A333867 -id:A333867 - cp's OEIS frontend

A342585 Inventory sequence: record the number of zeros thus far in the sequence, then the number of ones thus far, then the number of twos thus far and so on, until a zero is recorded; the inventory then starts again, recording the number of zeros.

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, 11, 11, 10

Offset: 1

Joseph Rozhenko, Mar 16 2021

To get started we ask: how many zero terms are there? Since there are no terms in the sequence yet, we record a '0', and having recorded a '0', we begin again: How many zero terms are there? There is now one 0, so we record a '1' and continue. How many 1's are there? There's currently one '1' in the sequence, so we record a '1' and continue. How many 2's are there? There are no 2's yet, so we record a '0', and having recorded a 0, we begin again with the question "how many zero terms are there?" And so on.
a(46) = 0 because no 8's appear before it; but note a higher number, namely 9, has appeared. - Michael S. Branicky, Mar 16 2021
A similar situation occurs at n=124, where 14 has not yet appeared in the sequence, although 15 has appeared.
Reminiscent of Van Eck's sequence A181391. - N. J. A. Sloane, May 02 2021
From Jan Ritsema van Eck, May 02 2021: (Start)
The first 1000 terms seem to grow more or less in saw-tooth fashion with the largest terms (= the number of 0's), as well as the distance between the 0's, both approximately equal to the inverse triangular numbers A003056 (see attached graph #1).
But the picture changes when we go out to 10000 terms. Around the 1700th term, the 1's become more frequent than the 0's and the largest values are consistently somewhat larger than the inverse triangular numbers. Around the 2500th term the 2's become the most frequent number. Also after some 4000 terms, the largest values become much larger than the inverse triangular numbers. See graph #2. (End)
Comment on the colored plot of the first 1000467 terms, from Hans Havermann, May 02 2021: (Start)
If one is drawing a points-joined graph, it will obscure some of the inherent large-number dynamics. To get around that, this plot joins the points with a green line, superimposing the actual points in blue. This plot was created by Mathematica.
Your browser will likely compress the very large image to window size, so click on it to expand.
The points fall into linear features of the various counts of the various integers. The count for each integer changes as we move towards infinity and hence crosses over (changes place with) other counts unpredictably.
I decided to chart (see the blue text) the twenty largest counts at the rightmost spike which runs from the zero at 997010 to the zero at 1000467. These largest values are for the counts of integers 2 to 21 and appear at a(997013) for the 2-count; a(997014) for the 3-count, ..., and a(997032) for the 21-count.
The counts are 15275, 26832, 40162, 48539, 56364, 54372, 53393, 43588, 37288, 27396, 22425, 16735, 13099, 11460, 9466, 8386, 7191, 6478, 5777, and 5208, respectively. In my text they are sorted largest-to-smallest and written "count @ integer-being-counted": 56364 @ 6, 54372 @ 7, 53393 @ 8, 48539 @ 5, ... 5208 @ 21. (End)
A useful view may be gained by plotting the sequence against itself with an offset. Using the "Plot 2" link in the web page footer, enter "A342585" as sequences 1 and 2. Select "Plot Seq2(n+shift) vs Seq1(n)" and "Draw line segments". Start with "1" as the shift. The sequence appears somewhat like a fan, the first 4 or 5 sectors showing clearly, later sectors overlying each other. Larger shift values effectively compress early sectors into the vertical axis, making later sectors more visible. - Peter Munn, May 08 2021
For a version where a row ends not at the first zero, but rather at the last zero, see A347317. - N. J. A. Sloane, Sep 10 2021
For n around 2.5*10^9, the upper envelope of the sequence seems to be growing roughly like n/50, or maybe like O(n/log(n)). - N. J. A. Sloane, Feb 10 2023

			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)

Rémy Sigrist, Table of n, a(n) for n = 1..25000
Brady Haran and N. J. A. Sloane, "A Number Sequence with Everything" (the Inventory Sequence A342585), Numberphile video, November 2022.
Hans Havermann, Colored plot of 1000467 terms [See Comments for a description of this plot]
Hugo Pfoertner, Listening to the first 100000 terms of A342585, YouTube video from "Talabfahrer".
Hugo Pfoertner, Hear 1 million terms of A342585, YouTube video from "Talabfahrer", alternative audio conversion.
Luc Rousseau, AWK program for A342585
Rémy Sigrist, Scatterplot of the first 10^6 terms
Rémy Sigrist, Scatterplot of the first 10^7 terms
Rémy Sigrist, Scatterplot of the first 10^8 terms
Rémy Sigrist, Table of n, a(n) for n = 1..100000
Rémy Sigrist, PARI program for A342585
Jan Ritsema van Eck, Graph #1: 1000 terms (blue) and inverse triangular numbers A003056 (orange)
Jan Ritsema van Eck, Graph #2: 10000 terms (blue) and inverse triangular numbers A003056 (orange)
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 21.
Index entries for sequences related to the inventory sequence

