A353092 -id:A353092 - cp's OEIS frontend

A352799 Inventory sequence of binary weights.

0, 1, 1, 0, 2, 3, 1, 0, 3, 4, 2, 0, 4, 7, 2, 1, 0, 5, 9, 4, 1, 0, 6, 11, 5, 2, 0, 7, 12, 7, 4, 0, 8, 14, 7, 6, 0, 9, 14, 9, 7, 0, 10, 14, 11, 10, 0, 11, 14, 12, 12, 0, 12, 14, 15, 13, 1, 0, 13, 15, 15, 15, 4, 0, 14, 16, 15, 16, 5, 0, 15, 18, 17, 16, 6, 0, 16

Offset: 0

David James Sycamore, Apr 03 2022

nonn

Record the number of terms with binary weight zero, then successively record those with weights 1,2,... (including in the count the weights of new terms as they are recorded), until reaching a weight w for which there are zero terms with that weight, whereupon record a zero term. Repeat.

			a(0) = 0 because at the start there are no terms, therefore zero terms with binary weight zero.
a(1) = 1 because the first term (0) has binary weight zero and there is just one such term.
a(2) = 1 since a(1) = 1 has weight 1, and there is only one term with this weight.
a(3) = 0 since there are no terms with weight 2. Reset the count to zero weight and repeat.
a(4) = 2 because now there are 2 terms (a(0), a(3)) which have weight 0. And so on.
As an irregular triangle the sequence begins:
  0;
  1,  1, 0;
  2,  3, 1, 0;
  3,  4, 2, 0;
  4,  7, 2, 1, 0;
  5,  9, 4, 1, 0;
  6, 11, 5, 2, 0;

Michael S. Branicky, Table of n, a(n) for n = 0..10000
Michael De Vlieger, Log log scatterplot of a(n), n = 0..1128633, showing a(n) = 0 instead as 1/2 for visibility.

Cf. A000120, A342585, A347317, A347738, A345730, A347791, A348967, A353092.

Block[{a, c, j, k, m}, a[1] = c[] = 0; j = c[0] = 1; Do[k = 0; While[c[k] > 0, j++; Set[m, c[k]]; Set[a[j], m]; c[If[m < 1, 0, DigitCount[m, 2, 1] ] ]++; k++]; j++; Set[a[j], 0]; c[0]++, 25]; Array[a, j] ] (* _Michael De Vlieger, Jun 25 2025 *)

from collections import Counter
def w(n): return bin(n).count("1")
def aupton(nn):
    num, alst, inventory = 0, [0], Counter([0])
    for n in range(1, nn+1):
        c = inventory[num]
        num = 0 if c == 0 else num + 1
        alst.append(c)
        inventory.update([w(c)])
    return alst
print(aupton(76)) # Michael S. Branicky, Apr 03 2022

a(45) and beyond from Michael S. Branicky, Apr 03 2022

A362372 Inventory of powers. Initialize the sequence with '1'. Then record the number of powers of 1 thus far, then do the same for powers of 2 (2, 4, 8, ...), powers of 3, etc. When the count is zero, do not record a zero; rather start the inventory again with the powers of 1.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 5, 1, 1, 7, 1, 1, 9, 1, 2, 10, 2, 2, 10, 4, 2, 1, 1, 12, 6, 2, 1, 1, 1, 1, 16, 8, 2, 2, 1, 1, 1, 1, 1, 2, 21, 12, 2, 2, 1, 1, 1, 1, 1, 2, 26, 15, 2, 2, 1, 1, 1, 1, 1, 2, 31, 18, 2, 2, 1, 1, 1, 1, 1, 2, 36, 21, 2, 2, 1, 2, 1, 1, 1

Offset: 0

Damon Lay, Apr 17 2023

A variant of the inventory sequence, A342585.
The graph exhibits sharp jumps followed by a rapid decline forming a periodic hockey stick pattern. Larger-scale, near-linear structures also appear.
Periodic patterns in the relative frequency of any given number also are present. For example, perform a rolling count of the number of times 2 appears in the previous 40 entries.
Open question: will all positive integers appear in the sequence?

			As an irregular triangle, the table begins:
   1;
   1;
   2, 1;
   3, 1, 1;
   5, 1, 1;
   7, 1, 1;
   9, 1, 2;
  10, 2, 2;
  10, 4, 2, 1, 1;
  12, 6, 2, 1, 1, 1, 1;
  16, 8, 2, 2, 1, 1, 1, 1, 1, 2;
  ...
Initialize the sequence with '1'.
Powers of 1 are counted in the first column, powers of 2 in the second, powers of 3 in the third, etc.

Michael S. Branicky, Table of n, a(n) for n = 0..10000 (first 4330 terms from Damon Lay)

Cf. A342585 and similar variants thereof: A345730, A347791, A348218, A352799, A353092.

from collections import Counter
from sympy import divisors, perfect_power
def powers_in(n):
    t = perfect_power(n) # False for n == 1
    return [n] if not t else [t[0]**d for d in divisors(t[1])]
def aupton(nn):
    num, alst, inventory = 1, [1], Counter([1])
    while len(alst) <= nn:
        c = inventory[num]
        if c == 0: num = 1
        else: num += 1; alst.append(c); inventory.update(powers_in(c))
    return alst
print(aupton(100)) # Michael S. Branicky, May 05 2023

cp's OEIS Frontend

A352799 Inventory sequence of binary weights.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python

Extensions