A373416 -id:A373416 - cp's OEIS frontend

A323712 Number of the following described shuffles required to return a deck of n cards to its original state. Create two piles by alternating a card from the top of the deck left-right-left-right until the deck is exhausted. Then, placing the left pile on top of the right pile constitutes one shuffle.

1, 1, 3, 4, 4, 6, 6, 3, 9, 5, 5, 12, 12, 4, 12, 8, 8, 9, 9, 6, 6, 22, 22, 20, 20, 9, 27, 28, 28, 10, 10, 5, 15, 12, 12, 36, 36, 12, 12, 20, 20, 7, 7, 12, 36, 46, 46, 42, 42, 8, 24, 52, 52, 20, 20, 9, 9, 29, 29, 60, 60, 6, 18, 12, 12, 33, 33, 22, 66, 70, 70, 18

Offset: 1

David Lovler, Jan 24 2019

nonn

The card shuffling procedure is the same for even n and odd n.
Here are a few conjectures.
a(n) <= n for all n.
a(p)=a(p-1) and a(p)|p-1 when p is a prime >= 5.
a(n)=a(n-1) and a(n)|n-1 for nonprimes 341=31*11 and 22369621=8191*2731 and probably other pseudoprimes of the form p*((p+2)/3) where p is a Mersenne prime and (p+2)/3 is prime.
n cards are returned to their original state after n shuffles when n=1, 3, 4, 6, 9, 12, 22, 27, 28, 36, 46, 52, 60, 70, 78, 81, ... (A373461) . These values of n are either of the form p-1 where p is an odd prime number or 3^i, i >= 0.
a(c) is relatively small (compared with nearby values) when c is a Catalan number.
a(2n+1)=3*a(2n) or a(2n+1)=a(2n) for all n.

			For n=4, {a1,a2,a3,a4}-->{a3,a1,a4,a2}-->{a4,a3,a2,a1}-->{a2,a4,a1,a3}-->{a1,a2,a3,a4}, so a(4)=4.
For n=5, {a1,a2,a3,a4,a5}-->{a5,a3,a1,a4,a2}-->{a2,a1,a5,a4,a3}-->{a3,a5,a2,a4,a1}-->{a1,a2,a3,a4,a5}, so a(5)=4.

Alois P. Heinz, Table of n, a(n) for n = 1..1024

Cf. A024222, A022998, A163776, A373416 (fixed points).

pileShuf := proc(L::list)
    local i,n,shf ;
    shf := [] ;
    n := nops(L) ;
    if type(n,'odd') then
        for i from n to 1 by -2 do
            shf := [op(shf),op(i,L)] ;
        end do:
        for i from n-1 to 2 by -2 do
            shf := [op(shf),op(i,L)] ;
        end do:
    else
        for i from n-1 to 1 by -2 do
            shf := [op(shf),op(i,L)] ;
        end do:
        for i from n to 2 by -2 do
            shf := [op(shf),op(i,L)] ;
        end do:
    end if;
    shf ;
end proc:
A323712 := proc(n)
    local L,itr,isord,i;
    L := [seq(i,i=1..n)] ;
    for itr from 1 to n! do
        L := pileShuf(L) ;
        isord := true ;
        for i from 1 to nops(L) do
            if op(i,L) <> i then
                isord := false ;
                break ;
            end if;
        end do:
        if isord then
            return itr ;
        end if;
    end do:
    -1 ;
end proc:
seq(A323712(n),n=1..50) ; # R. J. Mathar, Aug 02 2024

perm(n, vn) = {my(va = List(), vb = List()); for (k=1, n, if (k % 2, listput(va, vn[k]), listput(vb, vn[k]));); Vec(concat(Vecrev(va), Vecrev(vb)));}
a(n) = {my(vn = vector(n,k,k), vs = perm(n, vn), nb = 1); while (vs != vn, vs = perm(n, vs); nb++); nb;} \\ Michel Marcus, Feb 06 2019

a(2^m) = m if m is odd, a(2^m) = 2m if m is even. - Alois P. Heinz, Feb 15 2019

A263458 Deal a pack of n cards into two piles and gather them up, n/2 times. All n such that this reverses the order of the deck.

Original entry on oeis.org

4, 6, 12, 22, 28, 30, 36, 46, 52, 60, 70, 78, 100, 102, 108, 126, 148, 150, 156, 166, 172, 180, 190, 196, 198, 222, 228, 238, 262, 268, 270, 276, 292, 310, 316, 348, 358, 366, 372, 382, 388, 396, 420, 430, 438, 460, 462, 478, 486, 502, 508, 540, 556, 598

Offset: 1

Christian Perfect, Oct 19 2015

nonn

This seems to be A003628(n)-1; that is, each element of this sequence is one less than a prime congruent to 5 or 7 modulo 8.

			Take a deck of 52 playing cards. Deal it into two piles, then pick up the first pile and put it on top of the other. Do this 26 times. The order of the deck is reversed, so 52 belongs to this sequence.
6 is in the sequence because the 3 shuffles are [1, 2, 3, 4, 5, 6] -> [5, 3, 1, 6, 4, 2] -> [4, 1, 5, 2, 6, 3] -> [6, 5, 4, 3, 2, 1], original reversed. 8 is not in the sequence because the 4 shuffles are [1, 2, 3, 4, 5, 6, 7, 8] -> [7, 5, 3, 1, 8, 6, 4, 2] -> [4, 8, 3, 7, 2, 6, 1, 5] -> [1, 2, 3, 4, 5, 6, 7, 8] -> [7, 5, 3, 1, 8, 6, 4, 2], not the original reversed. - _R. J. Mathar_, Aug 02 2024

Cf. A003628, A373416.

isA263458 := proc(n)
    local L,itr ;
    L := [seq(i,i=1..n)] ;
    for itr from 1 to n/2 do
        L := pileShuf(L) ; # function code in A323712
    end do:
    for i from 1 to nops(L) do
        if op(-i,L) <> i then
            return false ;
        end if;
    end do:
    true ;
end proc:
n := 1;
for k from 2 do
    if isA263458(k) then
        printf("%d %d\n",n,k) ;
        n := n+1 ;
    end if;
end do: # R. J. Mathar, Aug 02 2024

from itertools import cycle
def into_piles(r,deck):
    packs = [[] for i in range(r)]
    for card, pack in zip(range(1,deck+1),cycle(range(r))):
        packs[pack].insert(0,card)
    out = sum(packs,[])
    return Permutation(out)
def has_reversing_property(deck):
    p = power(into_piles(2,deck), deck/2)
    return p==into_piles(1,deck)
[i for i in range(2,400,2) if has_reversing_property(i)]

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