OFFSET
1,3
COMMENTS
A162319 is the same array with the lengths of base 1 numbers in the top row.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..10296 (Covers bases n = 1..144)
EXAMPLE
From Michael De Vlieger, Jan 02 2015: (Start)
Array read by antidiagonals begins:
1;
1, 2;
1, 1, 2;
1, 1, 2, 3;
1, 1, 1, 2, 3;
1, 1, 1, 2, 2, 3;
1, 1, 1, 1, 2, 2, 3;
1, 1, 1, 1, 2, 2, 2, 4;
1, 1, 1, 1, 1, 2, 2, 2, 4;
...
Array adjusted such that the rows represent base n and the columns m:
m
1 2 3 4 5 6 7 8 9 10
------------------------------
base 2: 1, 2, 2, 3, 3, 3, 3, 4, 4, (4);
base 3: 1, 1, 2, 2, 2, 2, 2, 2, (3, 3);
base 4: 1, 1, 1, 2, 2, 2, 2, (2, 2, 2);
base 5: 1, 1, 1, 1, 2, 2, (2, 2, 2, 2);
base 6: 1, 1, 1, 1, 1, (2, 2, 2, 2, 2);
base 7: 1, 1, 1, 1, (1, 1, 2, 2, 2, 2);
base 8: 1, 1, 1, (1, 1, 1, 1, 2, 2, 2);
base 9: 1, 1, (1, 1, 1, 1, 1, 1, 2, 2);
base 10: 1, (1, 1, 1, 1, 1, 1, 1, 1, 1);
...
For n = 12, a(12) is found in the second position in row 5 in the array read by antidiagonals. This equates to m = 2, base n = 5. The number m = 2 in base n = 5 requires 1 digit, thus a(12) = 1.
For n = 20, a(20) is found in the fifth position in row 6 in the array read by antidiagonals. This equates to m = 5, base n = 3. The number m = 5 in base n = 3 requires 2 digits, thus a(20) = 2. (End)
MATHEMATICA
a162320[n_] := Block[{t = {}, i, j}, For[i = 1, i <= n, i++, For[j = i, j > 1, j--, AppendTo[t, Floor@Log[j, i - j + 1] + 1]]]; t]]; a162320[14] (* Michael De Vlieger, Jan 02 2015 *)
CROSSREFS
KEYWORD
AUTHOR
Leroy Quet, Jul 01 2009
STATUS
approved