A082763 Roman numeral contains an asymmetric symbol (L).

40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152

Offset: 1

Author

Rick L. Shepherd, May 21 2003

Keywords

• base
• easy
• nonn

Comments

I,V,X,C,D,M - and even the vinculum (bar) and apostrophus (backwards "C") - are each symmetric: horizontally, vertically, or both.

Numbers containing a 4 when decimally encoded with A061493. - Reinhard Zumkeller, Apr 14 2013

Examples

			40 = XL, 89 = LXXXIX, 140 = CXL.
		

Links

• Nathaniel Johnston, Table of n, a(n) for n = 1..2000 (complete up to 3999)
• Gerard Schildberger, The first 3999 numbers in Roman numerals
• Eric Weisstein's World of Mathematics, Roman Numeral.

Crossrefs

Cf. A006968 (Roman numerals main entry), A078715 (Palindromic Roman numerals).

Programs

Haskell

a082763 n = a082763_list !! (n-1)
a082763_list = filter (containsL . a061493) [1..3999] where
   containsL x = d == 4 || x > 0 && containsL x' where
                 (x',d) = divMod x 10
-- Reinhard Zumkeller, Apr 14 2013

Maple

with(StringTools): for n from 1 to 152 do if(Search("L", convert(n, roman)) > 0)then printf("%d, ", n): fi: od: # Nathaniel Johnston, May 18 2011

Mathematica

Select[Range[200],StringCases[RomanNumeral[#],"L"]!={}&] (* Harvey P. Dale, Jun 10 2023 *)

PARI

/* "%" use below is actually identical to lift(Mod(n-1,50)) */ /* (n-1)50 could be used for integer division below */ /* instead of floor, but the OEIS sometimes loses  */ /* characters depending upon where on a submitted line they are. */ a(n)=floor((n-1)/50)*100+40+(n-1)%50 for(n=1,125,print1(a(n),","))

Formula

a(n+50) = a(n) + 100 for n >= 1 [a(n+L) = a(n) + C for n >= I], a(1) = 40 [a(I) = XL], a(n+1) = a(n) + 1 for 1 <= n <= 49 [a(n+I) = a(n) + I for I <= n <= XLIX]; so a(n) = floor((n-1)/50)*100 + 40 + ((n-1)(mod 50)) for n >= 1 [a(n) = floor((n-I)/L)*C + XL + ((n-I)(mod L)) for n >= I].