cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A036788 Length of Roman notation for n <= length of decimal representation.

Original entry on oeis.org

1, 5, 10, 11, 15, 20, 40, 50, 51, 55, 60, 90, 100, 101, 102, 104, 105, 106, 109, 110, 111, 115, 120, 140, 150, 151, 155, 160, 190, 200, 201, 205, 210, 250, 300, 400, 401, 405, 410, 450, 500, 501, 502, 504, 505, 506, 509, 510, 511, 515, 520, 540, 550, 551, 555
Offset: 1

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Author

Keywords

Comments

The Roman numeration system used here is the naive one taught in primary school. This sequence, like many others involving numeration systems, is neither well-defined nor interesting for large values of n. - N. J. A. Sloane, Jul 03 2008

Examples

			15 = XV has length 2 in both notations.
		

Crossrefs

Programs

  • Haskell
    a036788 n = a036788_list !! (n-1)
    a036788_list = [x | x <- [1..], a006968 x <= a055642 x]
    -- Reinhard Zumkeller, Apr 20 2013
  • Maple
    for n from 1 to 3999 do if(length(convert(n, roman)) <= length(n))then printf("%d, ", n): fi: od: # Nathaniel Johnston, May 18 2011
  • Mathematica
    Select[Range[560],StringLength[IntegerString[#,"Roman"]]<= IntegerLength[ #]&] (* Harvey P. Dale, Dec 18 2011 *)

Formula

A006968(a(n)) <= A055642(a(n)). - Reinhard Zumkeller, Apr 20 2013

Extensions

Corrected and extended by Larry Reeves (larryr(AT)acm.org), Sep 25 2000

A002904 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.

Original entry on oeis.org

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, 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, 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
Offset: 1

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Author

Keywords

Comments

From M. F. Hasler, Feb 21 2020: (Start)
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.
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.
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).
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)
There are 1411 nonzero terms; see Branicky link. - Michael S. Branicky, Jan 09 2023
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

Examples

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

Crossrefs

See A092302 and A121305 for other versions.

Formula

From Michael S. Branicky, Jan 09 2023: (Start)
a(100..999) = 0, due to "dd".
a(k*1000) = 500 for k in {1..4, 10, 14, 20..24, 40..44};
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
a(n) = 0 for all other 1100 <= n <= 999999, due to "id", "vd", "xd" or "dd".
a(n) = 0 for n >= 10^6, due to "ll".
(End)

Extensions

a(35) and beyond from Michael S. Branicky, Jan 09 2023
Showing 1-2 of 2 results.