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.

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

Offset: 1

Author

Eric Angelini and Rémy Sigrist, Sep 22 2018

Keywords

• nonn
• word

Comments

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A210415.

Programs

Perl

See Links section.

Formula

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