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.

Previous Showing 11-13 of 13 results.

A105690 A simple "Fractal Jump Sequence" (FJS).

Original entry on oeis.org

2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 1, 2
Offset: 2

Views

Author

Eric Angelini, May 04 2005

Keywords

Comments

See A105397 for definition of Fractal Jump Sequence.
This is how to construct the sequence: start with 2 on rows a and b; put 2 empty spaces behind the 2 on row a; choose any two digits and put them on row b under the 2 empty spaces of row a; go back to row a and add the same two digits but each one with its according spaces (1 must always be followed by 1 space on row a and 2 must always be followed by 2 spaces); go back to row b and add under the next available spaces of a the digits necessary so to have the same succession of digits in rows b and a. The sequence builds itself automatically. The row (c) is obtained by "pushing" (a) into (b) -- [the first digit of a and b melt in a single copy of themselves]. Row (c) is the FJS sequence above.
(a)..2..1.2..1.1.2..1.2..1.1.1.2..2.....
(b)..212.1.12.1.2.11.1.22.1.1.1.12.22...
----------------------------------------
(c)..21211212111221111222111111212222...

Examples

			To build such sequences one has only to choose the first digit d and the d digits to put under the d spaces of row (a).
		

A110909 First "Fractal Jump Sequence" (FJS) having its first ten digits different one from another, then the next ten, then the next ten, etc.

Original entry on oeis.org

9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 4, 2, 5, 6, 3, 7, 8, 9, 4, 0, 1, 2, 3, 5, 6, 7, 8, 9, 0, 6, 1, 2, 3, 4, 5, 8, 7, 9, 0, 1, 2, 3, 4, 5, 8, 6, 7, 9, 2, 3, 5, 6, 7, 0, 1, 8, 4, 9, 0, 1, 3, 2, 4, 6, 5, 7, 8, 9, 0, 1, 6, 2, 4, 5, 7, 8, 9, 3, 0, 1, 2, 7, 3, 4, 5, 6, 8, 9, 0, 8, 1, 2, 3, 4, 5, 6, 7, 9, 9, 0, 1, 2, 3
Offset: 9

Views

Author

Noelle Clou (keynews.tv(AT)skynet.be), Oct 03 2005

Keywords

Comments

See A105397 for definition of Fractal Jump Sequence.

Crossrefs

Cf. A105473.

Extensions

Terms computed by Gilles Sadowski.

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

Previous Showing 11-13 of 13 results.