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
KEYWORD
nonn,word
AUTHOR
Eric Angelini and Rémy Sigrist, Sep 22 2018
STATUS
approved