Decorate Pascal's Triangle

Question

Although what is a Pascal's triangle is well-known and we already can generate it, the task is now different:
Output \$n\$ first lines of the Pascal's triangle as colored bricks.
Color number is computed as maximal power of \$d\$, that divides the needed-to-color element or \$7\$ if power is more than \$7\$.
Numbered colors are (\$0\$-based): default, red, green, yellow, blue, magenta, cyan, white (default color is 0th, red is 1st, .. white is 7th).

• Any space-filling (I mean not .,' or " or smth small) character
  can be used for bricks, even '\u2588'.
• Whitespace before or after the bricks are optional.
• Brick horizontal length (\$1\$ or \$2\$) is your choice.
• You may choose any convenient output method -- ansi text or graphics or HTML or whatever
• If you choose output letters or color numbers instead of coloring the triangle itself, you should specify in the text of your answer which character corresponds to which color, but output wouldn't look so nice)

Input Integers \$n\$ and \$d>1\$ in any convenient method.
This is code-golf, the shortest code in each language wins.

Answer 1

C (gcc),  157 153 143 139  135 bytes

Saved several bytes thanks to @AlexeyBurdin
Saved  4  8 more bytes thanks to @ceilingcat

Outputs with ANSI colors.

long p,a,s,c,L;f(n,d){for(a=0;c=a++-n;puts(""))for(;p=c<a;printf(c<0?" ":"\e[3%dm@@",L),s=++c<1?1:s*a/c-s)for(L=0;p*=d,L<7&s%p<1;L++);}

Try it online! (no colors on TIO)

Example output for \$f(40,2)\$

Formula

At row \$r\$ and column \$c-1\$, we have:

$$P_{r,c-1}={r\choose c-1}=\frac{r!}{(c-1)!(r-c+1)!}$$

On the same row and the next column \$c\$, we have:

$$\begin{align}P_{r,c}={r\choose c}&=\frac{r!}{c!(r-c)!}\\ &=\frac{r-c+1}{c}\times\frac{r!}{(c-1)!(r-c+1)!}\\ &=\frac{r-c+1}{c}\times P_{r,c-1}\\ &=\frac{r+1}{c}P_{r,c-1}-P_{r,c-1}\end{align}$$

which is translated as s = s * a / c - s in the C code (with \$a=r+1\$).

Commented

long p, a, s, c, L;               // declare a few 64-bit integers
f(n, d) {                         // n = number of rows, d = colorization parameter
  for(a = 0; a++ < n; puts(""))   // for a = 1 to n, with a linefeed added after each
    for(                          // iteration:
      c = a + ~n;                 //   for c = a - n - 1 to a - 1:
      p = c < a;                  //
      printf(                     //     update the output after each iteration:
        c < 0 ?                   //       if c is negative:
          " "                     //         just append a space
        :                         //       else:
          "\e[3%dm@@",            //         append the ANSI color code, followed by '@@'
        L                         //       set the 2nd digit of the color code
      ),                          //
      s = ++c < 1 ? 1             //     increment c; set s to 1 while c is less than 1
                  : s * a / c - s //     then, update s to s * a / c - s
    )                             //
      for(                        //       compute the color L:
        L = 0;                    //         start with L = 0
        p *= d,                   //         multiply p by d
        L < 7 & s % p < 1;        //         stop if L = 7 or p does not divide s
        L++                       //         increment L
      );                          //
}                                 //

Answer 2

Python 2, 158 134 132 bytes

n,d=input()
o=r=[1]
exec"print''.join([`i`for i in range(8)if c%d**i<1][-1]*2for c in r).center(2*n);r=o+map(sum,zip(r,r[1:]))+o;"*n

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22 bytes from tips by Jonathan Allan.

Uses 01234567 as the 8 characters instead of 8 colors.