Records: A347305 and A348782.
Cf. A181391, A343878, A343880, A347299, A347315, A347316, A347317.
Other inventory-type sequences: A030717, A174382, A333867, A358066, A357443, A356784.
A012257 (cf. also A011784) reverses the inventory process.
See A347062, A347738, A355916, A355917, A355918, A357317 for variants.

# 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

PARI
```
\\ 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

A030717 The first list after the following procedure: starting with a list [1] and an empty list, repeatedly add the distinct values already in the first list in ascending order to the second list and add the corresponding frequencies of those values to the first list.

Original entry on oeis.org

1, 1, 2, 2, 1, 3, 2, 3, 3, 1, 4, 3, 3, 4, 3, 5, 1, 5, 3, 6, 2, 1, 6, 4, 7, 2, 2, 1, 7, 6, 7, 3, 2, 2, 1, 8, 8, 8, 3, 2, 3, 3, 8, 9, 11, 3, 2, 3, 3, 3, 8, 10, 15, 3, 2, 3, 3, 4, 1, 1, 10, 11, 18, 4, 2, 3, 3, 5, 1, 1, 1, 1, 14, 12, 20, 5, 3, 3, 3, 5, 1, 2, 2, 1

Offset: 1

Clark Kimberling

The clarifying comment that follows refers to the old name, which was: Row 1, where, at stage k>1, write i in row 1 and j in row 2, where i is the number of j's in row 1, for j=1,2,...,m, where m=max number in row 1 from stages 1 to k-1; state 1 is 1 in row 1.

Numbers j for which the count is 0 are omitted, cf. A333867 for the corresponding sequence where they are included. - Sean A. Irvine, Apr 08 2020

			First list begins:
  1;
  1;
  2;
  2, 1;
  3, 2;
  3, 3,  1;
  4, 3,  3;
  4, 3,  5, 1;
  5, 3,  6, 2, 1;
  6, 4,  7, 2, 2, 1;
  7, 6,  7, 3, 2, 2, 1;
  8, 8,  8, 3, 2, 3, 3;
  8, 9, 11, 3, 2, 3, 3, 3;

Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Sean A. Irvine)
Sean A. Irvine, Java program (github)

Cf. A030718 (2nd list), A030719 (row lengths), A006920.
Cf. A030707 (repeatedly adds values and frequencies from both lists).
Cf. A333867 (0 counts included, but not the counts of 0s).
Cf. A030723, A030724, A030725, A030726, A358423, A358424, A358425.

t = {{1}}; Do[AppendTo[t, BinCounts[#, {1, Max[#] + 1}] &[Flatten[t]]], {25}];
DeleteCases[Flatten[t], 0]  (* Peter J. C. Moses, Apr 09 2020 *)

More terms from Franklin T. Adams-Watters, Dec 14 2006
Rolled back to original definition and data by Sean A. Irvine, Apr 08 2020
Name revised in line with A030777 by Peter Munn, Oct 11 2022

A079668 Start with 1; at n-th step, write down what is in the sequence so far.

Original entry on oeis.org

1, 1, 1, 3, 1, 4, 1, 0, 2, 1, 3, 1, 0, 6, 1, 1, 2, 2, 3, 1, 4, 2, 0, 10, 1, 3, 2, 3, 3, 2, 4, 0, 5, 1, 6, 4, 0, 12, 1, 6, 2, 6, 3, 3, 4, 1, 5, 2, 6, 0, 7, 0, 8, 0, 9, 1, 10, 8, 0, 15, 1, 8, 2, 8, 3, 5, 4, 2, 5, 5, 6, 1, 7, 1, 8, 1, 9, 2, 10, 0, 11, 1, 12, 10

Offset: 1

N. J. A. Sloane Jan 26 2003

After 1 1 1 3 1, we see "4 1's, 0 2's and 1 3", so next terms are 4 1 0 2 3 1.

			Row n lists all terms written at the n-th step:
  1;
  1, 1;
  3, 1;
  4, 1,  0, 2, 1, 3;
  1, 0,  6, 1, 1, 2, 2, 3, 1, 4;
  2, 0, 10, 1, 3, 2, 3, 3, 2, 4, 0, 5, 1, 6;
  4, 0, 12, 1, 6, 2, 6, 3, 3, 4, 1, 5, 2, 6, 0, 7, 0, 8, 0, 9, 1, 10;
  ...

Alois P. Heinz, Table of n, a(n) for n = 1..20377 (first 55 steps)

For other versions see A051120, A055186, A079686.

Cf. A030717, A333867.

b:= proc(n) option remember; `if`(n<1, 0, b(n-1)+add(x^i, i=T(n))) end:
T:= proc(n) option remember; `if`(n=1, 1, (p->
      seq([coeff(p, x, i), i][], i=ldegree(p)..degree(p)))(b(n-1)))
    end:
seq(T(n), n=1..10);  # Alois P. Heinz, Aug 24 2025

s={1}; Do[s=Flatten[{s,{Count[s,#],#}&/@Range[Min[s],Max[s]]}],{20}];s (* Peter J. C. Moses, Mar 21 2013 *)

