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 A123060 Least positive integer k such that n has the same number of characters in base k and in Roman numeral representation, or 0 if no such k exists. 1
 1, 1, 1, 3, 6, 3, 2, 2, 4, 11, 4, 3, 2, 3, 4, 3, 0, 2, 3, 5, 3, 0, 2, 0, 3, 0, 2, 0, 3, 4, 3, 0, 2, 0, 3, 0, 2, 0, 0, 7, 4, 3, 0, 3, 4, 3, 0, 2, 3, 51, 8, 4, 3, 4, 8, 4, 3, 0, 4, 8, 4, 3, 0, 3, 5, 3, 0, 0, 3, 5, 3, 0, 0, 0, 3, 0, 0, 2, 0, 3, 3, 0, 2, 0, 3, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 LINKS Nathaniel Johnston, Table of n, a(n) for n = 1..3999 Gerard Schildberger, The first 3999 numbers in Roman numerals. FORMULA a(n) = min{k: StringLength(n base k) = StringLength(Roman(n))}, or 0 if no such k exists. a(n) = min{k: A006968(n) = 1 + floor(log_b(n))}, or 0 if no such k exists. EXAMPLE a(1) = 1 since Roman(1) = I and 1(base 1) have the same (1) number of characters. a(4) = 3 since Roman(4) = IV and 11(base 3) have the same (2) number of characters. a(8) = 2 since Roman(8) = VIII and 1000(base 2) have the same (4) number of characters. a(10) = 11 since Roman(10) = X and X(base 11) have the same (1) number of characters. a(11) = 4 since Roman(11) = XI and 23(base 4) have the same (2) number of characters. a(12) = 3 since Roman(12) = XII and 110(base 3) have the same (3) number of characters. a(17) = 0 because Roman(17) = XVII has 4 characters, while 17 = 10001(base 2) has 5 characters and 17 = 122(base 3) has 3 characters. a(30) = 4 because Roman(30) = XXX has 3 characters, as do 110(base 5) and 132(base 4), but Min{4,5} = 4. MAPLE A123060 := proc(n) local k, l, r: if(n<=3)then return 1:fi: r:=length(convert(n, roman)): for k from 2 to n+1 do l:=nops(convert(n, base, k)): if(l = r)then return k: elif(l

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Last modified April 23 11:13 EDT 2024. Contains 371905 sequences. (Running on oeis4.)