Seidel Triangle

Question

The Seidel Triangle is a mathematical construction similar to Pascal's Triangle, and is known for it's connection to the Bernoulli numbers.

The first few rows are:

      1
      1  1
   2  2  1
   2  4  5  5
16 16 14 10 5
16 32 46 56 61 61

Each row is generated as follows:

If the row number is even (1-indexed):

• Bring down the first item of the previous row

• Every next item is the sum of the previous item and the item above it

• Duplicate the last item

If the row number is odd:

• Bring down the last item of the previous row

• Going backwards, each item is the sum of the previous item and the item above it

• Duplicate what is now the first item.

Basically, we construct the triangle in a zig-zag pattern:

    1
    v
    1 > 1
        v
2 < 2 < 1
v
2 > 4 > 5 > 5

For more information, see the Wikipedia page on Bernoulli numbers.

The Challenge:

Given n, either as a function argument or from STDIN, print or return either the nth row of the Seidel triangle or the first n rows. You may use either 0 or 1 indexing.

You do not need to handle negative or non-integer input (nor 0, if 1-indexed). You do not have to handle outputs larger than 2147483647 = 2^31 - 1

As this is code-golf, do this in as few bytes as possible.

Examples:

In these examples the return value is the nth row, 0-indexed.

Input   ->  Output

0           1
1           1 1
2           2 2 1
6           272 272 256 224 178 122 61
13          22368256 44736512 66750976 88057856 108311296 127181312 144361456 159575936 172585936 183194912 191252686 196658216 199360981 199360981

Answer 1

Brain-Flak, 66 bytes

<>(())<>{({}[()]<(()[{}]<<>{(({}<>{}))<>}>)>)}{}{{}<>{({}<>)<>}}<>

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Row is 0-indexed.

# Push 1 (the contents of row 0) on other stack; use implicit zero as parity of current row
<>(())<>

# Do a number of times equal to input:
{({}[()]<

  # Subtract the row parity from 1
  (()[{}]<

    # For each entry in old row:
    <>{

      # Add to previous entry in new row and push twice
      (({}<>{}))<>

    }

  >)

>)}{}

# If row parity is odd:
{{}

  # Reverse stack for output
  <>{({}<>)<>}

# Switch stacks for output
}<>

Answer 2

JavaScript (SpiderMonkey), 67 bytes

This code abuses the sort() method and doesn't work on all engines.

Rows are 0-indexed.

f=(n,a=[1],r)=>n--?f(n,[...a.map(n=>k+=n,k=0),k].sort(_=>n|r),!r):a

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How?

We conditionally reverse an array by using the sort() method with a callback function that ignores its parameters and returns either 0 or a positive integer. Don't try this at home! This only works reliably on SpiderMonkey.

let A = [1,2,3,4,5] and B = [1,2,3,4,5,6,7,8,9,10,11]

             | SpiderMonkey (Firefox)  | V8 (Chrome)             | Chakra (Edge)
-------------+-------------------------+-------------------------+------------------------
A.sort(_=>0) | 1,2,3,4,5               | 1,2,3,4,5               | 1,2,3,4,5
A.sort(_=>1) | 5,4,3,2,1               | 5,4,3,2,1               | 1,2,3,4,5
B.sort(_=>0) | 1,2,3,4,5,6,7,8,9,10,11 | 6,1,3,4,5,2,7,8,9,10,11 | 1,2,3,4,5,6,7,8,9,10,11
B.sort(_=>1) | 11,10,9,8,7,6,5,4,3,2,1 | 6,11,1,10,9,8,7,2,5,4,3 | 1,2,3,4,5,6,7,8,9,10,11

Note that V8 is probably using different sort algorithms depending on the length of the array (less or more than 10 elements).

