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

1,8

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.

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

Example

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, ...

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 *)

Prog

(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
(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

Crossrefs

Cf. A007376.

Cf. A061493; A222581 (run lengths).

Sequence in context: A019173 A230192 A172359 * A021647 A181668 A247824

Adjacent sequences: A093793 A093794 A093795 * A093797 A093798 A093799

Keyword

nonn,base,easy,fini,full

Author

Reinhard Zumkeller, May 18 2004

Status

approved