A342585 -id:A342585 - cp's OEIS frontend

A347738 A variant of the inventory sequence: record the number of terms >= 0 thus far in the sequence, then the number of terms >= 1 thus far, then the number of terms >= 2 thus far, and so on, until a zero is recorded; the inventory then starts again, recording the number of terms >= 0, etc.

0, 1, 1, 0, 4, 3, 2, 2, 1, 0, 10, 8, 6, 5, 5, 5, 3, 2, 2, 1, 1, 0, 22, 19, 15, 12, 11, 11, 9, 9, 10, 10, 9, 6, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 0, 46, 42, 35, 28, 24, 23, 21, 20, 21, 21, 19, 17, 16, 16, 17, 18, 18, 17, 15, 13, 11, 10, 7, 6, 5, 4, 4, 4, 4, 3, 3

Offset: 0

David James Sycamore, Sep 12 2021

Sequence starts off as A342585 but diverges after a(4). The effect is to introduce some numbers earlier in this sequence than in the original, and to stretch out the incidences of zero terms by the fact that the term immediately following a zero is now the total number of prior terms, rather than the total number of prior zero terms.

In A342585 zeros occur at positions 1,4,8,14,20,28,... (see A343880) whereas in this version they occur at positions 1,4,10,22,46,... (which is A033484, as is easily proved by induction).

			As an irregular triangle this begins:
   0;
   1,  1,  0;
   4,  3,  2,  2,  1,  0;
  10,  8,  6,  5,  5,  5, 3, 2,  2,  1, 1, 0;
  22, 19, 15, 12, 11, 11, 9, 9, 10, 10, 9, 6, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 0;
  46, ...
(for row lengths see A003945)

Michael De Vlieger, Table of n, a(n) for n = 0..24573 (rows 0 <= k <= 12 when considered as an irregular triangle)
Michael De Vlieger, log-log scatterplot of a(n) for 1 <= n <= 49150 (ignoring zeros).

Cf: A342585, A033484, A003945, A343880, A003945 (row lengths), A347324 (row sums).

A347326 has a version of this in which the rows have been normalized.

a[n_] := a[n] = Block[{t}, t = If[a[n - 1] == 0, 0, b[n - 1] + 1]; b[n] = t; Sum[If[a[j] >= t, 1, 0], {j, n - 1}]]; b[1] = a[1] = 0; Array[a, 77] (* Michael De Vlieger, Sep 12 2021, after Jean-François Alcover at A342585 *)

def aupton(nn):
    num, gte_inventory, alst, bigc = 0, [1], [0], 0
    while len(alst) < nn+1:
        c = gte_inventory[num] if num <= bigc else 0
        num = 0 if c == 0 else num + 1
        for i in range(min(c, bigc)+1):
            gte_inventory[i] += 1
        for i in range(bigc+1, c+1):
            gte_inventory.append(1)
        bigc = len(gte_inventory) - 1
        alst.append(c)
    return alst
print(aupton(76)) # Michael S. Branicky, Sep 19 2021

Offset changed to 0 by N. J. A. Sloane, Sep 12 2021

A032531 An inventory sequence: triangle read by rows, where T(n, k), 0 <= k <= n, records the number of k's thus far in the flattened sequence.

Original entry on oeis.org

0, 1, 1, 1, 3, 0, 2, 3, 1, 2, 2, 4, 3, 3, 1, 2, 5, 4, 4, 3, 1, 2, 6, 5, 5, 3, 3, 1, 2, 7, 6, 7, 3, 3, 2, 2, 2, 7, 9, 9, 3, 3, 2, 3, 0, 3, 7, 10, 13, 3, 3, 2, 4, 0, 2, 4, 7, 12, 15, 5, 4, 2, 5, 0, 2, 1, 5, 8, 14, 15, 6, 6, 4, 5, 1, 2, 1, 0, 6, 10, 15, 15, 7, 7, 5, 7, 1, 2, 2, 0, 1, 7, 12, 17

Offset: 0

Dmitri Papichev (Dmitri.Papichev(AT)iname.com)

