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A019567
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Order of the Mongean shuffle permutation of 2n cards: a(n) is least number m for which either 2^m + 1 or 2^m - 1 is divisible by 4n + 1.
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4
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1, 2, 3, 6, 4, 6, 10, 14, 5, 18, 10, 12, 21, 26, 9, 30, 6, 22, 9, 30, 27, 8, 11, 10, 24, 50, 12, 18, 14, 12, 55, 50, 7, 18, 34, 46, 14, 74, 24, 26, 33, 20, 78, 86, 29, 90, 18, 18, 48, 98, 33, 10, 45, 70, 15, 24, 60, 38, 29, 78, 12, 84, 41, 110, 8, 84, 26, 134, 12, 46, 35, 36, 68, 146
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OFFSET
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0,2
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COMMENTS
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Write down 1, then 2 to left, 3 to right, 4 to left, ..., getting [ 2n,2n-2,...,4,2,1,3,5,...,2n-1 ]; the sequence 2,3,6,4,6,10,14,5,18,10,12,21,26,9,... gives order of permutation sending 1 to 2n, 2 to 2n-2, ..., 2n to 2n-1.
Equivalently, the sequence 2,3,6,4,6,10,14,5,18,10,12,21,26,9,... gives the number of Mongean shuffles needed to return a deck of 2n cards (n=1,2,3,...) to its original order.
It appears that a(n) = order((-1)^(n+1)*2 in Z_{2n+1}) / f with f=1 when n==2 (mod 3) and for n = 0, 19, 21, 30,33, 52, 55, 61, 63, 70, ..., f=2 else. I don't know how to characterize the "exceptional" n's. - M. F. Hasler, Mar 31 2019
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REFERENCES
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A. P. Domoryad, Mathematical Games and Pastimes, Pergamon Press, 1964; see pp. 134-135.
W. W. Rouse Ball, Mathematical Recreations and Essays, 11th ed. 1939, p. 311
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LINKS
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FORMULA
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EXAMPLE
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Illustrating the initial terms:
n 4n+1 2^m+1 2^m-1 m
0 1 1 1
1 5 5 2
2 9 9 3
3 13 5*13 6
4 17 17 4
5 21 3*21 6
6 25 41*25 10
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MAPLE
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for m from 1 do
if modp(2^m-1, 4*n+1) =0 or modp(2^m+1, 4*n+1)=0 then
return m ;
end if;
end do;
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MATHEMATICA
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a[n_] := For[m=1, True, m++, If[AnyTrue[{-1, 1}, Divisible[2^m+#, 4n+1]&], Return[m]]];
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PROG
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(PARI) A019567(n, z=Mod(2, 4*n+1))=for(m=1, oo, bittest(5, lift(z^m+1))&&return(m)) \\ M. F. Hasler, Mar 31 2019
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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Comments corrected by Mikko Nieminen, Jul 26 2007, who also provided the Domoryad reference
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STATUS
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approved
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