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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.
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%I #24 Nov 08 2021 06:44:23

%S 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,

%T 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,

%U 9,24,13,5,2,0,13,9,28,14,6,2,0,14,9,29,18,7

%N 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.

%C 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.

%C 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.

%H Michael De Vlieger, <a href="/A347791/b347791.txt">Table of n, a(n) for n = 0..10000</a> (rows 0 <= n <= 836, flattened)

%H Michael De Vlieger, <a href="/A347791/a347791.png">Log-log scatterplot of row n</a> for 1 <= n <= 2^16 (1199674 terms), with color function showing k, black represents the number of 0s, red 1s, orange primes, etc.

%e 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.

%e As an irregular triangle the sequence begins:

%e 0;

%e 1, 1, 0;

%e 2, 2, 2, 0;

%e 3, 2, 5, 0;

%e 4, 2, 7, 1, 0;

%e 5, 3, 10, 2, 0;

%e 6, 3, 12, 3, 1, 0;

%e 7, 4, 14, 5, 1, 0;

%e 8, 5, 16, 5, 2, 1, 0;

%e 9, 6, 18, 7, 3, 1, 0;

%e 10, 7, 21, 9, 3, 1, 0;

%e 11, 8, 23, 10, 4, 1, 0; etc.

%t 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 *)

%Y Cf. A001222, A342585, A347738.

%K nonn,tabf

%O 0,5

%A _David James Sycamore_, Sep 13 2021