Old name: a(n) = number of a(i) for 0<=iA002262(n).
This sequence is a variation of the Inventory sequence A342585. The same rules apply except that in this variation each row ends after k terms, where k is the current row count which starts at 1. The behavior up to the first 1 million terms is similar to A342585 but beyond that the most common terms do not increase, likely due to the rows being cut off after k terms thus numbers such as 1 and 2 no longer make regular appearances. Larger number terms do increase and overtake the leading early terms, and it appears this pattern repeats as n increases. See the linked images. - Scott R. Shannon, Sep 13 2021
The complexity of this sequence derives from the totals being updated during the calculation of each row. If each row recorded an inventory of only the earlier rows, we would get the much simpler A025581. - Peter Munn, May 06 2023

			Triangle begins:
  0;
  1, 1;
  1, 3, 0;
  2, 3, 1, 2;
  2, 4, 3, 3, 1;
  2, 5, 4, 4, 3, 1;
  2, 6, 5, 5, 3, 3, 1;
  2, 7, 6, 7, 3, 3, 2, 2;
  2, 7, 9, 9, 3, 3, 2, 3, 0;
  ...

Scott R. Shannon, Table of n, a(n) for n = 0..20000.
Scott R. Shannon, Image of the first 10^6 terms.
Scott R. Shannon, Image of the first 10^7 terms.
Scott R. Shannon, Image of the first 10^8 terms.
Index entries for sequences related to inventory sequences

Cf. A002262, A025581, A342585.

A002262 := proc(n)
    n - binomial(floor(1/2+sqrt(2*(1+n))), 2);
end proc:
A032531 := proc(n)
    option remember;
    local a,piv,i ;
    a := 0 ;
    piv := A002262(n) ;
    for i from 0 to n-1  do
        if procname(i) = piv then
            a := a+1 ;
        end if;
    end do:
    a ;
end proc:
seq(A032531(n),n=0..100) ; # R. J. Mathar, May 08 2020

A002262[n_] :=  n - Binomial[Floor[1/2 + Sqrt[2*(1 + n)]], 2];
A032531[n_] := A032531[n] = Module[{a, piv, i}, a = 0; piv = A002262[n]; For[i = 0, i <= n-1, i++, If[A032531[i] == piv, a++]]; a];
Table[A032531[n], {n, 0, 100}] (* Jean-François Alcover, Mar 25 2024, after R. J. Mathar *)

from math import comb, isqrt
from collections import Counter
def idx(n): return n - comb((1+isqrt(8+8*n))//2, 2)
def aupton(nn):
    num, alst, inventory = 0, [0], Counter([0])
    for n in range(1, nn+1):
        c = inventory[idx(n)]
        alst.append(c)
        inventory[c] += 1
    return alst
print(aupton(93)) # Michael S. Branicky, May 07 2023

New name from Peter Munn, May 06 2023

A354606 a(1) = 1; for n > 1, a(n) is number of terms in the first n-1 terms of the sequence that have the same number of divisors as a(n-1).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 3, 4, 1, 4, 2, 5, 6, 1, 5, 7, 8, 2, 9, 3, 10, 3, 11, 12, 1, 6, 4, 4, 5, 13, 14, 5, 15, 6, 7, 16, 1, 7, 17, 18, 2, 19, 20, 3, 21, 8, 9, 6, 10, 11, 22, 12, 4, 7, 23, 24, 1, 8, 13, 25, 8, 14, 15, 16, 2, 26, 17, 27, 18, 5, 28, 6, 19, 29, 30, 2, 31, 32, 7, 33, 20, 8, 21, 22, 23, 34

Offset: 1

Scott R. Shannon, Jul 08 2022

After 250000 terms the most common number of divisors of all terms are 4, 8, 2, 12, 16 divisors. These correspond to the uppermost five lines of the attached image. It is unknown if these stay the most common or are passed by numbers with more divisors as n gets arbitrarily large.

See A355606 for the indices where a(n) = 1.

			a(6) = 2 as a(5) = 3 which has two divisors, and the total number of terms in the first five terms with two divisors is two, namely a(3) = 2 and a(5) = 3.

Scott R. Shannon, Table of n, a(n) for n = 1..10000
Scott R. Shannon, Image of the first 250000 terms. The green line is y = n.
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20.

