OFFSET
1,2
COMMENTS
The longest sequence is finite for all n. We can deduce this, because we know from the work of Saias that A337125(m)/m * log m is bounded, where A337125(m) is the length of the longest simple path in the divisor graph of {1,...,m}. See the comment in A337125 giving constraints on its terms.
The sequence of row lengths starts 2, 6, 25, 97.
LINKS
Peter Munn, Rows n = 1..4 of triangle, flattened
E. Saias, Applications des entiers à diviseurs denses, Acta Arithmetica, 83, 3 (1998), 225-240.
FORMULA
For n >= 1, 1 <= k <= row length(n), T(n,k) <= n * k.
For n >= 1, 1 <= k < row length(n), max(T(n,k+1)/T(n,k), T(n,k)/T(n,k+1)) is in A000040.
EXAMPLE
For n = 1, the only sequences of distinct positive integers that have their k-th term not exceeding 1*k = k, are those whose n-th term is k. The longest such sequence in which the smaller of adjacent terms divides the larger, giving a prime, is (1, 2), since 3/2 is 1.5. So row 1 has length 2, with T(1,1) = 1, T(1,2) = 2.
Table begins:
1, 2;
1, 3, 6, 2, 4, 8;
1, 2, 4, 12, 6, 18, 9, 3, 15, 30, 10, 5, 35, 7, 21, 42, 14, 28, 56, 8, 16, 48, 24, 72, 36;
...
CROSSREFS
KEYWORD
nonn,hard,tabf
AUTHOR
Peter Munn, Dec 28 2020
STATUS
approved