A222581 Run lengths of digits when concatenating Roman numerals less than 4000, cf. A093796.

7, 3, 1, 1, 2, 1, 4, 3, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 5, 1, 2, 2, 2, 3, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 3, 2, 1, 7, 1, 3, 2, 3, 3, 3, 1, 1, 3, 1, 3, 1, 1, 3, 1, 2, 3, 1, 3, 3, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1

Offset: 1

Author

Reinhard Zumkeller, Apr 14 2013

Keywords

• nonn
• base
• fini
• full

Comments

See A078715 for a discussion on the Roman 4M-problem;

a(n) <= 7, that is, the longest run of consecutive equal digits in A093796 has length = 7; see also example.

Examples

			The 3999 Roman numerals of all numbers less than 4000 consist of 30000 digits; there are 19770 runs of consecutive equal digits: a(19770) = 1 is the last term of this sequence;
a(1)=a(52)=7, there are two runs with length 7: the first is "IIIIIII" which is the prefix of the concatenation of I, II, III and IV, the second is "XXXXXXX" which is contained in the concatenation of XXIX, XXX and XXXI;
a(1022)=a(14573)=6, there are also two runs with length 6: the first is "CCCCCC" which is a prefix of the concatenation of CCC and CCCI, the second is "MMMMMM" which is a prefix of the concatenation of MMM and MMMI;
a(30)=5, there is just one run with length 5: "XXXXX" which is contained in the concatenation of XIX, XX and XXI;
a(7)=a(644)=a(1359)=a(9375)=a(19194)=4, there are five runs with length 4: "IIII", two times "CCCC" and "MMMM", they occur in concatenations of (VIII, IX), (CC, CCI), (CCCXC, CCCXCI), (MM, MMI), (MMMCM, MMMCMI), respectively.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..19770
• Eric Weisstein's World of Mathematics, Roman Numerals
• Wikipedia, Roman numerals

Crossrefs

Cf. A006968.

Programs

Haskell

import Data.List (group)
a222581 n = a222581_list !! (n-1)
a222581_list = map length $ group a093796_list

Mathematica

A222581full = Map[Length, Split[Flatten[FromRomanNumeral[Characters[RomanNumeral[ Range[3999]]]]]]]; A222581full[[;;100]] (* Paolo Xausa, Mar 03 2024 *)

Python

from itertools import groupby
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():
    return [len(list(g)) for k, g in groupby("".join(r(i) for i in range(1, 4000)))]
print(afull()[:90]) # Michael S. Branicky, Mar 03 2024