Cf. A355606, A000005, A124056, A342585.

nn = 120; c[] := 0; a[1] = j = 1; Do[k = ++c[DivisorSigma[0, j]]; Set[{a[n], j}, {k, k}], {n, 2, nn}]; Array[a, nn] (* _Michael De Vlieger, Dec 12 2024 *)

from sympy import divisor_count
from collections import Counter
def f(n): return divisor_count(n)
def aupton(nn):
    an, fan, alst, inventory = 1, 1, [1], Counter([1])
    for n in range(2, nn+1):
        an = inventory[fan]
        fan = f(an)
        alst.append(an)
        inventory.update([fan])
    return alst
print(aupton(86)) # Michael S. Branicky, Jul 09 2022

A356348 a(0) = 0; for n > 0, a(n) is the number of preceding terms having the same digit sum as a(n-1).

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 11, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 2, 10, 12, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 3, 10, 13, 4, 5, 4, 6, 4, 7, 4, 8, 4, 9, 4, 10, 14, 5, 6, 5, 7, 5, 8, 5, 9, 5, 10, 15, 6, 7, 6, 8, 6, 9, 6, 10, 16, 7, 8, 7, 9, 7, 10, 17, 8, 9

Offset: 0

Scott R. Shannon, Oct 15 2022

			a(21) = 2 as a(20) = 11 which has a digit sum of 2, and there has been two previous terms with a digit sum of two: a(3) = 2 and a(20) = 11.

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Scott R. Shannon, Image of n=0..1000000.

Cf. A007953, A137671 (base 2), A342585.

nn = 94; c[] = 0; a[0] = k = 0; c[0] = 1; Do[a[n] = c[k]; k = Total@ IntegerDigits[a[n]]; c[k]++, {n, nn}]; Array[a, nn] (* _Michael De Vlieger, Oct 15 2022 *)

from itertools import islice
from collections import Counter
def sd(n): return sum(map(int, str(n)))
def agen(): # generator of terms
    an, s, inventory = 0, 0, Counter([0])
    while True:
        yield an; an = inventory[s]; s = sd(an); inventory[s] += 1
print(list(islice(agen(), 95))) # Michael S. Branicky, Oct 21 2022

A347062 Record the number of zero terms having a following term, then the number of terms equal to 1 having a following term, then 2, 3, etc. until recording a zero, whereupon the count is reset.

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 1, 0, 3, 3, 1, 2, 0, 4, 4, 2, 2, 2, 0, 5, 4, 5, 2, 3, 2, 0, 6, 4, 7, 3, 4, 2, 1, 1, 0, 7, 6, 8, 4, 5, 2, 2, 2, 1, 0, 8, 7, 11, 4, 6, 3, 3, 3, 2, 0, 9, 7, 12, 7, 7, 3, 3, 6, 2, 1, 0, 10, 8, 13, 9, 7, 3, 4, 7, 3, 2, 1, 1, 1, 1, 0, 11, 12, 14, 11, 8

Offset: 0

David James Sycamore, Oct 16 2021

Inventory sequence recording the number of existing terms immediately following occurrences of a zero term, then the number immediately following occurrences of 1, then 2, and so on until another zero is recorded, after which the count is reset. Inclusion of a term in any count requires it to have been followed by another term first, therefore lead terms are not included in the current count.
The scatterplot exhibits trajectories attributable to the register of occurrences of the immediately-preceding term m, c(m), and irregular periodicity of nondecreasing length. - Michael De Vlieger, Oct 16 2021
Suggested by A342585.

			At first there are no terms, thus none following a zero, so a(0) = 0.
After a(0) = 0 the count is reset, and since there are still no terms following a zero, a(1) = 0. The count is now reset again and we have one term a(1) = 0 which follows a zero term, so a(2) = 1.
We now have 0,0,1 and because no term yet follows 1, a(3) must be 0 (the lead term here is 1 but it is not counted).
The count is now reset and there are two terms (a(1) and a(2)) which follow a zero term, thus a(4) = 2; etc.
As an irregular triangle the sequence begins:
0;
0;
1, 0;
2, 1, 1, 0;
3, 3, 1, 2, 0;
4, 4, 2, 2, 2, 0;
5, 4, 5, 2, 3, 2, 0;
6, 4, 7, 3, 4, 2, 1, 1, 0;
...

