cp's OEIS Frontend

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.

%I A348967 #13 Nov 24 2021 01:25:11
%S A348967 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,
%T A348967 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,
%U A348967 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
%N 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.
%C A348967 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.
%C A348967 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.
%H A348967 Scott R. Shannon, <a href="/A348967/a348967.png">Image of the first 100000 terms</a>.
%e A348967 a(1) = 0 as there have been no pairs so far in the sequence.
%e A348967 a(2) = 1 as there has been one pair with a difference of 0: |a(1) - a(0)|.
%e A348967 a(3) = 1 as there has been one pair with a difference of 1: |a(2) - a(1)|.
%e A348967 a(4) = 0 as there has been no pairs with a difference of 2. The count now resets to 0.
%e A348967 a(5) = 2 as there has been two pairs with a difference of 0: |a(1) - a(0)|, |a(3) - a(2)|.
%e A348967 a(6) = 2 as there has been two pairs with a difference of 1: |a(2) - a(1)|, |a(4) - a(3)|.
%e A348967 a(7) = 1 as there has been one pair with a difference of 2: |a(5) - a(4)|.
%o A348967 (Python)
%o A348967 from collections import Counter
%o A348967 def aupton(terms):
%o A348967     num, alst, inventory = 0, [0, 0], Counter([0])
%o A348967     for n in range(2, terms+1):
%o A348967         c = inventory[num]
%o A348967         num = 0 if c == 0 else num + 1
%o A348967         alst.append(c)
%o A348967         inventory.update([abs(alst[-2] - alst[-1])])
%o A348967     return alst
%o A348967 print(aupton(93)) # _Michael S. Branicky_, Nov 05 2021
%Y A348967 Cf. A342585, A348966 (pair sums).
%K A348967 nonn
%O A348967 0,6
%A A348967 _Scott R. Shannon_, Nov 05 2021