%I #16 Sep 19 2021 19:26:22

%S 0,0,0,1,0,1,2,2,3,4,0,2,4,5,5,5,7,8,8,9,9,10,0,3,7,10,11,11,13,13,12,

%T 12,13,16,18,19,19,19,20,20,20,20,21,21,21,22,0,4,11,18,22,23,25,26,

%U 25,25,27,29,30,30,29,28,28,29,31,33,35,36,39,40,41,42,42,42,42,43,43,43,43,43,43,43,44,44,44,44,44,44,44,45,45,45,45,46

%N A347738 with rows normalized by subtracting each term in a row from the first term in the row.

%C As a result of the normalization, each row starts at 0 and is nondecreasing.

%C There was a possibility that the new rows would appear to be converging to something, although that is not apparent at present.

%H Michael S. Branicky, <a href="/A347326/b347326.txt">Table of n, a(n) for n = 0..24573</a> (rows 0 <= k <= 13 when considered as an irregular triangle)

%e Row 2 of A347738 is [4,3,2,2,1,0], and subtracting each term from the first term, 4, we get row 2 of the present sequence, [0, 1, 2, 2, 3, 4].

%e The first few normalized rows are:

%e [0],

%e [0, 0, 1],

%e [0, 1, 2, 2, 3, 4],

%e [0, 2, 4, 5, 5, 5, 7, 8, 8, 9, 9, 10],

%e [0, 3, 7, 10, 11, 11, 13, 13, 12, 12, 13, 16, 18, 19, 19, 19, 20, 20, 20, 20, 21, 21, 21, 22],

%e [0, 4, 11, 18, 22, 23, 25, 26, 25, 25, 27, 29, 30, 30, 29, 28, 28, 29, 31, 33, 35, 36, 39, 40, 41, 42, 42, 42, 42, 43, 43, 43, 43, 43, 43, 43, 44, 44, 44, 44, 44, 44, 44, 45, 45, 45, 45, 46],

%e ...

%o (Python)

%o def aupton(nn):

%o num, gte_inventory, bigc, row, alst = 0, [1], 0, [], [0]

%o while len(alst) < nn + 1:

%o c = gte_inventory[num] if num <= bigc else 0

%o row.append(c)

%o num += 1

%o if c == 0:

%o num = 0

%o alst.extend([row[0] - row[i] for i in range(len(row))])

%o row = []

%o for i in range(min(c, bigc)+1):

%o gte_inventory[i] += 1

%o for i in range(bigc+1, c+1):

%o gte_inventory.append(1)

%o bigc = len(gte_inventory) - 1

%o return alst

%o print(aupton(92)) # _Michael S. Branicky_, Sep 19 2021

%Y Cf. A347738.

%K nonn,tabf

%O 0,7

%A _N. J. A. Sloane_, Sep 13 2021