Pascal's tree-angle

J, 67 bytes

(1785951#:~13#7)(0(,~' '#~12-<[email protected]:@#)@":[email protected][}.}.)&>1;2}.<@(!{:)\i.14

The pascal's triangle part is adapted from @ephemient's answer here.

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How it works


The first 14 rows of the triangle.


Drop the first two rows, prepend 1.


0 0 0 0 0 2 1 1 1 5 5 6 6 stored in base 7.

":[email protected][}.}.)&>1

Unbox, drop the first N and last N elements from each row, where N is the corresponding item from the base 7 array before.


Convert each row to string and …

12-<[email protected]:@#

Count its length, halve and round it down.

,~' '#~

Prepend that amount of spaces before each row.

Python 2, 118 bytes

exec"a=n/6+n/10*4+(n==6);print' '.join(str(((2**n+1)**n>>n*k)%2**n)for k in range(a,n+1%n-a)).center(23);n+=1;"*13

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Expresses binomial coefficients inline like in my tip here with binom(n,k)=((2**n+1)**n>>n*k)%2**n. The number of entries cut off on each side for the n'th row (one-indexed) is expressed as n/6+n/10*4+(n==6). For the first row, one additional entry on the right is cut off.

Here's a slightly more readable version without an exec so that the syntax highlighting works:

119 bytes

while n<14:a=n/6+n/10*4+(n==6);print' '.join(str(((2**n+1)**n>>n*k)%2**n)for k in range(a,n+1%n-a)).center(23);n+=1

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05AB1E, 26 bytes


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13L            # Push a list in the range [1,13]
   ε           # Map each value `y` to:
    D          #  Duplicate the value `y`
     Ý         #  Pop and push a list in the range [0,`y`]
      c        #  Take the binomial coefficient of `y` with each value in this list
               #  (we now have the 0-based `y`'th Pascal row)
    •6hö¢ðU•   #  Push compressed integer 6655111200000
            R  #  Reverse it to "0000021115566"
    Nè         #  Index the map-index into it to get the `N`'th digit
      F        #  Loop that many times:
       ¦¨      #   Remove both the first and last item of the current Pascal row list
   ]           # Close both the inner loop and map
    »          # Join each inner list by spaces, and then each string by newlines
     ¦         # Remove the leading 1 on the very first line
      .c       # Left-focused centralize the newline-delimited string
               # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why •6hö¢ðU• is 6655111200000.