1
$\begingroup$

A sequence $a_i$ ($i=1,\ldots$) is lexicographically smaller than sequence $b_i$ if either $a_1 < b_1$, or $a_j = b_j$ for $j=1,\ldots, k$ and $a_{k+1} < b_{k+1}$.

If I asked for the lexicographically smallest sequence of natural numbers not in the OEIS, then I think it would start $1,1,1,1,1,1,1,1,1,1,1,2$—eleven $1$'s followed by a $2$—because A055642 starts with ten $1$'s followed by a $2$.

But what about integer sequences? After seeing @RossMillikan's answer, what I should ask for is the largest of all those sequences smaller than any sequence in the OEIS.

Of course once identified, it could be added to the OEIS.

$\endgroup$
3
  • 2
    $\begingroup$ It probably shouldn't - lest all matter of self-referential and changing issues arise. $\endgroup$ – Parcly Taxel Oct 20 '18 at 20:24
  • $\begingroup$ @ParclyTaxel: My final remark is tongue-in-cheek. :-) $\endgroup$ – Joseph O'Rourke Oct 20 '18 at 20:25
  • 2
    $\begingroup$ For the reason Ross Millikan explains in his answer, there is no lexicographically smallest sequence after $(1, 1, 1, \ldots)$. It is, however, meaningful to ask what sequence is lexicographically second smallest. One candidate is oeis.org/A160338, Height (maximum absolute value of coefficients) of the n-th cyclotomic polynomial: $a_1 = a_2 = \cdots = a_{104} = 1$, but $a_{105} = 2$. $\endgroup$ – Travis Willse Oct 20 '18 at 20:51
3
$\begingroup$

Given that you ask for naturals which do not include $0$, the first sequence lexicographically is OEIS A000012, which is all $1$'s. There is no sequence which is the next one lexicographically after this. You suggest starting with eleven $1$'s and a $2$, but then I suggest starting with twelve $1$'s and a $2$, then someone else will suggest a hundred $1$'s and a $2$, and so on.

The same problem occurs for integer sequences. Given any sequence that is missing, there is a lexicographically earlier one missing.

$\endgroup$
1
  • $\begingroup$ I see I asked incorrectly: I meant the maximum of all those smaller than the smallest. May not be repairable now... $\endgroup$ – Joseph O'Rourke Oct 20 '18 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.