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Untuk tantangan ini, 'metasequence' akan didefinisikan sebagai urutan angka di mana tidak hanya angka itu sendiri akan meningkat, tetapi juga kenaikan, dan kenaikan akan meningkat dengan nilai yang meningkat, dll.

Misalnya, metasequence tingkat 3 akan dimulai sebagai:

1 2 4 8 15 26 42 64 93 130 176

karena:

    1 2 3  4  5  6  7  8   9       >-|
      ↓+↑ = 7                        | Increases by the amount above each time
  1 2 4 7  11 16 22 29 37  46  >-| <-|
                                 | Increases by the amount above each time
1 2 4 8 15 26 42 64 93 130 176 <-|

Tantangan

Dengan bilangan bulat positif, hasilkan dua puluh item pertama dari metasequence dari tier itu.

Uji kasus

Input: 3Keluaran:[ 1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176, 232, 299, 378, 470, 576, 697, 834, 988, 1160 ]

Input: 1Keluaran:[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 ]

Input: 5Keluaran:[ 1, 2, 4, 8, 16, 32, 63, 120, 219, 382, 638, 1024, 1586, 2380, 3473, 4944, 6885, 9402, 12616, 16664 ]

Input: 13Keluaran:[ 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16383, 32752, 65399, 130238, 258096, 507624 ]

Seperti yang Anda menyadari, pertama $t+1$ item dari setiap urutan tingkat $t$ adalah yang pertama $t+1$ kekuatan dari 2 ...

Aturan

• Celah standar berlaku
• Ini adalah kode-golf , jadi jawaban tersingkat dalam byte menang

Jelly , 8 7 byte

20ḶcþŻS

Cobalah online!

   cþ       Table of binom(x,y) where:
20Ḷ           x = [0..19]
     Ż        y = [0..n]    e.g.  n=3 → [[1, 1, 1, 1, 1, 1,  …]
                                         [0, 1, 2, 3, 4, 5,  …]
                                         [0, 0, 1, 3, 6, 10, …]
                                         [0, 0, 0, 1, 4, 10, …]]

      S     Columnwise sum.           →  [1, 2, 4, 8, 15, 26, …]

Ini menggunakan @ wawasan alephalpha yang

Bahasa Wolfram (Mathematica) , 34 byte

0~Range~19~Binomial~i~Sum~{i,0,#}&

Cobalah online!

Metasequence tingkat $n$ adalah jumlah dari elemen $n+1$ pertama dari setiap baris segitiga Pascal.

Haskell , 34 byte

(iterate(init.scanl(+)1)[1..20]!!)

Menggunakan input yang diindeks 0 ( f 4tingkat pengembalian 5.)

Haskell , 36 byte

f 1=[1..20]
f n=init$scanl(+)1$f$n-1

Cobalah online! Menggunakan input yang diindeks 1 ( f 5tingkat pengembalian 5.)

Penjelasan

scanl (+) 1adalah fungsi yang mengambil jumlah parsial dari daftar, mulai dari (dan prepending) 1.

Sebagai contoh: scanl (+) 1 [20,300,4000]sama dengan [1,21,321,4321].

Ternyata tingkat $n$ hanya fungsi ini diterapkan $(n-1)$ kali ke daftar $[1,2,3,\dots]$ .

(Atau yang setara: $n$ kali ke daftar semua yang ada.)

Kami menggunakan salah satu initatau [1..20-n]untuk memperhitungkan daftar yang semakin lama setiap $1$ aplikasi.

Brain-Flak , 84 82 byte

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

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Beranotasi

<>               Switch to the off stack
((()()()()()){}) Push 10
{({}[((()))])}{} Make twice that many 1s
<>               Switch back
{                While ...
({}[(())]<       Subtract one from the input and push 1
<>               Switch
{                For every x on the stack
({}<>({}))<>     Remove x and add it to a copy of the other TOS
}                End loop
<>{}             Remove 1 element to keep it 20
{({}<>)<>}       Copy everything back to the other stack
>)}<>            End scopes and loops

Cobalah online!

R, 36 bytes

rowSums(outer(0:19,0:scan(),choose))

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Thanks to @Giuseppe for suggesting outer.

This is based on the approach @alephalpha described

Python 2, 69 58 55 bytes

Saved bytes thanks to ovs and Jo King; also, it works in Python 3 now as well.

m=lambda t:[1+sum(m(t-1)[:n])for n in range(~t and 20)]

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The math

Let $a(t,n)$ be the $n^{th}$ term (0-indexed) of the sequence at tier $t$. A little analysis leads to the following recurrence formula:

Working backwards, we define $a(0,n) = 1$ and $a(-1,n) = 0$ for all $n$. These definitions will simplify our base case.