Answer 3

05AB1E, 17 16 bytes

ÝεyÝcI7LmδÖO}».c

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Ý                 # range 0..input
 ε          }     # for each number y in that range:
  yÝ              #  range 0..y
    c             #  binomial coefficient (yields a row of the triangle)
     I            #  second input
      7L          #  range 1..7
        m         #  power (yields [input, input², ..., input**7])
         δÖ       #  double-vectorized divisible-by
           O      #  sum each inner list
»                 # join by newlines, joining sublists by spaces
 .c               # center

05AB1E, 22 bytes, no binomial coefficient built-in

1λ£0ªDÁ+}εI7LmδÖOºJ}.c

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Answer 4

Jelly, 23 bytes

1Ø0j+ƝƊ⁸СṖọ«7ịØaḤz⁶ZṚ€

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A full program taking the number of rows as its first argument and the power as its second. Outputs the triangle to STDOUT. Colours are represented by z for 0/default and a to g for 1 to 7+.

Answer 5

05AB1E, 24 21 bytes

FNNÝcεU7ÝR.ΔmXsÖ])».c

First input is the amount of rows n, second input the power-base d.
Outputs [0,7] instead of the colors, and only outputs a single digit for the bricks (with spaces).

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Explanation:

F                # Loop in the range [0, (implicit) input n):
 NN              #  Push the current loop-index twice
   Ý             #  Pop the top one, and create a list in the range [0, index]
    c            #  Calculate the binomial coefficient of the index and this ranged list
                 #   i.e. 10 choose [0,1,2,3,4,5,6,7,8,9,10]
                 #    → [1,9,36,84,126,126,84,36,9,1]
    ε            #  Map each value to:
     U           #   Pop and store the current value in variable `X`
      7Ý         #   Push a list in the range [0,7]
        R        #   Reverse it to make the range [7,0]
         .Δ      #   Find the first digit which is truthy for:
           m     #    Take the (implicit) input d to the power of the current digit
            Xs   #    Push variable `X` and swap the two values
              Ö  #    Check if `X` is divisible by this digit ** `d`
]                # Close the find_first; map; and loop
 )               # Wrap all lists into a list
  »              # Join each inner list by spaces, and then each string by newlines
   .c            # Centralize the lines by padding leading spaces
                 # (after which the result is output implicitly)

Answer 6

Japt, 41 36 32 29 25 bytes

_p1 ä+T}h[]à)m£_XuVpZa7}f

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      h   )  // run input times
       []à   // starting with empty 2d array (permutations of [])

_    }       // function that:
 p1          // appends 1
    T        // and prepends 0
  ä+         // sums consecutive elements
      

m            // for each line
 £           // for each element
           f // find first number (starting at 0) that return false this function:  
  _       }  // f(Z)
   Xu        // triangle element modulo
     Vp      // 2nd input raised to  
       Za8   // absolute difference 

Uses numbers [7...0] as colors Footer inverts colors to [0...7] for simpler output

Saved a lot thanks to @Shaggy

Answer 7

Charcoal, 36 bytes

Ｎθ⊞υ¹ＦＮ«Ｐ⭆υ×¹¹ΣＥ⁸¬﹪κＸθμ↙⊞υ⁰ＵＭυ⁺κ§υ⊖λ

Try it online! Link is to verbose version of code. Takes the number of rows as the second input. Output uses digits 1-8 to represent the colours (1 = default). Explanation:

Ｎθ

Input the base.

⊞υ¹

Start the first row with a single 1.

ＦＮ«

Loop over each row.

Ｐ⭆υ×¹¹ΣＥ⁸¬﹪κＸθμ↙

For each cell, determine its divisibility by the first 8 powers of the base, and take the sum. Then multiply by 11 to duplicate the digit. (Alternatively, casting to string and then duplicating the digit also works.) Don't move the cursor while printing, instead finish by moving the cursor down and left.

⊞υ⁰ＵＭυ⁺κ§υ⊖λ

Add another column to the row and calculate each row as the sum of the two cells above.

Answer 8

Ruby, 123 bytes

->n,d{(0..n).map{|w|' '*(n-w)+(0..w).map{|i|"\e[#{30+(b=[*1..w].combination(i).size).downto(0).find{|w|b%d**w<1}}m##"}*''}}

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