login
Lexicographically earliest sequence of distinct terms starting with a(1) = 1 ("one") such that the letter count of a(n) + the letter count of a(n+1) = the letter count of [a(n) + a(n+1)].
2

%I #23 Jan 03 2020 23:02:01

%S 1,10,2,20,3,30,4,9,5,40,6,7,50,8,60,41,72,42,71,53,61,51,63,54,59,55,

%T 64,49,65,57,56,58,67,46,66,47,100,11,92,21,200,12,91,22,95,18,110,13,

%U 210,15,96,16,90,14,99,19,94,32,81,68,85,36,82,31,93,34,300,17,400,23,500,24,600,25,98,26,700,27,800,28

%N Lexicographically earliest sequence of distinct terms starting with a(1) = 1 ("one") such that the letter count of a(n) + the letter count of a(n+1) = the letter count of [a(n) + a(n+1)].

%H Hans Havermann, <a href="/A330365/b330365.txt">Table of n, a(n) for n = 1..1000</a>

%e one + ten = eleven (6 letters on both sides of the equal sign);

%e ten + two = twelve (6 letters on both sides);

%e two + twenty = twenty-two (9 letters on both sides);

%e twenty + three = twenty-three (11 letters on both sides);

%e three + thirty = thirty-three (11 letters on both sides);

%e thirty + four = thirty-four (10 letters on both sides);

%e four + nine = thirteen (8 letters on both sides);

%e nine + five = thirteen (8 letters on both sides); etc.

%t len[n_]:=StringLength[StringReplace[IntegerName[n,"Words"],{" "-> "",","-> "","-"-> ""}]]; a[1]=1;

%t a[n_]:=a[n]=Module[{m=First[Complement[Range[n],a/@Range[n-1]]]},While[len[a[n-1]]+len[m]!=len[a[n-1]+m]||MemberQ[a/@Range[n-1],m]==True,m++];m];

%t a/@Range[100] (* _Ivan N. Ianakiev_, Dec 14 2019 *)

%Y Cf. A213006 (positive numbers that, spelled in English, are as long as the spelling of their digits).

%K nonn,word

%O 1,2

%A _Eric Angelini_ and _Hans Havermann_, Dec 12 2019