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A036746
Numbers with "long" representations in Roman notation: given by last n letters from ...MMMDCCCLXXXVIII.
7
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
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
KEYWORD
nonn,base,easy
AUTHOR
EXTENSIONS
Checked by Robert G. Wilson v, Aug 13 2008
Edited by M. F. Hasler, Jan 12 2015
STATUS
approved