A093796 Sequence of digit-values after concatenating the natural numbers < 4000 in Roman numeral representation.

1, 1, 1, 1, 1, 1, 1, 5, 5, 5, 1, 5, 1, 1, 5, 1, 1, 1, 1, 10, 10, 10, 1, 10, 1, 1, 10, 1, 1, 1, 10, 1, 5, 10, 5, 10, 5, 1, 10, 5, 1, 1, 10, 5, 1, 1, 1, 10, 1, 10, 10, 10, 10, 10, 1, 10, 10, 1, 1, 10, 10, 1, 1, 1, 10, 10, 1, 5, 10, 10, 5, 10, 10, 5, 1, 10, 10, 5, 1

Offset: 1

Author

Reinhard Zumkeller, May 18 2004

Keywords

• nonn
• base
• easy
• fini
• full

Comments

In other words, read the sequence of Roman numerals of natural numbers without spaces: I II III IV V VI VII VIII IX, ..., replacing I by 1, V by 5, X by 10, etc.

Examples

			I,II,III,IV,V,VI,VII,VIII,IX,X,XI,XII, ...
I,(I,I),(I,I,I),(I,V),V,(V,I),(V,I,I),(V,I,I,I),(I,X), ...
1,(1,1),(1,1,1),(1,5),5,(5,1),(5,1,1),(5,1,1,1),(1,10), ...
1,1,1,1,1,1,1,5,5,5,1,5,1,1,5,1,1,1,1,10,10,10,1,10,1,1, ...
		

References

• GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See Question 300(b), page 199.

Links

• Nathaniel Johnston, Table of n, a(n) for n = 1..30000 (complete up to 3999)
• Stephanus Gibbs, Roman Numeral and Date Conversion
• Eric Weisstein's World of Mathematics, Roman Numerals

Crossrefs

Cf. A007376.

Cf. A061493; A222581 (run lengths).

Programs

Haskell

import Data.List (unfoldr)
a093796 n = a093796_list !! n
a093796_list = concatMap (reverse . unfoldr r) $ map a061493 [1..3999]
   where r 0 = Nothing
         r x = Just ([0,1,5,10,50,100,500,1000] !! fromInteger d, x')
               where (x', d) = divMod x 10
-- Reinhard Zumkeller, Apr 14 2013

Maple

for n from 1 to 50 do r:=convert(n, roman): for j from 1 to length(r) do printf("%d, ", convert(r[j],arabic)): od: od: # Nathaniel Johnston, May 18 2011

Mathematica

A093796full = Flatten[FromRomanNumeral[Characters[RomanNumeral[Range[3999]]]]];
A093796full[[;;100]] (* Paolo Xausa, Mar 03 2024 *)

Python

def f(s, k):
    return s[:2] if k==4 else (s[1]*(k>=5)+s[0]*(k%5) if k<9 else s[0]+s[2])
def r(n):
    m, c, x, i = n//1000, (n%1000)//100, (n%100)//10, n%10
    return "M"*m + f("CDM", c) + f("XLC", x) + f("IVX", i)
def afull():
    v = {"I":1, "V":5, "X":10, "L":50, "C":100, "D":500, "M":1000}
    ans = []
    for i in range(1, 4000): ans.extend([v[d] for d in r(i)])
    return ans
print(afull()[:80]) # Michael S. Branicky, Mar 04 2024