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Minimum sum of digits for any number of length n digits in fractional base 4/3.
1

%I #29 Dec 17 2023 03:14:42

%S 0,3,5,6,6,8,8,9,10,10,11,11,11,11,13,14,16,17,17,17,18,19,21,22,22,

%T 23,24,26,26,26,27,28,29,29,29,29,29,29,31,33,34,35,36,37,38,38,38,39,

%U 39,41,41,42,42,43,43,45,45,46,46,48,50,50,52,52,52,52,53,55

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

%C 0 is taken to be 1 digit long so a(1) = 0.

%C Terms can be derived from A364779 by a(n) = s for the smallest s where k = A364779(s) is >= n digits long (noting that stripping trailing 0's from k suffices to show numbers with sum of digits s exist at each length down to where sum s-1 exists).

%H Kevin Ryde, <a href="/A364751/b364751.txt">Table of n, a(n) for n = 1..209</a>

%H Kevin Ryde, <a href="/A364751/a364751_1.pdf">Mean Digit Plot</a>

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

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

%Y Cf. A024631 (base 4/3), A244041 (sum of digits), A364779 (largest with sum).

%Y Cf. A363758 (maximum sum).

%K nonn,base

%O 1,2

%A _Kevin Ryde_, Sep 07 2023