%I #22 Sep 29 2018 06:09:46
%S 5,11,80,4,7,9,22,24,29,32,41,300,51,58,75,79,86,95,101,107,109,116,
%T 118,120,127,128,140,301,146,149,155,159,162,168,171,173,177,183,188,
%U 191,197,203,204,208,214,216,221,222,226,232,236,242,248,252,257,259
%N The sequence gives the distinct positions, not necessarily in order, of all letters E in the concatenation of the English names (without spaces or hyphens) of its terms. This is the lexicographically earliest such sequence.
%C This sequence has similarities with A210415: here we consider the positions of E's, there the positions of 1's.
%H Rémy Sigrist, <a href="/A319537/b319537.txt">Table of n, a(n) for n = 1..10000</a>
%H Rémy Sigrist, <a href="/A319537/a319537_1.pl.txt">Perl program for A319537</a>
%F Apparently, a(n+1) > a(n) for any n > 28.
%e The sequence starts with 5, 11, 80, 4, 7, 9, 22.
%e The corresponding concatenated English names are:
%e FIVEELEVENEIGHTYFOURSEVENNINETWENTYTWO
%e This must be read as:
%e The 5th letter of the concatenation is an E; the 11th letter is an E; the 80th letter too; the 4th letter too; and so are the 7th, the 9th, the 22nd, etc.
%e The sequence was built trying always to find the smallest integer that does not lead to a contradiction. Thus we could not start with ONE as the first letter would not be an E but an O; TWO also fails as the second letter is not an E but a W; THREE fails for the same reason (R instead of an expected E); FOUR fails again (R instead of E); FIVE is ok as it will be possible to put an E in position 5 in the sequence (either with EIGHT, ELEVEN, EIGHTEEN, EIGHTY, etc.).
%e This means that a(2) must begin with an E; we try EIGHT but EIGHT fails as the 8th letter of the sequence would not be an E but the H of EIGHT itself. ELEVEN fits, because there will be a way to extend the sequence with an 11th letter being E; a(3) cannot be EIGHTEEN as the 18th letter of the sequence would be the N of EIGHTEEN itself; thus a(3) = EIGHTY; a(4) = FOUR as this is the smallest number not leading to a contradiction (the 4th letter of the sequence is indeed the E of FIVE); a(5) = SEVEN as the 7th letter of the sequence is precisely the middle E of ELEVEN, etc.
%e We see that the sequence uses a lot of backtracking - making this kind of sequence quite hard to compute.
%o (Perl) See Links section.
%Y Cf. A210415.
%K nonn,word
%O 1,1
%A _Eric Angelini_ and _Rémy Sigrist_, Sep 22 2018