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 1-5 of 5 results.

A093371 Start with any initial string of n numbers s(1), ..., s(n), with s(1) = 2, other s(i)'s = 2 or 3 (so there are 2^(n-1) starting strings). The rule for extending the string is this as follows: To get s(n+1), write the string s(1)s(2)...s(n) 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. Then a(n) = number of starting strings for which k = 1.

Original entry on oeis.org

1, 1, 2, 3, 6, 10, 20, 37, 74, 143, 286, 562, 1124, 2230, 4460, 8884, 17768, 35465, 70930, 141720, 283440, 566600, 1133200, 2265843, 4531686, 9062261, 18124522, 36246826, 72493652, 144982872
Offset: 1

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Author

N. J. A. Sloane, Apr 28 2004

Keywords

Comments

See A122536 for many more terms. - N. J. A. Sloane, Oct 25 2012

Crossrefs

Equals A122536/2. - N. J. A. Sloane, Sep 25 2012
Different from, but easily confused with, A007148 and A045690.

Formula

a(n) = 2^(n-1) - A093370(n).

Extensions

More terms from N. J. A. Sloane, Sep 26 2012

A094004 a(n) = (conjectured) length of longest string that can be generated by a starting string of 2's and 3's of length n, using the rule described in the Comments lines.

Original entry on oeis.org

1, 4, 5, 8, 9, 14, 15, 66, 68, 70, 123, 124, 125, 132, 133, 134, 135, 136, 138, 139, 140, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 200, 201, 202, 203, 204, 205, 206, 207, 209, 250, 251, 252, 253
Offset: 1

Views

Author

N. J. A. Sloane, May 31 2004. Revised by N. J. A. Sloane, Sep 17 2012

Keywords

Comments

Start with an initial string 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 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 if k >=2, but if k=1 you must stop (without writing down the 1).
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. - N. J. A. Sloane, Sep 17 2012

Examples

			a(3) = 5, using the starting string 3,2,2, which extends to 3,2,2,2,3, of length 5.
a(4) = 8, using the starting string 2,3,2,3, which extends to 2,3,2,3,2,2,2,3 of length 8.
a(8) = 66: start = 23222323, end = 232223232223222322322232223232223222322322232223232223222322322332.
a(22) = 142: start = 2322322323222323223223: see A116909 for trajectory.
		

Crossrefs

Extensions

a(27)-a(30) from Allan Wilks, Jul 29 2004
a(31)-a(36) from Benjamin Chaffin, Apr 09 2008
a(37)-a(44) (computed in 2008) from Benjamin Chaffin, Dec 04 2009
a(45)-a(48) from Benjamin Chaffin, Dec 18 2009
a(49)-a(50) from Benjamin Chaffin, Dec 26 2009
a(51)-a(52) from Benjamin Chaffin, Jan 10 2010
a(53)-a(80) from Benjamin Chaffin, Jan 10 2012

A093370 Start with any initial string of n numbers s(1), ..., s(n), with s(1) = 2, other s(i)'s = 2 or 3 (so there are 2^(n-1) starting strings). The rule for extending the string is this as follows: To get s(n+1), write the string s(1)s(2)...s(n) 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. Then a(n) = number of starting strings for which k > 1.

Original entry on oeis.org

0, 1, 2, 5, 10, 22, 44, 91, 182, 369, 738, 1486, 2972, 5962, 11924, 23884, 47768, 95607, 191214, 382568, 765136, 1530552, 3061104, 6122765, 12245530, 24492171, 48984342, 97970902, 195941804, 391888040
Offset: 1

Views

Author

N. J. A. Sloane, Apr 28 2004

Keywords

Examples

			For n=2 there are 2 starting strings, 22 and 23 and only the first has k > 1.
For n=4 there are 8 starting strings, but only 5 have k > 1, namely 2222, 2233, 2322, 2323, 2333.
		

Crossrefs

Formula

Equals A121880(n)/2, or 2^(n-1) - A122536(n)/2.
a(n)/2^(n-1) seems to converge to a number around 0.73.

Extensions

More terms from Sarah Nibs, via A122536, Sep 18 2006

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

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 1-5 of 5 results.