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Array read by antidiagonals: a(n,m) = the number of digits of m is when written in base n. The top row is the number of digits for each m in base 1.
5

%I #25 Aug 31 2017 21:59:24

%S 1,1,2,1,2,3,1,1,2,4,1,1,2,3,5,1,1,1,2,3,6,1,1,1,2,2,3,7,1,1,1,1,2,2,

%T 3,8,1,1,1,1,2,2,2,4,9,1,1,1,1,1,2,2,2,4,10,1,1,1,1,1,2,2,2,3,4,11,1,

%U 1,1,1,1,1,2,2,2,3,4,12,1,1,1,1,1,1,2,2

%N Array read by antidiagonals: a(n,m) = the number of digits of m is when written in base n. The top row is the number of digits for each m in base 1.

%C A162320 is the array without the base 1 number lengths, and with the lengths of base 2 numbers in the top row.

%H Michael De Vlieger, <a href="/A162319/b162319.txt">Table of n, a(n) for n = 1..10440</a> (covers bases 1..144)

%e From _Michael De Vlieger_, Jan 02 2015: (Start)

%e Array read by antidiagonals begins:

%e 1;

%e 1, 2;

%e 1, 2, 3;

%e 1, 1, 2, 4;

%e 1, 1, 2, 3, 5;

%e 1, 1, 1, 2, 3, 6;

%e 1, 1, 1, 2, 2, 3, 7;

%e 1, 1, 1, 1, 2, 2, 3, 8;

%e 1, 1, 1, 1, 2, 2, 2, 4, 9;

%e 1, 1, 1, 1, 1, 2, 2, 2, 4, 10;

%e ...

%e Array adjusted such that the rows represent base n and the columns m:

%e m

%e 1 2 3 4 5 6 7 8 9 10

%e ------------------------------

%e base 1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10;

%e base 2: 1, 2, 2, 3, 3, 3, 3, 4, 4, (4);

%e base 3: 1, 1, 2, 2, 2, 2, 2, 2, (3, 3);

%e base 4: 1, 1, 1, 2, 2, 2, 2, (2, 2, 2);

%e base 5: 1, 1, 1, 1, 2, 2, (2, 2, 2, 2);

%e base 6: 1, 1, 1, 1, 1, (2, 2, 2, 2, 2);

%e base 7: 1, 1, 1, 1, (1, 1, 2, 2, 2, 2);

%e base 8: 1, 1, 1, (1, 1, 1, 1, 2, 2, 2);

%e base 9: 1, 1, (1, 1, 1, 1, 1, 1, 2, 2);

%e base 10: 1, (1, 1, 1, 1, 1, 1, 1, 1, 1);

%e ...

%e 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 = 4. The number m = 2 in base n = 4 requires 1 digit, thus a(12) = 1.

%e For n = 14, a(14) is found in the fourth position in row 5 in the array read by antidiagonals. This equates to m = 4, base n = 2. The number m = 4 in base n = 2 requires 3 digits, thus a(14) = 3. (End)

%t Table[Function[k, If[k == 1, m, IntegerLength[m, k]]][k - m + 1], {k, 13}, {m, k}] // Flatten (* _Michael De Vlieger_, Aug 31 2017 *)

%Y Cf. A162320.

%K base,nonn,tabl

%O 1,3

%A _Leroy Quet_, Jul 01 2009