The code

We define a function m(t) that returns the first 20 elements of the sequence at tier t. If t is nonnegative, we use the recursive formula above; if t is -1, we return an empty list. The empty list works as a base case because the result of each recursive call is sliced ([:n]) and then summed. Slicing an empty list gives an empty list, and summing an empty list gives 0. That's exactly the result we want, since tier $-1$ should behave like a constant sequence of all $0$'s.

m=lambda t:                     # Define a function m(t):
 [          ]                   # List comprehension
     for n in range(         )  # for each n from 0 up to but not including...
                    ~n and 20   # 0 if n is -1, else 20:
  1+sum(          )             # a(t,n) = 1 + sum of
              [:n]              # the first n elements of
        m(t-1)                  # the previous tier (calculated recursively)

dzaima/APL REPL, 14 bytes

(+\1,19↑)⍣⎕⍳20

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(+\1,19↑)⍣⎕⍳20
(       )⍣⎕     repeat the function below input times:
 +\               cumulative sum of
   1,             1 prepended to
     19↑          the first 19 items of the previous iteration
           ⍳20  starting with the first 20 integers

Pari/GP, 39 bytes

n->Vec(sum(i=1,n+1,(1/x-1)^-i)+O(x^21))

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Pari/GP, 40 bytes

n->Vec((1-(1/x-1)^-n++)/(1-2*x)+O(x^20))

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The generating function of the tier $n$ metasequence is:

Perl 6, 34 32 bytes

-2 bytes thanks to Jo King

{(@,{[\+] 1,|.[^19]}...*)[$_+1]}

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Explanation

{                              }  # Anonymous block
   ,                ...*  # Construct infinite sequence of sequences
  @  # Start with empty array
    {              }  # Compute next element as
     [\+]     # cumulative sum of
          1,  # one followed by
            |.[^19]  # first 19 elements of previous sequence
 (                      )[$_+1]  # Take (n+1)th element

Python 3.8 (pre-release), 62 bytes

f=lambda n:[t:=1]+[t:=t+n for n in(n and f(n-1)[:-1]or[0]*19)]

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Explanation

f=lambda n:     # funtion takes a single argument
     [t:=1]     # This evaluates to [1] and assigns 1 to t
                # assignment expressions are a new feature of Python 3.8
       +        # concatenated to
     [  ....  ] # list comprehension

# The list comprehesion works together with the
# assignment expression as a scan function:
[t := t+n for n in it]
# This calculates all partial sums of it 
# (plus the initial value of t, which is 1 here)

# The list comprehension iterates
# over the first 19 entries of f(n-1)
# or over a list of zeros for n=0
 for n in (n and f(n-1)[:-1] or [0]*19)

R (63 47 bytes)

function(n,k=0:19)2^k*pbeta(.5,pmax(k-n,0),n+1)

Online demo. This uses the regularised incomplete beta function, which gives the cumulative distribution function of a binomial, and hence just needs a bit of scaling to give partial sums of rows of Pascal's triangle.

Octave (66 46 bytes)

@(n,k=0:19)2.^k.*betainc(.5,max(k-n,1E-9),n+1)

Online demo. Exactly the same concept, but slightly uglier because betainc, unlike R's pbeta, requires the second and third arguments to be greater than zero.

Many thanks to Giuseppe for helping me to vectorise these, with significant savings.

Ruby, 74 bytes

a=->b{c=[1];d=0;b==1?c=(1..20).to_a: 19.times{c<<c[d]+(a[b-1])[d];d+=1};c}

Ungolfed version:

def seq num
    ary = [1]
    index = 0
    if num == 1
        ary = (1..20).to_a
    else
        19.times{ary << ary[index]+seq(num-1)[index]; index+=1}
    end
    return ary
end

Quite resource-intensive--the online version can't calculate the 13th metasequence.

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Wolfram Language (Mathematica), 42 bytes

Nest[FoldList[Plus,1,#]&,Range[21-#],#-1]&

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JavaScript (Node.js), 58 bytes

t=>Array(20).fill(t).map(g=(t,i)=>i--*t?g(t,i)+g(t-1,i):1)

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It is trivial to write down following recursive formula based on the description in question

05AB1E, 11 9 bytes

20LIF.¥>¨

0-indexed

Try it online or verify all test cases.

Explanation:

20L        # Create a list in the range [1,20]
   IF      # Loop the input amount of times:
     .¥    #  Get the cumulative sum of the current list with 0 prepended automatically
       >   #  Increase each value in this list by 1
        ¨  #  Remove the trailing 21th item from the list
           # (after the loop, output the result-list implicitly)

R, 59 49 bytes

f=function(n)`if`(n,Reduce(`+`,f(n-1),1,,T),1:20)

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Recursively Reduce with +, init=1 and accumulation=TRUE to avoid having to subset. Thanks to Criminally Vulgar for suggesting the recursive approach!

