

A319537


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.


1



5, 11, 80, 4, 7, 9, 22, 24, 29, 32, 41, 300, 51, 58, 75, 79, 86, 95, 101, 107, 109, 116, 118, 120, 127, 128, 140, 301, 146, 149, 155, 159, 162, 168, 171, 173, 177, 183, 188, 191, 197, 203, 204, 208, 214, 216, 221, 222, 226, 232, 236, 242, 248, 252, 257, 259
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

This sequence has similarities with A210415: here we consider the positions of E's, there the positions of 1's.


LINKS

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, Perl program for A319537


FORMULA

Apparently, a(n+1) > a(n) for any n > 28.


EXAMPLE

The sequence starts with 5, 11, 80, 4, 7, 9, 22.
The corresponding concatenated English names are:
FIVEELEVENEIGHTYFOURSEVENNINETWENTYTWO
This must be read as:
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.
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.).
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.
We see that the sequence uses a lot of backtracking  making this kind of sequence quite hard to compute.


PROG

(Perl) See Links section.


CROSSREFS

Cf. A210415.
Sequence in context: A192436 A154797 A161852 * A002359 A292754 A090518
Adjacent sequences: A319534 A319535 A319536 * A319538 A319539 A319540


KEYWORD

nonn,word


AUTHOR

Eric Angelini and Rémy Sigrist, Sep 22 2018


STATUS

approved



