A032531 -id:A032531 - cp's OEIS frontend

A107263 Position where n (presumably) appears the last time in A032531, or 0 if n keeps appearing.

20678, 95695, 120724, 133876, 1148205

Offset: 0

Ralf Stephan, May 15 2005

The sequence resulted from analysis of A032531(n), n<= 2*10^6.
We can only speak of provisional values and, in the absence of any proof, I am not sure how rigorous these results are for n > 2*10^6. - Herman Jamke (hermanjamke(AT)fastmail.fm), Nov 03 2006
I extended the analysis of A032531(n) to all n<= 10^7. Same comments apply considering the new limit and, of course, the uniqueness of Stephan's sequence remains as always only a conjecture since there's no proof that the sequence should be anything different from the zero sequence for all, most or even any of the terms - Herman Jamke (hermanjamke(AT)fastmail.fm), Nov 08 2006
Search extended to the first 10^9 terms of A032531. - Andrew Howroyd, Nov 12 2024

Cf. A107262.

\\ Warning: conjectured sequence.
accrows(n)={my(v=vector(n), p=vector(n), idx=0); for(i=1, n, for(j=1, i, my(t=v[j]+1); if(t<=n, v[t]++; p[t]=idx); idx++)); p}
\\ search 5000 rows of A032531 (first binomial(5000+1,2) = 12502500 terms).
{ accrows(5000)[1..31] } \\ Andrew Howroyd, Nov 12 2024

Corrected and extended by Herman Jamke (hermanjamke(AT)fastmail.fm), Nov 03 2006, Nov 08 2006

A107261 Number of a(i), 0<=iA053615(n).

Original entry on oeis.org

0, 0, 2, 0, 1, 1, 3, 2, 2, 1, 3, 3, 3, 3, 3, 6, 0, 6, 3, 3, 4, 3, 3, 10, 1, 0, 1, 10, 3, 5, 5, 5, 3, 12, 1, 3, 2, 3, 1, 14, 4, 7, 5, 7, 4, 14, 3, 4, 2, 2, 2, 4, 5, 15, 7, 7, 5, 7, 7, 15, 5, 7, 2, 7, 0, 8, 2, 7, 5, 15, 9, 7, 6, 7, 9, 15, 5, 9, 3, 11, 1, 3, 1, 11, 3, 9, 5, 18, 9, 9, 6, 9, 9, 18, 5

Offset: 0

Ralf Stephan, May 15 2005

nonn

Andrew Howroyd, Table of n, a(n) for n = 0..10000

Cf. A032531, A107262.

\\ here b(n)=A053615(n)
b(n)={my(t=(sqrtint(4*n) + 1)\2); t-abs(t^2-n)}
seq(n)={my(f=vector(sqrtint(n)+1), a=vector(n+1)); for(i=0, n, my(k=f[1+b(i)]); a[1+i]=k; if(k<#f, f[1+k]++)); a} \\ Andrew Howroyd, Nov 12 2024

A347794 The "Look and Say" version of the Inventory sequence A342585.

Original entry on oeis.org

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

Offset: 0

Scott R. Shannon, Sep 13 2021

This sequences uses the same rules as the Inventory sequence A342585 except here each term is described using the "Look and Say" method of A005150. See the Examples below. Start with a(0) = 0.

			a(1) = 1 and a(2) = 0, as after a(0) there is "one 0".
a(3) = 1 and a(4) = 1, as after a(0)..a(2) there is "one 1".
a(5) = 0 and a(6) = 2, as after a(0)..a(4) there are "zero 2's". As a number with zero occurrences has appeared, the count resets back to counting zeros.
a(7) = 3 and a(8) = 0, as after a(0)..a(6) there are "three 0's".
a(9) = 3 and a(10) = 1 as after a(0)..a(8) there are "three 1's".
a(11) = 1 and a(12) = 2 as after a(0)..a(10) there is "one 2".
a(13) = 2 and a(14) = 3 as after a(0)..a(12) there are "two 3's".
a(15) = 0 and a(16) = 4 as after a(0)..a(14) there are "zero 4's". As a number with zero occurrences has appeared, the count resets back to counting zeros.

Scott R. Shannon, Image of the first 10 million terms.

Cf. A342585, A005150, A347317, A032531, A181391.

A348218 Variation on the Inventory Sequence A342585: record the number of previous terms which are divisible by the iterating number, starting at 1, until 0 is recorded, then restart the iterating number from 1. See the Comments.

