A289281 Square array whose rows m >= 2 hold the limit under iterations of the morphism { x -> (x, ..., x+k-1) if k|x ; x -> x+1 otherwise }, starting with (0); read by falling antidiagonals.
0, 1, 0, 2, 1, 0, 2, 2, 1, 0, 3, 2, 2, 1, 0, 2, 3, 3, 2, 1, 0, 3, 3, 2, 3, 2, 1, 0, 4, 3, 3, 4, 3, 2, 1, 0, 2, 4, 4, 2, 4, 3, 2, 1, 0, 3, 5, 3, 3, 5, 4, 3, 2, 1, 0, 4, 3, 4, 4, 2, 5, 4, 3, 2, 1, 0, 4, 4, 4, 5, 3, 6, 5, 4, 3, 2, 1, 0, 5, 5, 5, 3, 4, 2, 6, 5, 4, 3, 2, 1, 0, 2, 3, 6, 4, 5, 3, 7, 6, 5, 4, 3, 2, 1, 0, 3, 4, 7, 5, 6, 4, 2, 7, 6, 5, 4, 3, 2, 1, 0, 4, 5, 4, 5, 3, 5, 3, 8, 7, 6, 5, 4
Offset: 2
Examples
The array starts (first row: m=2) [ 0 1 2 2 3 2 3 4 2 3 4 4 5 2 3 4 4 5 4 5 6 2 3 4 4 ...] [ 0 1 2 2 3 3 3 4 5 3 4 5 3 4 5 5 6 3 4 5 5 6 3 4 5 ...] [ 0 1 2 3 2 3 4 3 4 4 5 6 7 4 4 5 6 7 4 5 6 7 6 7 8 ...] [ 0 1 2 3 4 2 3 4 5 3 4 5 5 6 7 8 9 4 5 5 6 7 8 9 5 ...] [ 0 1 2 3 4 5 2 3 4 5 6 3 4 5 6 6 7 8 9 10 11 4 5 6 6 ...] [ 0 1 2 3 4 5 6 2 3 4 5 6 7 3 4 5 6 7 7 8 9 10 11 12 13 ...] [ 0 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 8 9 10 11 ...] [ 0 1 2 3 4 5 6 7 8 2 3 4 5 6 7 8 9 3 4 5 6 7 8 9 9 ...] [ 0 1 2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 10 3 4 5 6 7 8 ...] [ 0 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 10 11 3 4 5 6 ...] [ 0 1 2 3 4 5 6 7 8 9 10 11 2 3 4 5 6 7 8 9 10 11 12 3 4 ...] [ 0 1 2 3 4 5 6 7 8 9 10 11 12 2 3 4 5 6 7 8 9 10 11 12 13 ...] ... It is easy to prove that row m starts with (0, ..., m-1; 2, ..., m; 3, ..., m; m, ..., 2m-1; ...).
Links
- Kerry Mitchell, Table of n, a(n) for n = 2..10012
Programs
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PARI
A289281_row(n=30,k=2,a=[0])={while(#a
Comments