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A361819
Irregular triangle read by rows where T(n,k) is the distance which number A361660(n,k) moves in the process described in A361642.
1
2, 3, 3, 4, 2, 2, 4, 5, 3, 4, 3, 5, 6, 4, 2, 3, 3, 2, 4, 6, 7, 5, 3, 5, 2, 5, 3, 5, 7, 8, 6, 4, 2, 4, 4, 4, 4, 2, 4, 6, 8, 9, 7, 5, 3, 6, 3, 3, 3, 3, 6, 3, 5, 7, 9, 10, 8, 6, 4, 2, 5, 5, 2, 6, 2, 5, 5, 2, 4, 6, 8, 10, 11, 9, 7, 5, 3, 7, 4, 4, 5, 5, 4, 4, 7, 3, 5, 7, 9, 11
OFFSET
1,1
COMMENTS
Number A361660(n,k) moves to the right and then down and T(n,k) counts the steps in both.
All moves are T(n,k) >= 2 steps since a number moves at least one step right and one step down.
Row n has sum A002378(n-1) which is the total steps to move a column down to a row irrespective of the order of movement.
Each row is a palindrome (the same when reversed), since the moves in A361642 are exactly the reverse moves to send its row back to the starting column.
EXAMPLE
Irregular triangle T(n,k) begins:
n/k | 1 2 3 4 5 6 7 8 9
------------------------------------------------------
1 | (empty row)
2 | 2;
3 | 3, 3;
4 | 4, 2, 2, 4;
5 | 5, 3, 4, 3, 5;
6 | 6, 4, 2, 3, 3, 2, 4, 6;
7 | 7, 5, 3, 5, 2, 5, 3, 5, 7;
...
PROG
(MATLAB)
function a = A361819( max_row )
k = 1;
for r = 2:max_row
h = zeros(1, r); h(1) = r;
while max(h) > 1
j = find(h == max(h), 1, 'last' );
m = find(h < max(h)-1, 1, 'first' );
a(k) = (m-j) + (h(j)-h(m)) - 1;
h(j) = h(j) - 1; h(m) = h(m) + 1;
k = k+1;
end
end
end % Thomas Scheuerle, Mar 27 2023
CROSSREFS
Cf. A361642, A361660, A002541 (row lengths), A002378 (row sums).
Sequence in context: A357372 A316910 A251715 * A304758 A103893 A288159
KEYWORD
nonn,tabf
AUTHOR
Tamas Sandor Nagy, Mar 25 2023
STATUS
approved