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%I #26 Jul 22 2023 21:01:00
%S 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,
%T 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,
%U 27,9,7,25,5,23,21,39,9,7,25,5,23,3,21,19,37,3,21
%N 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.
%C 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.
%C The n-th row will have A141002(n) nodes. The leftmost border is A008344 and the rightmost is A000217.
%H Rémy Sigrist, <a href="/A360173/b360173.txt">Table of n, a(n) for n = 0..9395</a> (rows for n = 0..17 flattened)
%e 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.
%e The tree begins:
%e row
%e [n]
%e [0] 0
%e \
%e [1] 1
%e \
%e [2] ___3___
%e / \
%e / \
%e [3] 0 __6__
%e \ / \
%e [4] 4 2 10
%e \ \ / \
%e [5] 9 7 5 15
%p T:= proc(n) option remember; `if`(n=0, 0, map(x->
%p [`if`(x<n, [][], x-n), x+n][], [T(n-1)])[])
%p end:
%p seq(T(n), n=0..10); # _Alois P. Heinz_, Jan 30 2023
%o (Python)
%o def A360173_rowlist(row_n):
%o A = [[0]]
%o for i in range(0,row_n):
%o A.append([])
%o for j in range(0,len(A[i])):
%o x = A[i][j]
%o if x - i -1 >= 0:
%o A[i+1].append(x-i-1)
%o if x + i + 1 >= 0:
%o A[i+1].append(x+i+1)
%o return(A)
%o (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
%Y Cf. A000217, A008344, A141001, A141002.
%Y Row sums give A360229.
%K nonn,look,tabf,easy
%O 0,3
%A _John Tyler Rascoe_, Jan 28 2023
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