A348967 Variation on the Inventory Sequence A342585: record the number of occurrences of the pair difference of all adjacent terms until 0 is recorded, then restart the count from 0. Start with a(0) = 0. See the Comments.

0, 0, 1, 1, 0, 2, 2, 1, 0, 3, 4, 1, 2, 0, 3, 6, 2, 4, 1, 0, 3, 7, 3, 6, 3, 0, 3, 7, 3, 10, 5, 1, 0, 3, 8, 3, 11, 6, 4, 0, 3, 8, 4, 12, 8, 5, 0, 3, 8, 4, 14, 10, 7, 0, 3, 8, 4, 16, 12, 8, 0, 3, 8, 4, 17, 15, 9, 1, 2, 4, 0, 3, 9, 6, 19, 16, 9, 2, 4, 4, 0, 4, 9, 7, 20, 18, 10, 2, 4, 6, 0, 4, 9, 11

Offset: 0

Author

Scott R. Shannon, Nov 05 2021

Keywords

• nonn

Comments

This sequence is a variation of A342585. Here we record the number of previous occurrences of the pair differences of all adjacent terms until 0 is recorded, after which the pair difference count restarts at 0. For example the terms 0,0,2,1,3 contain one pair with a difference of 0 (0,0), one pair with a difference of 1 (2,1), and two pairs with a difference of 2 (0,2 and 1,3). See the Examples below.

After 20 million terms the largest term is a(19995157) = 2537781, which counts the occurrences of pairs with a difference of 1, while there are 5725 terms between zeros. It is likely the most common pair difference remains at 1 as n increases although this is unknown.

Examples

			a(1) = 0 as there have been no pairs so far in the sequence.
a(2) = 1 as there has been one pair with a difference of 0: |a(1) - a(0)|.
a(3) = 1 as there has been one pair with a difference of 1: |a(2) - a(1)|.
a(4) = 0 as there has been no pairs with a difference of 2. The count now resets to 0.
a(5) = 2 as there has been two pairs with a difference of 0: |a(1) - a(0)|, |a(3) - a(2)|.
a(6) = 2 as there has been two pairs with a difference of 1: |a(2) - a(1)|, |a(4) - a(3)|.
a(7) = 1 as there has been one pair with a difference of 2: |a(5) - a(4)|.
		

Links

• Scott R. Shannon, Image of the first 100000 terms.

Crossrefs

Cf. A342585, A348966 (pair sums).

Programs

Python

from collections import Counter
def aupton(terms):
    num, alst, inventory = 0, [0, 0], Counter([0])
    for n in range(2, terms+1):
        c = inventory[num]
        num = 0 if c == 0 else num + 1
        alst.append(c)
        inventory.update([abs(alst[-2] - alst[-1])])
    return alst
print(aupton(93)) # Michael S. Branicky, Nov 05 2021