from itertools import islice
def agen(): # generator of terms
    a = [1]
    yield 1
    while True:
        counts = []
        for d in range(min(a), max(a)+1):
            c = a.count(d)
            counts.extend([c, d])
        a += counts
        yield from counts
print(list(islice(agen(), 84))) # Michael S. Branicky, Aug 24 2025

More terms from Sean A. Irvine, Aug 24 2025

A174382 T(1,0)=0 and for n > 1, T(n,k) is the number of k's in rows 1 to n - 1.

Original entry on oeis.org

0, 1, 1, 1, 1, 3, 1, 4, 0, 1, 2, 6, 0, 1, 1, 3, 8, 1, 1, 1, 0, 1, 4, 12, 1, 2, 1, 0, 1, 0, 1, 6, 16, 2, 2, 2, 0, 1, 0, 1, 0, 0, 0, 1, 11, 19, 5, 2, 2, 0, 2, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 19, 22, 8, 2, 2, 1, 2, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 27, 28, 11, 2, 2, 1, 2, 0, 2, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 2, 0, 0, 1

Offset: 1

Paolo P. Lava & Giorgio Balzarotti, Mar 17 2010

Construction as in A333867 but starting with a 0 and including a count of 0s at the start of each row. [Edited by Peter Munn, Oct 11 2022]

See A342585 for a similarly defined sequence that has been analyzed more and has lists of other related sequences. - Peter Munn, Oct 08 2022

			0;
1; # one zero
1,1; # one zero, one one
1,3; # one zero, three ones
1,4,0,1; # one zero, four ones, zero twos, one three

Reinhard Zumkeller, Rows n = 1..25 of table, flattened

Cf. A240508 (row lengths).

Cf. A333867, A342585.

import Data.List (sort, group)
a174382 n k = a174382_tabf !! (n-1) !! k
a174382_row n = a174382_tabf !! (n-1)
a174382_tabf = iterate f [0] where
   f xs = g (xs ++ [0, 0 ..]) [0..] (map head zs) (map length zs)
     where g   _ [] = []
           g (u:us) (k:ks) hs'@(h:hs) vs'@(v:vs)
             | k == h = u + v : g us ks hs vs
             | k /= h = u : g us ks hs' vs'
           zs = group $ sort xs
-- Reinhard Zumkeller, Apr 06 2014

b:= proc(n) option remember; `if`(n<1, 0, b(n-1)+add(x^i, i=T(n))) end:
T:= proc(n) option remember; `if`(n=1, 0, (p->
      seq(coeff(p, x, i), i=0..degree(p)))(b(n-1)))
    end:
seq(T(n), n=1..12);  # Alois P. Heinz, Aug 25 2025

cp's OEIS Frontend

A126027 Row lengths for A333867.

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A253170 Row sums of table A333867.

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A342585 Inventory sequence: record the number of zeros thus far in the sequence, then the number of ones thus far, then the number of twos thus far and so on, until a zero is recorded; the inventory then starts again, recording the number of zeros.

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A030717 The first list after the following procedure: starting with a list [1] and an empty list, repeatedly add the distinct values already in the first list in ascending order to the second list and add the corresponding frequencies of those values to the first list.

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A079668 Start with 1; at n-th step, write down what is in the sequence so far.

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A174382 T(1,0)=0 and for n > 1, T(n,k) is the number of k's in rows 1 to n - 1.

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A006920 At each step, record how many 1's, 2's, etc. have been seen so far in the sequence.

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A139124 Starting from 1, at any step count the appearances of k, where k ranges from the highest number to 1.