Michael De Vlieger, Scatterplot of a(n) for n=0..2877 (rows 0..31)
Michael De Vlieger, Labeled scatterplot of a(n) for n=0..336 (rows 0..23), with trajectory of c(0) in black, c(1) in red, c(2) in orange, c(3) in yellow, c(4) in green, c(5) in blue, and c(6) in purple.
Michael De Vlieger, Scatterplot of a(n) for n=0..39555 (rows 0..255), showing trajectories with color function as stated above.

Cf. A342585, A347564.

Block[{c, k, m, n}, c[0] = 1; m = 0; {0, 0}~Join~Reap[Do[k = 0; While[IntegerQ[c[k]], Set[n, c[k]]; Sow[n]; If[IntegerQ@ c[m], c[m]++, c[m] = 1]; Set[m, n]; k++]; Sow[0]; If[IntegerQ@ c[m], c[m]++, c[m] = 1]; Set[m, 0], 11]][[-1, -1]]] (* Michael De Vlieger, Oct 16 2021 *)

A347791 Inventory sequence using prime divisors (with multiplicity): Record the number of terms thus far which are divisible by every prime, then the number of terms thus far not divisible by any prime, then the number divisible (once) by a single prime, then 2 (including with multiplicity), then 3 etc until a zero is recorded. Repeat after every zero term.

Original entry on oeis.org

0, 1, 1, 0, 2, 2, 2, 0, 3, 2, 5, 0, 4, 2, 7, 1, 0, 5, 3, 10, 2, 0, 6, 3, 12, 3, 1, 0, 7, 4, 14, 5, 1, 0, 8, 5, 16, 5, 2, 1, 0, 9, 6, 18, 7, 3, 1, 0, 10, 7, 21, 9, 3, 1, 0, 11, 8, 23, 10, 4, 1, 0, 12, 9, 24, 13, 5, 2, 0, 13, 9, 28, 14, 6, 2, 0, 14, 9, 29, 18, 7

Offset: 0

David James Sycamore, Sep 13 2021

Inspired by the original Inventory sequence (A342585). It follows from the definition that zero appears infinitely many times. Every nonzero number number appears because the number of zero terms increments at every extension of the sequence between consecutive zeros. Since the count for each number of divisors increments (eventually) as the sequence extends, it follows that every number appears infinitely many times.

The count of terms divisible by every prime is the number of zeros, the count of terms divisible by no prime is the number of 1s, the count of terms divisible by one prime (once) is the number of primes, etc.

			a(0)=0 because at this point there are zero terms divisible by every prime. a(1)=1 because there is now one term (0) which is divisible by every prime. a(2)=1 because there is now one term (1) with no prime divisor. a(3)=0 because at this point there is no term divisible by one prime, and so on.
As an irregular triangle the sequence begins:
   0;
   1,  1,  0;
   2,  2,  2,  0;
   3,  2,  5,  0;
   4,  2,  7,  1,  0;
   5,  3, 10,  2,  0;
   6,  3, 12,  3,  1,  0;
   7,  4, 14,  5,  1,  0;
   8,  5, 16,  5,  2,  1,  0;
   9,  6, 18,  7,  3,  1,  0;
  10,  7, 21,  9,  3,  1,  0;
  11,  8, 23, 10,  4,  1,  0; etc.

Michael De Vlieger, Table of n, a(n) for n = 0..10000 (rows 0 <= n <= 836, flattened)
Michael De Vlieger, Log-log scatterplot of row n for 1 <= n <= 2^16 (1199674 terms), with color function showing k, black represents the number of 0s, red 1s, orange primes, etc.

Cf. A001222, A342585, A347738.

Block[{a = {}, c, k, m}, c[-1] = 0; Do[k = -1; c[-1]++; AppendTo[a, 0]; While[IntegerQ[c[k]], AppendTo[a, c[k]]; Set[m, PrimeOmega[c[k]]]; If[IntegerQ[c[m]], c[m]++, Set[c[m], 1]]; k++], 10]; a] (* Michael De Vlieger, Sep 15 2021 *)

A357317 Inventory count sequence: record what you see and where it is located.

