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.

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A076876 Meandric numbers for a river crossing two parallel roads at n points.

Original entry on oeis.org

1, 1, 2, 3, 8, 14, 43, 81, 272, 538, 1920, 3926, 14649, 30694, 118489, 252939, 1002994, 2172830, 8805410, 19304190, 79648888, 176343390, 738665040, 1649008456, 6996865599, 15730575554, 67491558466, 152663683494, 661370687363, 1503962954930, 6571177867129
Offset: 0

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Author

N. J. A. Sloane and Jon Wild, Nov 26 2002

Keywords

Comments

a(n) = number of ways that a curve can start in the (-,-) quadrant, cross two parallel lines and end up in the (+,+) or (+,-) quadrant if n is even or head East between the two roads if n is odd.
A107321 is a lower bound. - R. J. Mathar, May 06 2006
It appears that for odd n, A076876(n) = A005316(n+1). And for even n, A076876(n) >= A005316(n+1). - Robert Price, Jul 27 2013.

Examples

			Let b(n) = A005316(n). Then a(0) = b(0), a(1) = b(1), a(2) = b(1) + b(2), a(3) = b(3) + b(2), a(4) = b(4) + 2*b(3) + 1, a(5) = b(5) + b(3)*b(2) + b(4) + 1.
Consider n=5: if we do not cross the second road there are b(5) = 8 solutions. If we cross the first road 3 times and then the second road twice there are b(3)*b(2) = 2 solutions. If we cross the first road once and the second road 4 times there are b(4) = 3 solutions. The only other possibility is to cross road 1, road 2 twice, road 1 twice and exit to the right.
For larger n it is convenient to give the vector of the number of times the same road is crossed. For example for n=6 the vectors and the numbers of possibilities are as follows:
[6] ...... 14
[5 1] ..... 8
[3 3] ..... 4
[3 2 1] ... 2
[1 5] ..... 8
[1 4 1] ... 3
[1 2 3] ... 2
[1 2 2 1] . 2
Total .... 43
		

Crossrefs

Extensions

More terms from R. J. Mathar, Mar 04 2007
a(12)-a(20) from Robert Price, Apr 15 2012
a(21)-a(40) from Andrew Howroyd, Dec 07 2015
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