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 1-4 of 4 results.

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

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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

A093369 a(n) = sum of lengths of strings that can be generated by any starting string of n 2's and 3's that starts with a 2, using the rule described in the Comments lines.

Original entry on oeis.org

1, 6, 14, 42, 98, 242, 552, 1394, 2935, 6471, 14006, 30060, 64223, 136914, 290224, 613509, 1292567, 2717311, 5696864, 11920124, 24889066, 51880008, 107954163, 224305440, 465388743, 964349526, 1995808823, 4125871527, 8520180124, 17577302639, 36228352911
Offset: 1

Views

Author

N. J. A. Sloane, Apr 28 2004

Keywords

Comments

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:
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. 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-1) starting strings.

Examples

			a(3) = 14: the starting string, final string and length are as follows:
222 2223 4
223 223 3
232 232 3
233 2332 4, for a total of 4+3+3+4 = 14.
		

Crossrefs

Extensions

a(21)-a(31) from Lars Blomberg, Jul 25 2017

A216813 Sum of tail length of S over all 2^n strings S consisting of n 2's and 3's.

Original entry on oeis.org

0, 3, 6, 18, 39, 96, 201, 582, 1220, 2590, 5345, 10919, 21859, 44167, 88629, 178050, 356598, 715084, 1431514, 2866876, 5736311, 11480839, 22966942, 45949687, 91910241, 183852468, 367726473, 735517466, 1471078571, 2942286009, 5884661772, 11769583511, 23539346216, 47079214312, 94158788295
Offset: 1

Views

Author

N. J. A. Sloane, Sep 18 2012 - Sep 21 2012, Oct 23 2012

Keywords

Comments

"Tail length" is defined in A216730.

Crossrefs

Formula

a(n) = A094005(n) - n*2^n.
Up to n=32, the average tail length a(n)/2^n seems to be approaching a number around 2.74.

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 1-4 of 4 results.