编写一个函数/程序，以众所周知的数字顺序输出n第一个n元素或第一个元素：
1, 2, 4, 8, 16 ...
哦，等等...我忘记了前几个数字：
1, 1, 1, 1, 2, 4, 8, 16 ...
哎呀，我会增加一些其他措施：
1, 1, 1, 1, 2, 4, 8, 16, 33, 69, 146, 312, 673, 1463, 3202, 7050, 15605, 34705 ...
这些数字是由（零索引）公式给出的广义加泰罗尼亚语数字：
哪里
这是OEIS A004149。
您可以选择是否要对序列进行零索引或一索引。当然，顺序必须相同，因此，如果对公式进行单索引，则必须重写该公式。
Answers:
{1,1,1,1,{sum @_[2..*]Z*@_[@_-4...0,0]}...*}匿名代码块，它返回值的惰性无限序列。这几乎实现了所描述的顺序，它的快捷方式是zip将所有第二个元素之后的所有元素乘以zip，然后将列表的倒数从第四个元素开始，最后添加一个额外的元素1。
{ } # Anonymous code block
...* # Create an infinite sequence
1,1,1,1, # Starting with four 1s
{ } # Where each new element is:
sum # The sum of
@_[2..*] # The second element onwards
Z* # Zip multiplied with
@_[@_-4...0 ] # The fourth last element backwards
,0 # And 1$ƒˆ¯Âø¨¨¨PO
输出第0个索引的第n个元素。
$ # push 1 and the input
ƒ # repeat (input+1) times
ˆ # add the top of the stack (initially 1) to the global array
¯ # push the global array
 # and a reversed copy of it
ø # zip the two together, giving a list of pairs
¨¨¨ # drop the last 3 pairs
P # take the product of each pair (or 1 if the list is empty)
O # take the sum of those products
# after the last iteration, this is implicitly output;
# otherwise, it's added to the global array by the next iteration
4Å1λ£₁λ¨Â¦¦s¦¦*O+
不小于现有的05AB1E答案，但是我想尝试新的05AB1E版本的递归功能作为自己的练习。也许可以打几个字节。编辑：确实可以，请参见下面@Grimy的05AB1E答案的递归版本，它是13 个字节。
说明：
λ # Create a recursive environment,
£ # to output the first (implicit) input amount of results after we're done
4Å1 # Start this recursive list with [1,1,1,1], thus a(0)=a(1)=a(2)=a(3)=1
# Within the recursive environment, do the following:
λ # Push the list of values in the range [a(0),a(n)]
¨ # Remove the last one to make the range [a(0),a(n-1)]
 # Bifurcate this list (short for Duplicate & Reverse copy)
¦¦ # Remove the first two items of the reversed list,
# so we'll have a list with the values in the range [a(n-3),a(0)]
s # Swap to get the [a(0),a(n-1)] list again
¦¦ # Remove the first two items of this list as well,
# so we'll have a list with the values in the range [a(2),a(n-1)]
* # Multiply the values at the same indices in both lists,
# so we'll have a list with the values [a(n-3)*a(2),...,a(0)*a(n-1)]
O # Take the sum of this list
₁ + # And add it to the a(n-1)'th value
# (afterwards the resulting list is output implicitly)@Grimy的 13 字节版本（如果您还没有，请确保对他的回答进行投票！）：
1λ£λ1šÂ¨¨¨øPO可以再次更改为基于0的索引或无限列表：1λèλ1šÂ¨¨¨øPOλλ1šÂ¨¨¨øPO
说明：
λ # Create a recursive environment,
£ # to output the first (implicit) input amount of results after we're done
1 # Start this recursive list with 1, thus a(0)=1
# Within the recursive environment, do the following:
λ # Push the list of values in the range [a(0),a(n)]
1š # Prepend 1 in front of this list
