A002904 - OEIS

%I #42 Nov 10 2024 23:04:06

%S 0,0,0,0,4,9,5,1,1,0,55,55,1,0,1,9,5,1,1,0,0,0,0,0,4,9,5,1,1,1,1,1,1,

%T 1,0,0,4,2,2,0,0,0,0,0,4,9,5,1,1,1,1,1,1,1,0,0,4,2,2,9,9,9,9,9,0,0,0,

%U 0,0,5,5,5,5,5,0,0,0,6,6,1,1,1,1,1,0,0,4,2,2,1,1,1,1,1,0,0,4,2,2,0

%N Delete all letters except c, d, i, l, m, v, x from the US English name of n, then read as Roman numeral if possible, otherwise 0.

%C From _M. F. Hasler_, Feb 21 2020: (Start)

%C There are several standards for Roman numerals, even though experts agree that none of them has ever been universally accepted, cf. references in Wikipedia article.

%C The current definition of the sequence uses "if possible", so whenever there is a possible interpretation according to any of these standards, it is acceptable.

%C Non-uniqueness could be caused by rare irregular variants which must be excluded in order to have a well-defined sequence. E.g., iixx will mean 18, not 22, although the standard notation for 18 is xviii (and xiix would also be used sometimes).

%C If required, the meaning of "possible" must be made more precise (e.g., allowing only "standard subtractive" or additionnally "standard additive" notation; this can be made completely explicit). (End)

%C There are 1411 nonzero terms; see Branicky link. - _Michael S. Branicky_, Jan 09 2023

%C The first term where British English differs from US is a(1001).  This gives 500 in US English, but is invalid in British English (1001 is one thousanD anD one). - _Paul Duckett_, Oct 24 2024

%H Michael S. Branicky, <a href="/A002904/a002904_1.txt">Table of n, a(n) for all nonzero a(n).</a>

%H Brady Haran and N. J. A. Sloane, <a href="https://www.youtube.com/watch?v=OeGSQggDkxI">What Number Comes Next?</a> (2018), Numberphile video.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Roman_numerals">Roman numerals</a>, as of Feb 19 2020.

%F From _Michael S. Branicky_, Jan 09 2023: (Start)

%F a(100..999) = 0, due to "dd".

%F a(k*1000) = 500 for k in {1..4, 10, 14, 20..24, 40..44};

%F a(k*1000+i) = (500 + a(i))*[a(i) != 0] for i in 1..99 for k in {1..4, 10, 14, 20..24, 40..44}; and

%F a(n) = 0 for all other 1100 <= n <= 999999, due to "id", "vd", "xd" or "dd".

%F a(n) = 0 for n >= 10^6, due to "ll".

%F (End)

%e To get a(5), write 5 = "five", delete 'f' & 'e' to get "iv", Roman for 4 = a(5).

%e To get a(35), write 35 = "thirty five", delete all but 'i's and 'v's, to get "iiv", which is not a legal Roman number*, so a(35) = 0 by definition of the sequence. (*It is never allowed to subtract more than one unit 'I' (or 'X' or 'C') from one of the symbols V, L, D representing 5 times a power of 10, as it is never allowed that one of V, L, D is subtracted from one of I, X, C, M.)

%Y See A092302 and A121305 for other versions.

%K nonn,word

%O 1,5

%A _J. H. Conway_

%E a(35) and beyond from _Michael S. Branicky_, Jan 09 2023