A348016 Record the number of terms with no proper divisors, then the number with one proper divisor, then two, three, etc., until reaching a zero term. After each zero term, repeat the count as before.

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

Offset: 0

Author

David James Sycamore, Sep 24 2021

Keywords

• nonn
• tabf

Comments

An inventory sequence counting the proper divisors of existing terms, where zero is taken to have no proper divisors (see A032741). After every occurrence of a zero term the incremental count of terms with 0,1,2,... proper divisors is repeated until another zero term is encountered.

Examples

			a(0) = 0 because at first there are no terms, therefore there are no terms with no proper divisors.
a(1) = 1 because now there is one term (a(0)) which has no proper divisors.
a(2) = 0 since there are no terms with one proper divisor.
a(3) = 3 since there are now three terms having just one proper divisor (0,1,0).
As an irregular triangle the sequence begins:
   0, 1, 0;
   3, 1, 0;
   5, 2, 0;
   6, 3, 0;
   7, 5, 0;
   8, 6, 0;
   9, 6, 4, 1, 0;
  11, 7, 2, 4, 0;
  etc.
		

Links

• David A. Corneth, Table of n, a(n) for n = 0..9999

Crossrefs

Cf. A032741, A342585, A345730, A347791.

Programs

PARI

first(n) = { t = 0; res = vector(n); l = List([1]); for(i = 2, n, for(i = #l + 1, t+1, listput(l, 0) ); res[i] = l[t + 1]; q = if(l[t + 1] == 0, 0, numdiv(l[t + 1]) - 1); for(i = #l + 1, q + 1, listput(l, 0) ); l[q + 1]++; if(res[i] == 0, t = 0 , t++ ) ); res } \\ David A. Corneth, Sep 25 2021

Python

from sympy import divisor_count
from collections import Counter
def f(n): return 0 if n == 0 else divisor_count(n) - 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([f(c)])
    return alst
print(aupton(84)) # Michael S. Branicky, May 07 2023

Extensions

Data corrected and extended by David A. Corneth, Sep 25 2021