 # Bifurcate the list (short for Duplicate & Reverse copy)
¨¨¨ # Remove (up to) the last three value in this reversed list
ø # Create pairs with the list we bifurcated earlier
# (which will automatically remove any trailing items of the longer list)
P # Get the product of each pair (which will result in 1 for an empty list)
O # And sum the entire list
# (afterwards the resulting list is output implicitly)n=100时间为0.65秒，但是当我禁用延迟加载时，它会在60秒后超时，即使是n=25。
输出n第1个索引的th项。
@Zí*Zz2)Ťx}g4Æ1
@Zí*Zz2)Ťx}g4Æ1 :Implicit input of integer U
@ :Function taking an array as an argument via parameter Z
Zí : Interleave Z with
Zz2 : Z rotated clockwise by 180 degrees (simply reversing would be a bye shorter but would modify the original array)
* : Reduce each pair by multiplcation
) : End interleave
Å : Slice off the first element
¤ : Slice off the first 2 elements
x : Reduce by addition
} :End function
g :Pass the following as Z, push the result back to it and repeat until it has length U
4Æ1 :Map the range [0,4) to 1s
:Implicit output of the last element
: f recursive dup 4 > if 0 over 3 do over 1- i - f i f * + loop else 1 then nip ;输出是n项，输入是1索引
编辑：通过切换到xnor的公式保存了17个字节。通过使用1索引保存了另外1个字节
: f \ start a new word definition
recursive \ mark that this word will be recursive
dup 4 > \ duplicate the input and check if it is greater than 4
if \ if it is:
0 over \ create an accumulator and copy n to top of stack
3 do \ start counted loop from 3 to n-1
over 1- i - f \ recursively calculate f(n-1-i)
i f \ recursively calculate f(i)
* + \ multiply results and add to accumulator
loop \ end the counted loop
else \ otherwise, if n < 5
1 \ put 1 on the stack
then \ end the if block
nip \ drop n from the stack
; \ end the word definitionＦ⁵⊞υ¹ＦＮ⊞υΣ✂Ｅ⮌υ×κ§υλ³→Ｉ§υ±⁴
在线尝试！链接是详细版本的代码。尽管在内部使用1索引进行计算，但打印0索引的第n个数字。说明：
Ｆ⁵⊞υ¹
从开始a[0] = a[1] = a[2] = a[3] = a[4] = 1。是的，这是1索引的，但是还有一个额外的第零值。这就是您的高尔夫代码。
ＦＮ
计算附加n条件。这是过分的，但是当时，它使找到所需的术语更加容易n<5。
⊞υΣ✂Ｅ⮌υ×κ§υλ³
对于每个术语，将下一个术语计算为到目前为止的术语总和乘以到目前为止的术语的反向，不包括三个术语。
→
这是一个禁止操作的方法，用于诱使Charcoal解析的2参数形式Slice，否则我将不得不使用一种不太复杂的方法来删除三个术语。
Ｉ§υ±⁴
输出第四项。
J*4]1VQ=+J+eJsPP*M.t,PJ_PJ0;<J
J*4]1VQ=+J+eJsPP*M.t,PJ_PJ0;<JQ # Full program, last Q = input (implicitly added)
J*4]1 # J = 4 * [1] (=[1,1,1,1])
VQ # for N in range(Q):
=+J # J +=
+eJ # J[-1] +
s # sum( )
*M # map(__operator_mul, )
.t 0 # transpose( , pad=0)
, # [ , ]
PJ # J[:-1]
_PJ # J[1::-1]
<JQ # J[::Q]
<@
g=(1:4).^0;for(i=3:(n=input('')))g(i+2)=g(4:i+1)*g(i-(2:i-1))';end;g(end)-2个字节，感谢Stewie Griffin。命令式方法再次胜过函数式递归方法。如下所示。
f(f=@(a)@(n){@()sum(arrayfun(@(k)a(a)(k)*a(a)(n-2-k),2:n-1)),1}{2-(n>3)}())Captcha在发布此信息时想验证我是人类。老实说，我不太确定。
n<4。
-1个字节感谢Jonathan。
int a(int n){int k=2,s=0;while(++k<n)s+=a(k)*a(n+~k);return s?s:1;}
a(n-1-k)是a(n+~k)？
a(++k)*a(n-k)可能起作用，并且从删除了2个字节for。但是我闻到了不确定的行为。
a(n-1-k)为a(n-k)，对吗？