cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A359385 The lexicographically earliest "Increasing Term Fractal Jump Sequence" that does not use the digit 0 in any terms.

Original entry on oeis.org

1, 2, 21, 22, 23, 112, 122, 132, 133, 134, 141, 221, 311, 2112, 2113, 3111, 21111, 31113, 31114, 31124, 31131, 34111, 41121, 42111, 43111, 111121, 111122, 112111, 112311, 131111, 211112, 211113, 1111311, 1111312, 3111311, 3111312, 4111131, 4111132, 4141111
Offset: 1

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Author

Tyler Busby, Dec 29 2022

Keywords

Comments

The rules of an "Increasing Term Fractal Jump Sequence" are described in A105647.
The digit zero is omitted as it can create situations where the next term can require a 0 in the first digit.

Examples

			The sequence and the "kept" digits begin
  1, 2, 21, 22, 23, 112, 122, 132, 133, ...
  ^     ^    ^      ^ ^    ^    ^    ^
  1     2    2      1 2    2    2    3
		

Crossrefs

A359611 The lexicographically earliest "Increasing Term Fractal Jump Sequence".

Original entry on oeis.org

1, 2, 20, 22, 100, 200, 201, 1000, 20000, 20001, 110000, 2000000, 2000001, 110100000, 200000000, 200000001, 1101001000000, 2000000000020, 2000000010101, 10100010000000, 20000000000002, 20020000000001, 101001010010000, 100000000200000000000000
Offset: 1

Views

Author

Tyler Busby, Jan 06 2023

Keywords

Comments

The rules of an "Increasing Term Fractal Jump Sequence" are described in A105647.
We define a "forced" digit in Fractal Jump Sequences as a digit that is required to be a specific value by a digit that occurred previously in the sequence. This is in opposition to digits that could have any value selected for them without breaking the Fractal Jump Sequence rules. In the diagram below, the digits with carets below them are the forced digits.
To find a(n), increment a(n-1) until all of the forced digits that will positionally occur in a(n) satisfy their forced values. Then, to avoid leading zeros in a(n+1), if there are forced zeros immediately following the candidate a(n), continue to increment until it is the same number of digits longer as there are consecutive forced zeros, and continue to increment until the candidate a(n) once again satisfies all forcing criteria (including the new zeros).
The only digits that appear in this sequence are 0, 1, and 2, even though no numerals are arbitrarily restricted from appearing.

Examples

			The sequence and the "kept"/"forced" digits begin
  1, 2, 20, 22, 100, 200, 201, 1000, 20000, ...
  ^     ^    ^    ^  ^    ^    ^ ^^  ^  ^^
  1     2    2    0  2    2    1 00  2  00
In the case of computing a(5), we have a 22 for a(4), so we would normally increment to 23, as there is nothing forcing the next two digits. However, since there is a 0 forcing the following digit, we must increment to the smallest number that satisfies this forced 0 (as we can't have leading zeros in a(6)).
		

Crossrefs

Showing 1-2 of 2 results.