

A049206


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


1



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
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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)^[(n1)/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 6card deck. The maximum mean distance between cards during these 4 shuffles and cuts, s1s4, 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



