A348016 - OEIS

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,4

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.

LINKS

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

EXAMPLE

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.

PROG

(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