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A349526
Modified lexicographic ordering of all pairs i,j with 1 <= i <= j; every pair i,j of positive integers occurs exactly once.
3
1, 1, 2, 2, 1, 3, 2, 3, 3, 1, 4, 2, 4, 3, 4, 4, 1, 5, 2, 5, 3, 5, 4, 5, 5, 1, 6, 2, 6, 3, 6, 4, 6, 5, 6, 6, 1, 7, 2, 7, 3, 7, 4, 7, 5, 7, 6, 7, 7, 1, 8, 2, 8, 3, 8, 4, 8, 5, 8, 6, 8, 7, 8, 8, 1, 9, 2, 9, 3, 9, 4, 9, 5, 9, 6, 9, 7, 9, 8, 9, 9, 1, 10, 2, 10, 3
OFFSET
1,3
COMMENTS
Concatenate segments: 1 1, then 1 2 2 2, then 1 3 2 3 3 3, etc., so that the general segment is 1 n 2 n ... n n. This is followed by 1; thus, not only does every i,j with i <= j occur, but so does every i,j with i >= j. So far, the procedure leaves A349520. Now, for each number that occurs three times in succession, remove the third occurrence, leaving the present sequence, which has the property that every pair i,j of positive integers occurs exactly once.
The pair n,1 occurs as a(n^2), a(n^2+1).
Is this a duplicate of A329949? - R. J. Mathar, Jan 06 2022
MATHEMATICA
t = {1, 1}; Do[t = Join[t, Riffle[Range[n], n], {n}], {n, 2, 10}];
u = Flatten[Partition[t, 2]];
v = Table[n (n + 1), {n, 1, 10}];
Delete[u, Map[{#} &, v]]
PROG
(Python)
def auptoj(maxj):
alst = []
for j in range(1, maxj+1):
for i in range(1, j+1):
if i != j: alst.extend([i, j])
else: alst.append(i)
return alst
print(auptoj(10)) # Michael S. Branicky, Nov 21 2021
KEYWORD
nonn
AUTHOR
Clark Kimberling, Nov 21 2021
STATUS
approved