Print this tree:
1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
15 20 15
7 21 35 35 21 7
8 28 56 70 56 28 8
9 36 84 126 126 84 36 9
252
462 462
924
1716 1716
This is code-golf, so shortest code in chars wins!
(1785951#:~13#7)(0(,~' '#~12-<.@-:@#)@":-@[}.}.)&>1;2}.<@(!{:)\i.14
The pascal's triangle part is adapted from @ephemient's answer here.
<@(!{:)\i.14
The first 14 rows of the triangle.
1;2}.
Drop the first two rows, prepend 1.
(1785951#:~13#7)
0 0 0 0 0 2 1 1 1 5 5 6 6 stored in base 7.
":-@[}.}.)&>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.
0(…)@":
Convert each row to string and …
12-<.@-:@#
Count its length, halve and round it down.
,~' '#~
Prepend that amount of spaces before each row.
n=1
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
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
n=1
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
13LεDÝc•6hö¢ðU•RNèF¦¨]»¦.c
Explanation:
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.
_=>[a=[1],..."000021115566"].map((v,n)=>(s=(a=[i=1,...a.map(v=>v+~~a[i++])]).slice(v,n+2+!n-v).join` `).padStart(23+s.length>>1)).join`
`
Ｆ³⊞υ﹪ι²Ｆ”)⧴→↨w﹪f”«ＵＭυ⁺κ§υ⊖λ⊞υ⁰Ｆ›ⅉ⁰≔⪫✂υＩι±Ｉι¹ ι⟦⁺× ⊘⁻²⁴Ｌιι
Try it online! Link is to verbose version of code. Explanation:
Ｆ³⊞υ﹪ι²
Start by preparing a list 0 1 0. (Charcoal's indexing is cyclic, so rather than letting it wrap I need to provide a safety margin.)
Ｆ”)⧴→↨w﹪f”«
Loop over the compressed string 111113222667 which represents the number of entries to slice off each side (except on the first iteration).
ＵＭυ⁺κ§υ⊖λ
Add each element to the next element (so the first element remains 0).
⊞υ⁰
Append another 0 to the end. (This is done here so that it can be easily sliced off again.)
Ｆ›ⅉ⁰≔⪫✂υＩι±Ｉι¹ ι
Except on the first loop, slice off the given number of entries from the start and the end (this has to be nonzero to work, which is why we already pushed the 0 only to slice it off again, and also why we have an extra 0 at the start, which otherwise wouldn't be necessary). On the first loop, the 1 just ends up being the output string.
⟦⁺× ⊘⁻²⁴Ｌιι
Centre the string in a width of 24 (23 would round the sixth and eighth lines down which is not what we want) and output each string on its own line.
/h/ 3//g/ 2//f/ 1//e/aa//d/
a//c/ //b/ea//a/c /bc1bc
b1gfbdac1hhfecdaf 4 6 4fe dc1 5f0f0 5facdac15g0f5e d 7g1h5h5g1 7a d8g8 56 70 56g8 8c
9h6 84f26f26 84h6 9
bg52b dac462 462ec
b 924b daf716f716e
66551112000090.digits.map{|i|n=0;$*.map!{|j|n+n=j}<<1;puts ($*[i..~i]*' ').center 23if i<9}
Iterates over the digits of 66551112000090 (from the right), one digit for each row of the tree. Each digit, with the exception of the 9, specifies the number of elements to be omitted from the beginning and end of the corresponding row of Pascal's triangle. The 9 is a distinct digit used to suppress printing of the second row of the triangle.
The Pascal's triangle code is based on this answer by @manatwork, golfed further by abusing the predefined array $*. $* normally stores the list of command-line options. Since we use no command-line options, $* is initialised to the empty array.
{i}{i}{i}ii{c}c{i}{i}dddc{d}{d}iii{c}c{d}{d}ddc{i}{i}iiccccccccc{i}{i}dddc{d}{d}iiic{i}{i}ddc{d}{d}iic{i}{i}dddc{d}{d}iiiccccccccc{d}{d}ddc{i}{i}iicccccccc{i}{i}dddc{d}{d}iiic{i}{i}dc{d}{d}ic{i}{i}dc{d}{d}ic{i}{i}dddc{d}{d}iiicccccccc{d}{d}ddc{i}{i}iiccccccc{i}{i}dddc{d}{d}iiic{i}{i}c{d}{d}c{i}{i}iic{d}{d}ddc{i}{i}c{d}{d}c{i}{i}dddc{d}{d}iiiccccccc{d}{d}ddc{i}{i}iiccccc{i}{i}dddc{d}{d}iiic{i}{i}ic{d}{d}dc{i}{i}dddcdc{d}ddddddc{i}{i}dddcdc{d}ddddddc{i}{i}ic{d}{d}dc{i}{i}dddc{d}{d}iiiccccc{d}{d}ddc{i}{i}iicccccccc{i}{i}dddciiiic{d}{d}dc{i}{i}ddcddc{d}ddddddc{i}{i}dddciiiic{d}{d}dccccccc{d}{d}ddc{i}{i}iicccc{i}{i}iiic{d}{d}dddc{i}{i}ddcdc{d}{d}iiic{i}{i}dciic{d}{d}dc{i}{i}dciic{d}{d}dc{i}{i}ddcdc{d}{d}iiic{i}{i}iiic{d}{d}dddcccc{d}{d}ddc{i}{i}iiccc{i}{i}iiiic{d}{d}ddddc{i}{i}ddciiiiiic{d}{d}ddddc{i}{i}icic{d}{d}ddc{i}{i}iiic{d}iiic{d}ddddddc{i}{i}icic{d}{d}ddc{i}{i}ddciiiiiic{d}{d}ddddc{i}{i}iiiic{d}{d}ddddcc{d}{d}ddc{{i}ddddd}dddc{d}{d}dddddc{i}{i}dciiic{d}{d}ddc{i}{i}iiiicddddc{d}{d}c{i}{i}dddciciiiic{d}{d}ddc{i}{i}dddciciiiic{d}{d}ddc{i}{i}iiiicddddc{d}{d}c{i}{i}dciiic{d}{d}ddc{i}{i}iiiiic{{d}iiiii}iiic{i}{i}ii{c}{i}{i}ddciiicdddc{d}{d}ii{c}{d}{d}ddc{i}{i}iicccccccc{i}{i}ciicddddc{d}{d}iic{i}{i}ciicddddc{d}{d}iicccccccc{d}{d}ddc{i}{i}ii{c}{i}{i}iiiiic{d}iiiciic{d}{d}{c}{d}{d}ddc{i}{i}iiccccccc{i}{i}dddciiiiiicddddddciiiiic{d}{d}ddc{i}{i}dddciiiiiicddddddciiiiic{d}{d}ddccccccc
I tried to fail, and I succeeded!