Original entry on oeis.org

0, 1, 0, 0, 3, 0, 0, 2, 3, 1, 1, 1, 5, 0, 0, 2, 3, 5, 6, 4, 1, 1, 9, 10, 11, 1, 2, 7, 2, 3, 4, 8, 7, 0, 0, 2, 3, 5, 6, 13, 14, 7, 1, 1, 9, 10, 11, 20, 21, 25, 4, 2, 7, 15, 26, 28, 4, 3, 4, 8, 16, 29, 2, 4, 19, 30, 2, 5, 12, 17, 1, 6, 18, 1, 7, 27, 1, 8, 31, 1, 9, 22, 1, 10, 23, 1, 11, 24, 9, 0, 0, 2, 3

Offset: 0

Ctibor O. Zizka, Sep 29 2022

To get started we set a(0)=0. Now start with zero terms. We record ONE ZERO at position ZERO. The sequence is now 0,1,0,0. Now start again with zero terms. We record THREE ZEROS at positions 0,2,3. Next start with one terms. We record ONE ONE at position ONE. The sequence is now 0,1,0,0,3,0,0,2,3,1,1,1. Now start again with zero terms...one terms...two terms...three terms. The sequence is now 0,1,...,2,3,4,8. And so on.

			a(0) = 0
counting over a(0):
a(1) = 1, a(2) = 0, a(3) = 0  (count one zero at position 0)
counting over a(0)..a(3):
a(4) = 3, a(5) = 0, a(6) = 0, a(7) = 2, a(8) = 3, (count 3 zeros at positions 0,2,3)
a(9) = 1, a(10) = 1, a(11) = 1 (count 1 one at position 1)
counting over a(0)..a(11):
a(12) = 5, a(13) = 0, a(14) = 0, a(15) = 2, a(16) = 3, a(17) = 5, a(18) = 6 (count 5 zeros at positions 0,2,3,5,6)
a(19) = 4, a(20) = 1, a(21) = 1, a(22) = 9, a(23) = 10, a(24) = 11 (count 4 ones at positions 1,9,10,11)
a(25) = 1, a(26) = 2, a(27) = 7 (count 1 two at position 7)
a(28) = 2, a(29) = 3, a(30) = 4, a(31) = 8 (count 2 threes at positions 4,8)
counting over a(0)..a(31):
etc.

Thomas Scheuerle, Table of n, a(n) for n = 0..4895
Thomas Scheuerle, a(0..10000) as scatter plot

Cf. A342585.

function a = A357317( min_length )
    a = 0;
    while length(a) < min_length
        b = a; c = unique(a);
        for m = 1:length(c)
            nextnumber = c(m);
            indices = find(a == nextnumber);
            b = [b length(indices) nextnumber indices-1];
        end
        a = b;
    end
end % Thomas Scheuerle, Sep 29 2022

A362031 a(1) = 1; for n > 1, a(n) is number of terms in the first n-1 terms of the sequence that have the same number of prime factors, counted with multiplicity, as a(n-1).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 3, 4, 1, 4, 2, 5, 6, 3, 7, 8, 1, 5, 9, 4, 5, 10, 6, 7, 11, 12, 2, 13, 14, 8, 3, 15, 9, 10, 11, 16, 1, 6, 12, 4, 13, 17, 18, 5, 19, 20, 6, 14, 15, 16, 2, 21, 17, 22, 18, 7, 23, 24, 3, 25, 19, 26, 20, 8, 9, 21, 22, 23, 27, 10, 24, 4, 25, 26, 27, 11, 28, 12, 13, 29, 30, 14, 28

Offset: 1

Scott R. Shannon, Apr 06 2023

nonn

After 1 million terms the most common numbers for the number of prime factors of the terms are 3, 2, 4, and 5. These correspond to the uppermost four lines of the attached image. It is unknown if these stay the most common or are passed by numbers with more prime factor as n gets arbitrarily large.

See A362033 for the indices where a(n) = 1.

			a(6) = 2 as the number of prime factors of a(5) = A001222(a(5)) = A001222(3) = 1, and there are two previous terms, a(3) and a(5), that have one prime factor.
a(9) = 1 as the number of prime factors of a(8) = A001222(a(8)) = A001222(4) = 2, and there is only one term, a(8), that has two prime factors.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^16, with a color function representing Omega(a(n-1)), where black = 0, red = 1, orange = 2, ..., magenta = 14.
Scott R. Shannon, Image of the first 1 million terms.