Commented

f = (                     // f = recursive function taking:
  n,                      //   n   = row counter
  a = [1],                //   a[] = current row, initialized to [1]
  r                       //   r   = 'reverse' flag, initially undefined
) =>                      //
  n-- ?                   // decrement n; if it was not equal to zero:
    f(                    //   do a recursive call with:
      n,                  //     - the updated value of n
      [ ...a.map(n =>     //     - a new array:
          k += n, k = 0   //       - made of the cumulative sum of a[]
        ), k              //         with the last value appended twice
      ].sort(_ => n | r), //       - reversed if n is not equal to 0 or r is set
      !r                  //     - the updated flag r
    )                     //   end of recursive call
  :                       // else:
    a                     //   stop recursion and return a[]

Answer 3

Perl 6, 52 bytes

{(1,{[[\+] |.reverse,0]}...*)[$_].sort((-1)**$_*-*)}

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Answer 4

Jelly, 12 bytes

1;0SƤUƊ⁸¡U⁸¡

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Answer 5

Python 3, 98 91 bytes

from itertools import*
f=lambda n:n and[*accumulate(f(n-1)[::n&1or-1]+[0])][::n&1or-1]or[1]

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Switching to 0-based row numbering saved 7 bytes.

Answer 6

Haskell, 89 87 82 bytes

(cycle[r,id]!!)<*>s
r=reverse
s 0=[1]
s n=let a=zipWith(+)(0:a)$(r.s$n-1)++[0]in a

Just s prints the lines in the zig-zag order, the anonymous function on the first row reverses half of the rows.

Thanks to @nimi for saving 5 bytes!

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Answer 7

Python 2, 103 97 bytes

f=lambda n,r=[1],k=2:n and f(n-1,[sum(r[-2::-1][:i],r[-1])for i in range(k)],k+1)or r[::-(-1)**k]

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Non-recursive version (easier to read):

Python 2, 106 bytes

def f(n,r=[1],j=1):
 while n:
	a=r[-2::-1];r=r[-1:];n-=1;j=-j
	for x in a+[0]:r+=[r[-1]+x]
 return r[::-j]

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Surely, better is possible!

Answer 8

Python 3, 91 bytes

from numpy import *
s=lambda n:n and cumsum(s(n-1)[::n%2*2-1]+[0]).tolist()[::n%2*2-1]or[1]

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Answer 9

Julia 0.6, 85 bytes

r(l,n=cumsum(l))=[n...,n[end]]
w=reverse
f(n)=n<2?[1]:n%2<1?r(f(n-1)):w(r(w(f(n-1))))

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This is a recursive solution in Julia. Note that it has 1-based indexing. Hence the tests.

Ungolfed version, to understand the logic:

function new_row(last_row)
    new_row = cumsum(last_row)
    push!(new_row, new_row[end])
    return new_row
end


function triangle(n)
    if n == 1
        return [1]
    elseif mod(n,2) == 0
        return new_row(triangle(n-1))
    else
        return reverse(new_row(reverse(triangle(n-1))))
    end
end

Asa bonus, here's a non-recursive version, but this is longer:

w=reverse;c=cumsum
r(l,i)=i%2<1?c([l...,0]):w(c(w([0,l...])))
f(n,l=[1])=(for i=2:n l=r(l,i)end;l)

Answer 10

Husk, 16 bytes

!z*:1İ_¡ȯ↔S:→∫;1

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There may be a better way for arriving at the rows in the correct order. To get the correct output, this program reverses every other row.

Explanation

!z*:1İ_¡ȯ↔S:→∫;1
              ;1 array [1]
       ¡ȯ        apply the following functions repeatedly on it, collecting results:
             ∫   cumulative sum
          S:→    repeat the last element
         ↔       reverse it
                 this generates all the required rows
 z*              zip multiply that with:
   :1İ_          [1,-1,1,-1,1,-1,...]
                 which reverses every other row
!                Get row at input index

Husk, 16 bytes

?I↔%2¹!¡ȯ↔S:→∫;1

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Uses a ternary instead.

Answer 11

Haskell, 59 bytes

(k!!)<*>(iterate(r.scanr1(+).(0:))[1]!!)
r=reverse
k=r:id:k

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Answer 12

k4, 17 bytes

{x|:/x{|+\x,0}/1}

Adapted from @razetime's APL answer.

Answer 13

APL (Dyalog Unicode), ²² 18 bytes

{⊖⍣⍵{⌽+\⍵,0}⍣⍵,1} 

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Same method as my Husk answer.

-4 bytes from coltim.

Explanation

{⊖⍣⍵{⌽(⊢,⊢/)+\⍵}⍣⍵,1} ⍵ ← input(n)
                  ,1  convert 1 to an array [1]
    {          }⍣⍵    apply the following n times:
            +\⍵       cumulative sum
      (⊢,⊢/)          duplicate the last element
     ⌽                reverse
 ⊖⍣⍵                  then reverse n times