JavaScript (ES6),  68  67 bytes

f=(n,a=[...f+f])=>n--?f(n,[s=1,...a.map(x=>s-=~--x)]):a.slice(0,20)

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JavaScript (ES6), 63 bytes

NB: this version works for $n\le20$.

f=(n,a=[...Array(20-n)])=>n--?f(n,[s=1,...a.map(x=>s+=x||1)]):a

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J, 24 bytes

<:(1+/\@,])^:[(1+i.20)"_

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NOTE: Turns out this is a translation of dzaima's APL answer, though I actually didn't notice it before writing this.

explanation

<: (1 +/\@, ])^:[ (1+i.20)"_
<:                           NB. input minus 1 (left input)
                  (1+i.20)"_ NB. 1..20 (right input)
   (         )^:[            NB. apply verb in parens 
                             NB. "left input" times
   (1     , ])               NB. prepend 1 to right input
   (  +/\@   )               NB. and take scan sum

Ruby, 49 bytes

f=->n{n<1?[1]*20:[o=1]+f[n-1][0,19].map{|x|o+=x}}

Recursive definition: Tier 0 is 1,1,1,1... and each subsequent tier is 1 followed by a sequence whose first differences are the previous tier. Annoyingly this would give me 21 values if I didn't explicitly slice out the first 20; seems like there should be a way to shorten this by avoiding that.

Retina, 59 bytes

.+
19*$(_,

Replace the input with 19 1s (in unary). (The 20th value is 0 because it always gets deleted by the first pass through the loop.)

"$+"{`
)`

Repeat the loop the original input number of times.

(.+),_*
_,$1

Remove the last element and prefix a 1.

_+(?<=((_)|,)+)
$#2*

Calculate the cumulative sum.

_+
$.&

Convert to decimal.

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C# (Visual C# Interactive Compiler), 120 bytes

n=>{for(long i=-1,h=0,m=0;++i<20;Print(i<1?1:h))for(m=h=0;m<=n;)h+=f(i)/(f(m)*f(i-m++));long f(long a)=>a>1?a*f(a-1):1;}

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Based off of alephalpha's formula.

Rust, 135 bytes

fn t(m:u64)->Vec<u64>{let f=|y|(1..=y).fold(1,|a,n|a*n);(0..20).map(|i| (0..=u64::min(i,m)).fold(0,|a,x|a+f(i)/f(x)/f(i-x))).collect()}

used @alephalpha 's idea, like several others. there is no builtin factorial so that takes up at least 36 bytes, (plus dealing with negatives). no builtin choose, another 16 bytes. iterator->declared vector type, 20 bytes.. etc etc.

Ungolfed at play.rust-lang.org

R (60 59 bytes)

function(n)Reduce(function(p,q)2*p-choose(q-1,n),1:19,1,,1)

Online demo

Straightforward implementation of the observation

T(n,k) = 2 T(n-1,k) - binomial(n-1,k). - M. F. Hasler, May 30 2010

from OEIS A008949. The arguments to Reduce are the function (obviously), the array over which to map, the starting value, a falsy value (to fold from the left rather than the right), and a truthy value to accumulate the intermediate results in an array.

K (oK), 17 bytes

-1 byte thanks to ngn (switching from 0-indexed to 1-indexed)

{x(+\1,19#)/20#1}

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1-indexed

K (oK), 18 bytes

{x(+\1,19#)/1+!20}

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0-indexed

Jelly, 10 bytes

20RṖ1;ÄƲ⁸¡

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0-indexed.

Perl 5, 48 bytes

$x=1,@A=(1,map$x+=$_,@A[0..18])for 0..$_;$_="@A"

TIO

Japt, 15 bytes

0-indexed; replace h with p for 1-indexed.

ÈîXi1 å+}gNh20õ

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CJam (20 bytes)

1aK*{1\{1$+}/;]}q~*p

Online demo. This is a program which takes input from stdin and prints to stdout; for the same score an anonymous block (function) can be obtained as

{1aK*{1\{1$+}/;]}@*}

Dissection

This applies the definition literally:

1aK*      e# Start with an array of 20 1s
{         e# Loop:
  1\      e#   Push a 1 before the current list
  {1$+}/  e#   Form partial sums (including that bonus 1)
  ;]      e#   Ditch the last and gather in an array (of length 20)
}
q~*       e# Take input and repeat the loop that many times
p         e# Pretty print