A036746 Numbers with "long" representations in Roman notation: given by last n letters from ...MMMDCCCLXXXVIII.

1, 2, 3, 8, 18, 28, 38, 88, 188, 288, 388, 888, 1888, 2888, 3888

Offset

1,2

Comments

Also: Least number using n symbols when written as Roman numeral, cf. A006968 and A061493. One could prefix a conventional a(0)=0. - M. F. Hasler, Jan 12 2015

Links

Table of n, a(n) for n=1..15.

Matt Parker, How Roman numerals broke the official dog database, YouTube video (2022).

Gerard Schildberger, The first 3999 numbers in Roman numerals.

Index entries for linear recurrences with constant coefficients, signature (2,-1).

Formula

a(n) = min{ k>0 | A006968(k)=n }. - M. F. Hasler, Jan 12 2015

Mathematica

(* assign the 3999 Roman numerals to the variable 'lst' so that lst = {I, II, III, ... MMMCMXCVII, MMMCMXCVIII, MMMCMXCIX}, e.g. using the link to Schildberger's file, then *) f[1] = 1; f[n_] := Block[{k = 2}, While[ StringLength@ SymbolName@ lst[[k]] != n, k++ ]; k]; Array[f, 15] (* Checked by Robert G. Wilson v, Aug 13 2008; edited by M. F. Hasler, Jan 12 2015 *)

Prog

(Haskell)
import Data.List (inits, elemIndex)
import Data.Maybe (mapMaybe)
a036746 n = a036746_list !! (n-1)
a036746_list = map (+ 1) $ mapMaybe (`elemIndex` a061493_list)
   (map (read . reverse) $ tail $ inits $ reverse $ show $ a061493 3888)
-- Reinhard Zumkeller, Apr 14 2013
(PARI) A036746(n)=(n%4+8/9)\.1^(n\4) \\ M. F. Hasler, Jan 12 2015

Crossrefs

Cf. A061493, A006968, A003587.

Sequence in context: A096254 A091765 A368882 * A232165 A129955 A034066

Adjacent sequences: A036743 A036744 A036745 * A036747 A036748 A036749

Keyword

nonn,base,easy

Author

J. H. Conway

Extensions

Checked by Robert G. Wilson v, Aug 13 2008

Edited by M. F. Hasler, Jan 12 2015

Status

approved