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%I #15 Jan 03 2021 00:42:57
%S 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,
%T 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,
%U 3,5,3,0,0,0,3,0,0,2,0,3,3,0,2,0,3,0
%N 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.
%H Nathaniel Johnston, <a href="/A123060/b123060.txt">Table of n, a(n) for n = 1..3999</a>
%H Gerard Schildberger, <a href="/A006968/a006968.txt">The first 3999 numbers in Roman numerals</a>.
%F 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.
%e a(1) = 1 since Roman(1) = I and 1(base 1) have the same (1) number of characters.
%e a(4) = 3 since Roman(4) = IV and 11(base 3) have the same (2) number of characters.
%e a(8) = 2 since Roman(8) = VIII and 1000(base 2) have the same (4) number of characters.
%e a(10) = 11 since Roman(10) = X and X(base 11) have the same (1) number of characters.
%e a(11) = 4 since Roman(11) = XI and 23(base 4) have the same (2) number of characters.
%e a(12) = 3 since Roman(12) = XII and 110(base 3) have the same (3) number of characters.
%e 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.
%e a(30) = 4 because Roman(30) = XXX has 3 characters, as do 110(base 5) and 132(base 4), but Min{4,5} = 4.
%p 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<r)then return 0: fi: od: return 0: end: seq(A123060(n),n=1..86); # _Nathaniel Johnston_, May 18 2011
%Y Cf. A006968.
%K base,easy,nonn
%O 1,4
%A _Jonathan Vos Post_, Sep 26 2006
%E Extended and corrected by _Nathaniel Johnston_, May 18 2011
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