%N Number of numerals to represent n in a base b multiplicative grouping numeral system where b=10.
%C This sequence assumes that there are unique numerals for all powers of b, namely, 1,b,b^2,b^3,... and there are other unique numerals for 2,3,...,b-1, where b=10 is the base here. There is no symbol for zero.
%C As an artificial example, suppose that 1,2,3,...,9 have their normal meanings and that A,B,C represent 10,10^2=100,10^3=1000, respectively. Then 3407 (normal base 10) = 3C4B7, using five "numerals". Note that at most one numeral is necessary for the units (because the absence of any following A,B,C,..., implies units) whereas the other numerals often occur in pairs.
%C However, I infer from the Britannica article (since no relevant examples are given) that CBA, using only three numerals, rather than 1C1B1A represents 1110 (normal base 10) -- as the absence of paired numerals can imply 1's. At any rate, that's how this sequence is calculated.
%C One historical example, "the principal example" of a multiplicative grouping system, is the Chinese traditional notational system, where the Britannica source shows the symbols for 1 through 9 and corresponding to A,B,C above but does not suggest the existence of symbols for powers of 10 larger than 1000. (This Chinese system is customarily written vertically downward.) In contrast, the Chinese modern national and mercantile systems are both positional number systems having a zero (a circle).
%D Encyclopaedia Britannica, 1981 ed., Vol. 11, "Mathematics, History of", p. 648.
%e a(121) = 4 as 121 (normal base 10) = B2A1 with A and B as discussed above and B2A1 has four numerals.
%e a(A002275(n)) = n for n >= 1.
%e a(m*A002275(n)) = 2*n - 1 for 2 <= m <= 9 and n >= 1.
%Y Cf. A055640 (similar for ciphered base 10 numeral system), A002275 (repunits).
%A _Rick L. Shepherd_, Jun 12 2005