731

# Jelly, 18 17 bytes

Ḥ‘½+.Ḟ©ịØB2®cạµ¹¿


This is a monadic link that prints to STDOUT. It's return value is 0 and should be ignored.

Try it online!

# dc, 74 bytes

?sa[2k_1K/1 4/la2*+v+0k1/dlardd*+2/-sadd10<t9>ula0<s]ss[87+P]st[48+P]sulsx


This is awful.

?sa             stores the input
[2k             sets precision to 2 so dc doesn't truncate 1/4
_1K/1 4/la2*+v+ uses the quadratic formula to find k, the next value to print
0k1/d           truncates k to an integer
lardd*+2/-sa    subtracts kth triangular number from the input
dd10<t9>u       determines whether to print k as a letter or a digit
la0<s]ss        loops when a is greater than 0
[87+P]st        prints k as a letter
[48+P]su        prints k as a digit (not p, as that leaves a trailing newline)
lsx             starts the main loop


## JavaScript (ES6), 61 57 bytes

Saved 4 bytes thanks to @Arnauld

f=(n,p=q=0)=>n?p-~q>n?q.toString(36)+f(n-p):f(n,++q+p):''

• I had f=(n,t=0)=>n?t+1>n?t.toString(36)+f(n):f(n-++t,t):1 May 2, 2017 at 0:37
• @Arnauld Oh wow, that's way better. You should post it yourself... May 2, 2017 at 0:39
• Alright. In your version, would it be safe to do f=(n,p=q=0) and f(n,++q+p)? May 2, 2017 at 0:58
• @Arnauld Yes, thanks! May 2, 2017 at 1:25

# Java 7, 81 bytes

int i=0;String c(int n){return i<n?c(n-++i):Long.toString(i,36)+(n>(i=0)?c(n):"");}


Port from @Arnauld's amazing JavaScript (ES6) answer.
My own approach was almost 2x as long..

Try it here.

Explanation:

int i=0;                  // Temp integer i (on class level)
String c(int n){          // Method with integer parameter and String return-type
return i<n?             //  If i is smaller than the input integer
c(n-++i)              //   Return a recursive call with input minus i+1 (and raise i by 1 in the process)
:                      //  Else:
Long.toString(i,36)+  //   Return i as Base-36 +
(n>(i=0)?            //   (If the input integer is larger than 0 (and reset i to 0 in the process)
c(n)                //    Recursive call with the input integer
:                    //   Else:
"");                //    an empty String)
}                         // End of method


# Retina, 11510838 34 bytes

.+
$*¶ (\G¶|¶\1)+ 0$1
+T_w¶w_.¶


[Try it online!] (Includes test suite) Uses uppercase letters. Edit: Saved 70 74 bytes by shamelessly adapting @MartinEnder's answer to Is this number triangular? Explanation: The number is converted into unary, then the largest possible triangular number is repeatedly matched until the number is exhausted. Each match is then converted into base 36.

# PHP, 74 Bytes

for(;$n=&$argn;$t.=$i>10?chr($i+86):$i-1)for($i=0;$n>=++$i;)$n-=$i;echo$t;


Online Version

# R, 87 bytes

Originally, I tried to preset the possible Triangular numbers. This led to this code with 105 bytes:

pryr::f(n,{l=cumsum(1:35)
k=''
while(n){y=tail(which(l<=n),1)
n=n-l[y]
k=paste0(k,c(1:9,letters)[y])}
k})


This required more indexing so I tried the methodology from @Arnauld to reduce the bytes down to 87.

pryr::f(n,{k='';while(n){t=0;while(t<n){t=t+1;n=n-t};k=paste0(k,c(1:9,letters)[t])};k})
`

Both codes made use of the preset letters, since their I couldn't find a short way to convert to base 36.