%I #9 Mar 17 2020 21:17:45
%S 119,139,159,179,199,319,339,359,379,399,519,539,559,579,599,719,739,
%T 759,779,799,919,939,959,979,999,1119,1139,1159,1179,1199,1319,1339,
%U 1359,1379,1399,1519,1539,1559,1579,1599,1719,1739,1759,1779,1799,1919
%N Each letter appears an even number of times in the English names for 1 through n taken together (names without "and").
%H Robert Israel, <a href="/A103154/b103154.txt">Table of n, a(n) for n = 1..10000</a>
%F It appears that a(n+5)=a(n)+200. - _Robert Israel_, Mar 17 2020
%e Together the English words for 1 through 119 contain 40 d's, 218 e's, 56 f's, 22 g's, 64 h's, 100 i's, 4 l's, 144 n's, 62 o's, 64 r's, 44 s's, 150 t's, 32 u's, 36 v's, 22 w's, 22 x's and 80 y's. All counts are even so 119 is in the sequence.
%p letters:= [$"a".."z"]:
%p V:= Vector(26):
%p Res:= NULL: count:= 0:
%p for n from 1 while count < 100 do
%p S:= convert(n,english);
%p V:= V + <seq(StringTools:-CountCharacterOccurrences(S,letters[i]),i=1..26)>;
%p if andmap(type,V,even) then
%p count:= count+1; Res:= Res, n;
%p fi
%p od:
%p Res; # _Robert Israel_, Mar 17 2020
%K nonn,word
%O 1,1
%A _David W. Wilson_, Jan 24 2005