Original entry on oeis.org

1, 1, 0, 2, 1, 0, 4, 2, 0, 6, 4, 1, 2, 0, 10, 7, 1, 2, 1, 1, 1, 0, 17, 8, 1, 3, 1, 1, 1, 1, 0, 25, 9, 3, 3, 2, 1, 1, 1, 1, 1, 0, 35, 10, 5, 3, 5, 1, 2, 1, 1, 2, 0, 45, 13, 7, 3, 7, 1, 4, 1, 2, 2, 0, 55, 16, 8, 6, 8, 2, 4, 4, 2, 2, 1, 0, 66, 26, 10, 9, 9, 3, 4, 4, 4, 3, 2, 0, 77, 32, 14, 13, 9, 3

Offset: 0

Scott R. Shannon, Oct 07 2021

nonn

This sequence is a variation of A342585. Instead of iteratively counting the occurrences of each number starting from zero and then repeating when zero is recorded, we start the iterating number at 1 and count the previous terms that are divisible by that number. This number increases for each division until zero previous terms is recorded, when the iterating number is reset to one and the divisor count repeats. The sequence starts with a(0) = 1.

After 10^7 terms the largest value is a(9990262) = 9988674, which is a count of the terms >= 1. The largest value the iterating divisor has reached is 13249.

			a(1) = 1 as the iterating dividing number starts at 1, and so far there has been one term, a(0), which is divisible by 1.
a(2) = 0 as the dividing number has increased to two, but there have been no terms in the sequence so far that are divisible by 2. The dividing number is now reset to 1.
a(3) = 2 as there have been two terms, a(0) and a(1), that are divisible by 1.
a(4) = 1 as there has been one term, a(3), that is divisible by 2.
a(5) = 0 as there have been no terms divisible by 3. The dividing number is now reset to 1.

Cf. A342585, A027750, A347738, A345730, A032531, A181391.

A348288 Variation on the Inventory Sequence A342585: the same rules as A342585 are used except that the terms count the occurrences of the iterating number, treated as a string, in the string concatenation of all previous terms. See the Comments.

Original entry on oeis.org

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, 4, 2, 0, 9, 10, 8, 5, 11, 6, 6, 2, 2, 4, 5, 3, 0, 11, 15, 10, 6, 12, 8, 8, 2, 4, 4, 6, 5, 1, 1, 1, 1, 3, 0, 13, 23, 13, 10, 14

Offset: 0

Scott R. Shannon, Oct 10 2021

This is a variation of A342585. The same rules apply except that each number, as it iterates from counting zeros to counting the next number until zero occurrences are found, is treated as a string. The number of occurrences of this string is counted in the concatenation of all previous terms which is also treated as a string. For example when the sequence is adding the number of occurrences of 10, this is treated as the string '10', and thus any occurrence of '10' in the concatenation of all previous terms is counted. This would therefore count the term 1 followed by 0 as an occurrence of '10'. The strings are allowed to overlap, e.g., the number '111' would increment the count of 1's three times, the count of 11's two times, and the count of 111's one time.
Counting the occurrences of each number treated as a string leads to many more terms being found before a zero term is recorded. For example the iteration spanning the 5 millionth term has a(4989084) = 0, a(4989085) = 1059723, then a(5019089) = 2, a(5019090) = 0. Therefore at this stage every number, treated as a string, from 0 to 30005 has occurred at least once in the concatenation of all previous terms.
Unlike A342585 the number of occurrences of the smaller values does not seem to change order as n gets larger. After 5 million terms the order of most frequent occurrences is 1,2,3,4,5,6,7,8,9,0,11,12,10,21,13,14,15,16. It is unlikely the single-digit order changes although, considering that 21 appears high on the count for 2-digit numbers, the order of these and other 2-digit values may change as n gets larger. See the linked image.

			a(56) = 4. This is the first term that differs from A342585 as in that sequence no 10's have occurred after 55 terms. However in this sequence '10' can be formed by '1' followed by '0', and in the concatenation of terms a(0) to a(55) that has occurred four times, starting as a(2), a(12), a(26), a(35).

Scott R. Shannon, Image of the first 5 million terms.

Cf. A342585, A347738, A345730, A032531, A181391.

def count_overlaps(subs, s):
    c = i = 0
    while i != -1:
        i = s.find(subs, i)
        if i != -1: c += 1; i += 1
    return c
def aupton(terms):
    alst, astr, numtocount = [0], "0", 0
    for n in range(2, terms+1):
        c = count_overlaps(str(numtocount), astr)
        numtocount = 0 if c == 0 else numtocount + 1
        alst.append(c)
        astr += str(c)
    return alst
print(aupton(95)) # Michael S. Branicky, Oct 10 2021

A349042 Triangle read by rows in which row n >= 1 lists the count of 0's, ..., k's in all previous terms in the triangle. T(0,0) = 0, k is from [0..A049820(n)].

