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Maximum sum of digits for any number with n digits in fractional base 4/3.
3

%I #25 Mar 26 2024 11:16:18

%S 0,3,6,8,9,12,13,15,17,19,22,24,26,28,30,32,33,36,37,40,42,44,46,48,

%T 50,52,54,56,57,60,62,65,67,70,71,73,75,77,80,83,84,87,90,93,94,96,98,

%U 101,104,106,108,109,112,115,117,120,122,123,126,129,131,133,134

%N Maximum sum of digits for any number with n digits in fractional base 4/3.

%C This sequence is strictly increasing since if a(n) is attained by the sum of digits of k, then the final digit of k is 3 and (k - (k mod 3))*4/3 + 3 is the same digits with a new second-least significant 1, 2 or 3 inserted, and so a(n+1) >= a(n) + 1.

%C Terms can be derived from A357425 by a(n) = s for the largest s where A357425(s) has n digits in base 4/3.

%H Kevin Ryde, <a href="/A363758/b363758.txt">Table of n, a(n) for n = 0..207</a>

%H Kevin Ryde, <a href="/A363758/a363758_2.pdf">Mean Digit Plot</a> (and <a href="/A363758/a363758_1.tex">LaTeX source</a>).

%H <a href="/index/Ba#base_fractional">Index entries for sequences related to fractional bases</a>

%F a(n) = Max_{4*A087192(n-1) <= i < 4*A087192(n)} A244041(i), for n>=2.

%e For n=9, the numbers with 9 digits in base 4/3 are 60 to 79 and among them the maximum sum of digits is A244041(75) = 19 (those digits being 321023323), and so a(9) = 19.

%Y Cf. A024631 (base 4/3), A244041 (sum of digits).

%Y Cf. A357425 (smallest with sum s), A087192.

%K nonn,base

%O 0,2

%A _Kevin Ryde_, Jun 20 2023