A348218 Variation on the Inventory Sequence A342585: record the number of previous terms which are divisible by the iterating number, starting at 1, until 0 is recorded, then restart the iterating number from 1. See the Comments.

1, 1, 0, 2, 1, 0, 4, 2, 0, 6, 4, 1, 2, 0, 10, 7, 1, 2, 1, 1, 1, 0, 17, 8, 1, 3, 1, 1, 1, 1, 0, 25, 9, 3, 3, 2, 1, 1, 1, 1, 1, 0, 35, 10, 5, 3, 5, 1, 2, 1, 1, 2, 0, 45, 13, 7, 3, 7, 1, 4, 1, 2, 2, 0, 55, 16, 8, 6, 8, 2, 4, 4, 2, 2, 1, 0, 66, 26, 10, 9, 9, 3, 4, 4, 4, 3, 2, 0, 77, 32, 14, 13, 9, 3

Offset: 0

Author

Scott R. Shannon, Oct 07 2021

Keywords

• nonn

Comments

This sequence is a variation of A342585. Instead of iteratively counting the occurrences of each number starting from zero and then repeating when zero is recorded, we start the iterating number at 1 and count the previous terms that are divisible by that number. This number increases for each division until zero previous terms is recorded, when the iterating number is reset to one and the divisor count repeats. The sequence starts with a(0) = 1.

After 10^7 terms the largest value is a(9990262) = 9988674, which is a count of the terms >= 1. The largest value the iterating divisor has reached is 13249.

Examples

			a(1) = 1 as the iterating dividing number starts at 1, and so far there has been one term, a(0), which is divisible by 1.
a(2) = 0 as the dividing number has increased to two, but there have been no terms in the sequence so far that are divisible by 2. The dividing number is now reset to 1.
a(3) = 2 as there have been two terms, a(0) and a(1), that are divisible by 1.
a(4) = 1 as there has been one term, a(3), that is divisible by 2.
a(5) = 0 as there have been no terms divisible by 3. The dividing number is now reset to 1.
		

Crossrefs

Cf. A342585, A027750, A347738, A345730, A032531, A181391.