Original entry on oeis.org

0, 1, 1, 1, 3, 1, 4, 1, 5, 0, 1, 2, 6, 1, 2, 7, 2, 1, 1, 1, 2, 10, 4, 1, 2, 2, 11, 6, 1, 2, 1, 2, 2, 13, 9, 1, 2, 1, 2, 2, 15, 12, 1, 2, 1, 2, 1, 0, 1, 3, 19, 14, 2, 2, 1, 2, 3, 20, 17, 3, 2, 1, 2, 1, 0, 1, 1, 1, 4, 25, 19, 4, 4, 1, 2, 1, 0, 1, 1, 5, 29, 20, 4, 6, 2, 3, 1, 0, 1, 1, 1

Offset: 0

Ctibor O. Zizka, Nov 06 2021

For n >= 1 the n-th row length equals A049820(n) + 1. The same definition, but for k from [0..n] gives A032531.

			Triangle begins:
      k=0  1  2  3  4  5
  n=0:  0;
  n=1:  1;
  n=2:  1;
  n=3:  1, 3;
  n=4:  1, 4;
  n=5:  1, 5, 0, 1;
  n=6:  2, 6, 1;
  n=7:  2, 7, 2, 1, 1, 1;

Michael De Vlieger, Scatterplot of a(n) for n=0..31686, i.e., rows 0..256.
Michael De Vlieger, Scatterplot of a(n)for n=0..518562, i.e., rows 0..1024
Index entries for sequences related to the inventory sequence

Cf. A032531, A049820.

c[] = 0; Reap[Do[w = {}; Array[(Set[m, c[#]]; c[m]++; AppendTo[w, m]) &, If[n == 0, 1, n - DivisorSigma[0, n] + 1], 0]; Sow[w], {n, 0, 15}]][[-1, -1]] // Flatten (* _Michael De Vlieger, Nov 09 2021 *)

from collections import Counter
from sympy import divisor_count
def auptor(rows):
    alst, inventory = [0], Counter([0])
    for m in range(1, rows):
        for k in range(m-divisor_count(m)+1):
            c = inventory[k]; alst.append(c); inventory.update([c])
    return alst
print(auptor(16)) # Michael S. Branicky, Nov 07 2021

A359393 a(n) is the number of times A025581(n-1) (runs of k..0) occur among terms a(1..n-1).

Original entry on oeis.org

0, 0, 2, 1, 1, 2, 0, 2, 2, 3, 0, 1, 4, 3, 4, 0, 2, 2, 6, 3, 5, 1, 1, 2, 3, 7, 5, 5, 1, 1, 3, 2, 5, 8, 7, 5, 1, 2, 1, 5, 2, 5, 10, 9, 5, 1, 1, 2, 1, 8, 2, 5, 12, 12, 5, 1, 1, 2, 2, 1, 10, 2, 5, 15, 15, 5, 0, 2, 1, 2, 2, 1, 12, 2, 5, 19, 17, 6, 3, 0, 2, 1, 2, 2, 2, 13

Offset: 1

Tamas Sandor Nagy, Dec 29 2022

An inventory sequence variation.

			The recorded terms of the sequence:
                Zero 0's thus far
                |    Zero 1's thus far
                |    |  Two 0's thus far
                |    |  |    One 2's thus far
                |    |  |    |  One 1's thus far
                |    |  |    |  |  Two 0's thus far
                |    |  |    |  |  |
This sequence:  0,   0, 2,   1, 1, 2, ...
      A025581:  0,   1, 0,   2, 1, 0, ...

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..20000
Kevin Ryde, PARI/GP Code

Cf. A032531 (with runs increasing), A025581, A342585.

PARI
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cp's OEIS Frontend

A107263 Position where n (presumably) appears the last time in A032531, or 0 if n keeps appearing.

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A107261 Number of a(i), 0<=iA053615(n).

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A347794 The "Look and Say" version of the Inventory sequence A342585.

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A348218 Variation on the Inventory Sequence A342585: record the number of previous terms which are divisible by the iterating number, starting at 1, until 0 is recorded, then restart the iterating number from 1. See the Comments.

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A348288 Variation on the Inventory Sequence A342585: the same rules as A342585 are used except that the terms count the occurrences of the iterating number, treated as a string, in the string concatenation of all previous terms. See the Comments.

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A349042 Triangle read by rows in which row n >= 1 lists the count of 0's, ..., k's in all previous terms in the triangle. T(0,0) = 0, k is from [0..A049820(n)].

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A359393 a(n) is the number of times A025581(n-1) (runs of k..0) occur among terms a(1..n-1).

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