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 1-3 of 3 results.

A094005 a(n) = sum of lengths of strings that can be generated by any starting string of n 2's and 3's, using the rule described in the Comments lines.

Original entry on oeis.org

2, 11, 30, 82, 199, 480, 1097, 2630, 5828, 12830, 27873, 60071, 128355, 273543, 580149, 1226626, 2584822, 5433676, 11392986, 23838396, 49776503, 103755527, 215904926, 448602871, 930771041, 1928682932, 3991605129, 8251710234, 17040335019, 35154540729, 72456654860, 149208536983
Offset: 1

Views

Author

N. J. A. Sloane, May 31 2004

Keywords

Comments

Start with any initial string of n numbers s(1), ..., s(n), all = 2 or 3 (so there are 2^n starting strings). The rule for extending the string is this:
To get s(i+1), write the string s(1)s(2)...s(i) as xy^k for words x and y (where y has positive length) and k is maximized, i.e., k = the maximal number of repeating blocks at the end of the sequence so far (k is the curling number of s(1)s(2)...s(i)). Then s(i+1) = k if k >=2, but if k=1 you must stop (without writing down the 1).
a(n) = sum of final length of string, summed over all 2^n starting strings.
See A094004 for more terms. - N. J. A. Sloane, Dec 25 2012

Crossrefs

Formula

Equals A216813(n) + n*2^n. - N. J. A. Sloane, Sep 26 2012
A093369 is closely related.

Extensions

a(27)-a(31) from N. J. A. Sloane, Sep 19 2012

A217209 Irregular triangle read by rows: T(n,k) (n>=1, 1 <= k <= A217208(n)) = number of strings of n 2's and 3's having a tail of length k.

Original entry on oeis.org

2, 2, 1, 1, 4, 2, 2, 6, 5, 3, 1, 1, 12, 9, 6, 2, 3, 20, 18, 12, 6, 7, 0, 0, 0, 1, 40, 34, 25, 11, 14, 1, 0, 1, 2, 74, 71, 47, 24, 28, 1, 3, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1
Offset: 1

Views

Author

N. J. A. Sloane, Oct 01 2012

Keywords

Comments

See A217208 or A216730 for definition of tail.

Examples

			Rows 1 through 8 are:
2,
2, 1, 1,
4, 2, 2,
6, 5, 3, 1, 1,
12, 9, 6, 2, 3,
20, 18, 12, 6, 7, 0, 0, 0, 1,
40, 34, 25, 11, 14, 1, 0, 1, 2,
74, 71, 47, 24, 28, 1, 3, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1,
148, 139, 95, 48, 56, 6, 4, 3, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 3, 1, 1,
...
		

Crossrefs

Cf. A217208 (row lengths), A216813 (means), A122536 (first column), A217210 (second column), A216730, A094004, A090822.

A217208 a(n) = (conjectured) length of longest tail that can be generated by a starting string of 2's and 3's of length n before a 1 is reached, using the rule described in the Comments lines.

Original entry on oeis.org

0, 2, 2, 4, 4, 8, 8, 58, 59, 60, 112, 112, 112, 118, 118, 118, 118, 118, 119, 119, 119, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 131, 132, 132, 132, 132, 132, 132, 132, 132, 133, 173, 173, 173, 173
Offset: 1

Views

Author

N. J. A. Sloane, Sep 29 2012; revised Oct 02 2012

Keywords

Comments

Start with an initial string S of n numbers s(1), ..., s(n), all = 2 or 3. The rule for extending the string is this:
To get s(i+1), write the current string s(1)s(2)...s(i) as XY^k for words X and Y (where Y has positive length) and k is maximized, i.e. k = the maximal number of repeating blocks at the end of the sequence so far (k is the "curling number" of the string). Then set s(i+1) = k.
The "tail length" t(S) of S is defined as follows: start with S and repeatedly append the curling number (recomputing it at each step) until a 1 is reached; t(S) is the number of terms that are appended to S before a 1 is reached. If a 1 is never reached, set t(S)=oo .
The "Curling Number Conjecture" is that if one starts with any finite string and repeatedly extends it by appending the curling number k, then eventually one must reach a 1. This has not yet been proved.
The values shown for n >= 49 are only conjectures, because certain assumptions used to cut down the search have not yet been rigorously justified. However, we believe that ALL terms shown are correct.

Examples

			a(3) = 2, using the starting string 3,2,2, which extends to 3,2,2,2,3, of length 5.
a(4) = 4, using the starting string 2,3,2,3, which extends to 2,3,2,3,2,2,2,3 of length 8.
a(8) = 58: start = 23222323, end = 232223232223222322322232223232223222322322232223232223222322322332.
a(22) = 120: start = 2322322323222323223223: see A116909 for trajectory.
		

Crossrefs

a(n) = length of n-th row of A217209.
a(n) = A094004(n) - n.
Showing 1-3 of 3 results.