A049206 Maximum mean distance between cards during perfect faro shuffles, with cut, to return to original order in A024222.

0, 1, 1, 2, 3, 3, 3, 4, 5, 5, 5, 6, 7, 7, 7, 8, 9, 9, 9, 10, 11, 11, 11, 12, 13, 13, 13, 14, 15, 15, 15, 16, 17, 17, 17, 18, 19, 19, 19, 20, 21, 21, 21, 22, 23, 23, 23, 24, 25, 25, 25, 26, 27, 27, 27, 28, 29, 29, 29, 30, 31, 31, 31, 32, 33, 33, 33, 34, 35, 35, 35, 36, 37, 37, 37

Offset

1,4

Links

Table of n, a(n) for n=1..75.

Formula

Take difference between successive cards after each shuffle. Compute mean (if necessary, round to nearest integer). Retain until replaced by a higher mean in a succeeding shuffle.

(1/4) {2n + 2 - (-1)^[n/2] + (-1)^[(n-1)/2] }. - Ralf Stephan, Jun 10 2005

a(n)=A004525(n), n>1. [From R. J. Mathar, Oct 15 2008]

Example

Consider n=6. There are 4 shuffles to return to original order in a 6-card deck. The maximum mean distance between cards during these 4 shuffles and cuts, s1-s4, is 3, computed as follows: s1, 415263, cut, 263415; s2, 421653, cut 653421; s3, 462513, cut 513462; s4, 456123, cut, 123456. Mean distances: s1 15/5=3, maximum; s2 7/5=1.4; s3 13/5=2.6; s4 5/5; mean cumulative distance: 40/20=2.

Crossrefs

Cf. A024222, A024542.

Sequence in context: A287355 A194171 A004525 * A194247 A084767 A137580

Adjacent sequences: A049203 A049204 A049205 * A049207 A049208 A049209

Keyword

easy,nonn

Author

Enoch Haga

Status

approved