Cf. A362033, A001222, A354606, A000005, A124056, A342585.

nn = 120; c[] = 0; j = a[1] = c[0] = 1; m = 0; Do[Set[k, c[m]]; (Set[{a[n], j, m}, {k, k, #}]; c[#]++) &[PrimeOmega[k]], {n, 2, nn}]; Array[a, nn] (* _Michael De Vlieger, Apr 06 2023 *)

from sympy import factorint
from itertools import islice
from collections import Counter
def A362031gen(): # generator of terms
    an, c, d = 1, Counter(), dict()
    while True:
        yield an
        pf = d[an] if an in d else sum(factorint(an).values())
        c[pf] += 1
        an = c[pf]
print(list(islice(A362031gen(), 83))) # Michael S. Branicky, Apr 06 2023

from itertools import islice
from sympy import primeomega
def A362031_gen(): # generator of terms
    a, b, c = {}, {}, 1
    while True:
        yield c
        d = b[c] = b.get(c,primeomega(c))
        c = a[d] = a.get(d,0)+1
A362031_list = list(islice(A362031_gen(),20)) # Chai Wah Wu, Apr 08 2023

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

A364027 a(0) = 0, a(1) = 0; for n > 1, a(n) is the number of pairs of consecutive terms that sum to the same value as a(n-2) + a(n-1).

Original entry on oeis.org

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

Offset: 0

Scott R. Shannon, Jul 01 2023

The same number cannot occur four times in a row as the second pair in a triplet of the same numbers increments the appearance count of the first pair by one, so the fourth number is always one more than the previous three numbers.

The occurrences of three consecutive equal numbers is quite rare, only occurring thirteen times in the first 20 million terms. The last such triplet is a(3641208) = a(3641209) = a(3641210) = 1177. It is likely such triplets occur infinitely often although this is unknown.

			a(2) = 1 as there is one pair that sums to a(0) + a(1) = 0, namely a(0) + a(1).
a(4) = 1 as a(2) + a(3) = 1 + 1 = 2, and there has been one previous pair that also sums to 2, namely a(2) + a(3).
a(5) = 2 as a(3) + a(4) = 1 + 1 = 2, and there has been two previous pairs that also sums to 2, namely a(2) + a(3) and a(3) + a(4).

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Scott R. Shannon, Image of the first 20 million terms.

Cf. A306251, A364036 (do not include previous pair), A342585, A347062.

nn = 120; c[] := 0; a[0] = a[1] = i = j = 0; Do[Set[k, 1 + c[i + j]++]; i = j; j = a[n] = k, {n, 2, nn}]; Array[a, nn + 1, 0] (* _Michael De Vlieger, Jul 02 2023 *)

def A364027_gen(): # generator of terms
    a, b, ndict = 0, 0, {0:1}
    yield from (0,0)
    while True:
        a, b = b, ndict[a+b]
        yield b
        ndict[a+b] = ndict.get(a+b,0)+1 # Chai Wah Wu, Jul 02 2023

a(n) = A306251(n-2) for any n >= 3. - Rémy Sigrist, Apr 03 2025

cp's OEIS Frontend

A347738 A variant of the inventory sequence: record the number of terms >= 0 thus far in the sequence, then the number of terms >= 1 thus far, then the number of terms >= 2 thus far, and so on, until a zero is recorded; the inventory then starts again, recording the number of terms >= 0, etc.

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A032531 An inventory sequence: triangle read by rows, where T(n, k), 0 <= k <= n, records the number of k's thus far in the flattened sequence.

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A354606 a(1) = 1; for n > 1, a(n) is number of terms in the first n-1 terms of the sequence that have the same number of divisors as a(n-1).

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A356348 a(0) = 0; for n > 0, a(n) is the number of preceding terms having the same digit sum as a(n-1).

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A347062 Record the number of zero terms having a following term, then the number of terms equal to 1 having a following term, then 2, 3, etc. until recording a zero, whereupon the count is reset.

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A357317 Inventory count sequence: record what you see and where it is located.

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MATLAB

A362031 a(1) = 1; for n > 1, a(n) is number of terms in the first n-1 terms of the sequence that have the same number of prime factors, counted with multiplicity, as a(n-1).

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