This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
s = {1}; t = {{1}}; Do[AppendTo[t, BinCounts[#, {1, Max[#] + 1}] &[Flatten[t]]]; AppendTo[s, Max@ Flatten@ t], {58}]; s (* Michael De Vlieger, Nov 15 2022, after Peter J. C. Moses at A030717 *)
As an irregular triangle this begins: 0; 1, 1, 0; 2, 2, 2, 0; 3, 2, 4, 1, 1, 0; 4, 4, 4, 1, 4, 0; 5, 5, 4, 1, 6, 2, 1, 0; 6, 7, 5, 1, 6, 3, 3, 1, 0; 7, 9, 5, 3, 6, 4, 4, 2, 0; 8, 9, 6, 4, 9, 4, 5, 2, 1, 3, 0; 9, 10, 7, 5, 10, 6, 6, 3, 1, 4, 2, 0; 10, 11, 8, 6, 11, 6, 9, 3, 2, 5, 3, 2, 0; ... For row lengths see A347299. - _N. J. A. Sloane_, Aug 27 2021 From _David James Sycamore_, Oct 18 2021: (Start) a(1) is 0 because the count is reset, and as yet there is no zero term immediately following another term. a(2) = 1 since the count is reset, a(1) = 0 and a(0) precedes it. The count now increments to terms equal to 1. a(3) = 1 since a(2) = 1 and a(1) precedes it. a(4) = 0 because there is no term equal to 2 which is immediately preceded by another term. a(5) = 2 since the count is reset, a(1) = a(4) = 0 and a(0), a(3) respectively, precede them. (End)
# See Links section. - Luc Rousseau, May 02 2021
function [val,arr]=invSeq(N) % val = Nth term, arr = whole array up to N
k=0;
arr=zeros(1,N); % pre-allocate array
for i=1:N
an=sum((k==arr(2:i)));
arr(i)=an;
if an == 0
k = 0;
else
k=k+1;
end
end
val=arr(end);
end % Ben Cha, Nov 11 2022
a:= proc(n) option remember; local t;
t:= `if`(a(n-1)=0, 0, b(n-1)+1);
b(n):=t; add(`if`(a(j)=t, 1, 0), j=1..n-1)
end: b(1), a(1):= 0$2:
seq(a(n), n=1..120); # Alois P. Heinz, Mar 16 2021
a[n_] := a[n] = Module[{t}, t = If[a[n-1] == 0, 0, b[n-1]+1];
b[n] = t; Sum[If[a[j] == t, 1, 0], {j, 1, n-1}]];
b[1] = 0; a[1] = 0;
Array[a, 120] (* Jean-François Alcover, May 03 2021, after Alois P. Heinz *)
A342585_vec(N,c=[],i)=vector(N,j, while(#c<=i||#c<=c[i+1], c=concat(c,0)); c[i+=1]+if(c[1+c[i]]++&&!c[i]||j==1,i=0)) \\ M. F. Hasler, Nov 13 2021
\\ See Links section.
def calc(required_value_number):
values_lst = []
current_count = 0
new_value = 0
for i in range(required_value_number):
new_value = values_lst.count(current_count)
values_lst.append(new_value)
if new_value == 0:
current_count = 0
else:
current_count += 1
return new_value # Written by Gilad Moyal
from collections import Counter
def aupton(terms):
num, alst, inventory = 0, [0], Counter([0])
for n in range(2, terms+1):
c = inventory[num]
num = 0 if c == 0 else num + 1; alst.append(c); inventory.update([c])
return alst
print(aupton(84)) # Michael S. Branicky, Jun 12 2021
# Prints the first 10,068 terms
library("dplyr")
options(max.print=11000)
inventory <- data.frame(1, 0)
colnames(inventory) <- c("n", "an")
value_to_count = 0
n = 1
for(x in 1:128) # Increase the 128 for more terms. The number of terms
# given is on the order of x^1.9 in the region around 128.
{
status <- TRUE
while(status)
{
count <- length(which(inventory$an == value_to_count))
n = n + 1
inventory <- rbind(inventory, c(n, count))
status <- isTRUE(count != 0)
value_to_count = value_to_count + 1
}
value_to_count = 0
}
inventory # Damon Lay, Nov 10 2023
Stage 1: [
[3],
[]
]
Stage 2: [
[3, 1],
[3]
]
Stage 3: [
[3, 1, 2, 1],
[3, 3, 1]
]
Stage 4: [
[3, 1, 2, 1, 3, 1, 3],
[3, 3, 1, 3, 2, 1]
]
Stage 5: [
[3, 1, 2, 1, 3, 1, 3, 6, 2, 5],
[3, 3, 1, 3, 2, 1, 3, 2, 1]
]
Stage 6: [
[3, 1, 2, 1, 3, 1, 3, 6, 2, 5, 1, 1, 7, 4, 6],
[3, 3, 1, 3, 2, 1, 3, 2, 1, 6, 5, 3, 2, 1]
]
def a030777_list(generations)
rows = [[3], []]
(2..generations).each do
new_additions = rows.flatten.uniq.sort.reverse.map do |j|
[rows.flatten.count(j), j]
end
rows = rows.zip(new_additions.transpose).map { |r, n| r + n }
end
rows[0]
end # Peter Kagey, Apr 09 2020
Stage 1: [
[3],
[]
]
Stage 2: [
[3, 1],
[3]
]
Stage 3: [
[3, 1, 2, 1],
[3, 3, 1]
]
Stage 4: [
[3, 1, 2, 1, 3, 1, 3],
[3, 3, 1, 3, 2, 1]
]
Stage 5: [
[3, 1, 2, 1, 3, 1, 3, 6, 2, 5],
[3, 3, 1, 3, 2, 1, 3, 2, 1]
]
Stage 6: [
[3, 1, 2, 1, 3, 1, 3, 6, 2, 5, 1, 1, 7, 4, 6],
[3, 3, 1, 3, 2, 1, 3, 2, 1, 6, 5, 3, 2, 1]
]
def a030777_list(generations)
rows = [[3], []]
(2..generations).each do
new_additions = rows.flatten.uniq.sort.reverse.map do |j|
[rows.flatten.count(j), j]
end
rows = rows.zip(new_additions.transpose).map { |r, n| r + n }
end
rows[1]
end # Peter Kagey, Apr 09 2020
1; 1; 2; 2, 1; 3, 2; 3, 3, 1; 4, 3, 3;
import Data.List (sort, group)
a030717 n k = a030717_tabf !! (n-1) !! (k-1)
a030717_row n = a030717_tabf !! (n-1)
a030717_tabf = [1] : f [1] where
f xs = ys : f ((filter (> 0) ys) ++ xs) where
ys = h (group $ sort xs) [1..] where
h [] _ = []
h vss'@(vs:vss) (w:ws)
| head vs == w = (length vs) : h vss ws
| otherwise = 0 : h vss' ws
-- Reinhard Zumkeller, Dec 28 2014
t = {{1}}; Do[AppendTo[t, BinCounts[#, {1, Max[#] + 1}] &[Flatten[t]]], {30}];
Map[Length, t] (* A126027*)
Map[Total, t] (* A253170*)
Flatten[t] (* A333867 *) (* Peter J. C. Moses, Apr 09 2020 *)
Comments