

A108622


Number of numerals to represent n in a base b multiplicative grouping numeral system where b=10.


0



1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 2, 2, 2, 2, 2
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OFFSET

1,11


COMMENTS

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,...,b1, where b=10 is the base here. There is no symbol for zero.
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.
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.
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).


REFERENCES

Encyclopaedia Britannica, 1981 ed., Vol. 11, "Mathematics, History of", p. 648.


LINKS

Table of n, a(n) for n=1..105.


EXAMPLE

a(121) = 4 as 121 (normal base 10) = B2A1 with A and B as discussed above and B2A1 has four numerals.
a(A002275(n)) = n for n >= 1.
a(m*A002275(n)) = 2*n  1 for 2 <= m <= 9 and n >= 1.


CROSSREFS

Cf. A055640 (similar for ciphered base 10 numeral system), A002275 (repunits).
Sequence in context: A235125 A238418 A085576 * A112348 A262303 A127321
Adjacent sequences: A108619 A108620 A108621 * A108623 A108624 A108625


KEYWORD

base,nonn


AUTHOR

Rick L. Shepherd, Jun 12 2005


STATUS

approved



