A360173 Irregular triangle (an infinite binary tree) read by rows. The tree has root node 0, in row n=0. Each node then has left child m - n if nonnegative and right child m + n. Where m is the value of the parent node and n is the row of the children.

0, 1, 3, 0, 6, 4, 2, 10, 9, 7, 5, 15, 3, 15, 1, 13, 11, 9, 21, 10, 8, 22, 8, 6, 20, 4, 18, 2, 16, 14, 28, 2, 18, 0, 16, 14, 30, 0, 16, 14, 12, 28, 12, 10, 26, 10, 8, 24, 6, 22, 20, 36, 11, 9, 27, 9, 7, 25, 5, 23, 21, 39, 9, 7, 25, 5, 23, 3, 21, 19, 37, 3, 21

Offset: 0

Author

John Tyler Rascoe, Jan 28 2023

Keywords

• nonn
• look
• tabf
• easy

Comments

A node will have a left child only if the value of that child is greater than or equal to 0. But, each node will have a right child, since adding n will always be greater than 0.

The n-th row will have A141002(n) nodes. The leftmost border is A008344 and the rightmost is A000217.

Examples

			The binary tree starts with root 0 in row n = 0. In row n = 3, the parent node m = 3 has the first left child since 3 - 3 >= 0.
The tree begins:
row
[n]
[0]           0
               \
[1]             1
                 \
[2]            ___3___
              /       \
             /         \
[3]         0         __6__
             \       /     \
[4]           4     2      10
               \     \    /  \
[5]             9     7  5    15
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..9395 (rows for n = 0..17 flattened)

Crossrefs

Cf. A000217, A008344, A141001, A141002.

Row sums give A360229.

Programs

Maple

T:= proc(n) option remember; `if`(n=0, 0, map(x->
      [`if`(xAlois P. Heinz, Jan 30 2023

Mathematica

row[n_] := Module[{r = {0}}, For[h = 1, h <= n, h++, r = Flatten[If[#-h >= 0, {#-h, #+h}, {#+h}]& /@ r]]; r];
Table[row[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Apr 16 2025, after Rémy Sigrist *)

PARI

row(n) = { my (r=[0]); for (h=1, n, r=concat(apply(v->if (v-h>=0, [v-h,v+h], [v+h]), r))); return (r) } \\ Rémy Sigrist, Jan 31 2023

Python

def A360173_rowlist(row_n):
    A = [[0]]
    for i in range(0,row_n):
        A.append([])
        for j in range(0,len(A[i])):
            x = A[i][j]
            if x - i -1 >= 0:
                A[i+1].append(x-i-1)
            if x + i + 1 >= 0:
                A[i+1].append(x+i+1)
    return(A)