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

0,5

Comments

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.

Links

Michael S. Branicky, Table of n, a(n) for n = 0..10000

Example

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;

Prog

(Python)
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

Crossrefs

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

Sequence in context: A355487 A071501 A004572 * A082839 A130717 A137396

Adjacent sequences: A352796 A352797 A352798 * A352800 A352801 A352802

Keyword

nonn

Author

David James Sycamore, Apr 03 2022

Extensions

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

Status

approved