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A348389
Irregular triangle read by rows: row n gives for n >= 2 a concatenation of the finite sequences of the multiples of k, larger than k and not exceeding n, for k = 1, 2, ..., floor(n/2).
3
2, 2, 3, 2, 3, 4, 4, 2, 3, 4, 5, 4, 2, 3, 4, 5, 6, 4, 6, 6, 2, 3, 4, 5, 6, 7, 4, 6, 6, 2, 3, 4, 5, 6, 7, 8, 4, 6, 8, 6, 8, 2, 3, 4, 5, 6, 7, 8, 9, 4, 6, 8, 6, 9, 8, 2, 3, 4, 5, 6, 7, 8, 9, 10, 4, 6, 8, 10, 6, 9, 8, 10, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 4, 6, 8, 10, 6, 9, 8, 10, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 4, 6, 8, 10, 12, 6, 9, 12, 8, 12, 10, 12
OFFSET
2,1
COMMENTS
The length of row n is A002541(n).
The sum of row n is A348392(n).
The lengths of the sublists for these multiples of k in row n are given by T(n, k) = A348388(n, k), for n >= 2 and k = 1, 2, ..., floor(n/2).
FORMULA
The entries a(n, m) of row n, for n > = 2 and m = 1, 2, ..., A002541(n), are given by the concatenation of the sequences k*(2, 3, ..., t(n,k)) for k = 1, 2, ..., floor(n/2), with t(n, k) = floor((n-k)/k) + 1.
EXAMPLE
The irregular triangle a(n, m) begins: (the k-sublists are separated by a vertical bar)
n\m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 ...
-------------------------------------------------------------------------
2: 2
3: 2 3
4: 2 3 4|4
5: 2 3 4 5|4
6: 2 3 4 5 6|4 6|6
7: 2 3 4 5 6 7|4 6| 6
8: 2 3 4 5 6 7 8|4 6 8| 6| 8
9: 2 3 4 5 6 7 8 9| 4 6 8| 6 9| 8
10: 2 3 4 5 6 7 8 9 10| 4 6 8 10| 6 9| 8|10
11: 2 3 4 5 6 7 8 9 10 11| 4 6 8 10| 6 9| 8|10
12: 2 3 4 5 6 7 8 9 10 11 12| 4 6 8 10 12| 6 9 12| 8 12|10|12
13: 2 3 4 5 6 7 8 9 10 11 12 13| 4 6 8 10 12| 6 9 12| 8 12|10|12
...
MATHEMATICA
nrows=10; Table[Flatten[Table[Range[2k, n, k], {k, Floor[n/2]}]], {n, 2, nrows+1}] (* Paolo Xausa, Nov 23 2021 *)
CROSSREFS
KEYWORD
nonn,easy,tabf
AUTHOR
Wolfdieter Lang, Oct 31 2021
STATUS
approved