

A002326


Multiplicative order of 2 mod 2n+1.
(Formerly M0936 N0350)


153



1, 2, 4, 3, 6, 10, 12, 4, 8, 18, 6, 11, 20, 18, 28, 5, 10, 12, 36, 12, 20, 14, 12, 23, 21, 8, 52, 20, 18, 58, 60, 6, 12, 66, 22, 35, 9, 20, 30, 39, 54, 82, 8, 28, 11, 12, 10, 36, 48, 30, 100, 51, 12, 106, 36, 36, 28, 44, 12, 24, 110, 20, 100, 7, 14, 130, 18, 36, 68, 138, 46, 60, 28
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OFFSET

0,2


COMMENTS

In other words, least m > 0 such that 2n+1 divides 2^m1.
Number of riffle shuffles of 2n+2 cards required to return a deck to initial state. A riffle shuffle replaces a list s(1), s(2), ..., s(m) with s(1), s((i/2)+1), s(2), s((i/2)+2), ... a(1) = 2 because a riffle shuffle of [1, 2, 3, 4] requires 2 iterations [1, 2, 3, 4] > [1, 3, 2, 4] > [1, 2, 3, 4] to restore the original order.
Concerning the complexity of computing this sequence, see for example Bach and Shallit, p. 115, exercise 8.
It is not difficult to prove that if 2n+1 is a prime then 2n is a multiple of a(n). But the converse is not true. Indeed, one can prove that a(2^(2t1))=4t. Thus if n=2^(2t1), where, for any m > 0, t=2^(m1) then 2n is a multiple of a(n) while 2n+1 is a Fermat number which, as is well known, is not always a prime. It is an interesting problem to describe all composite numbers for which 2n is divisible by a(n).  Vladimir Shevelev, May 09 2008
For an algorithm of calculation of a(n) see author's comment in A179680.  Vladimir Shevelev, Jul 21 2010
From V. Raman, Sep 18 2012, Dec 10 2012: (Start)
If 2n+1 is prime, then the polynomial (x^(2n+1)+1)/(x+1) factors into 2n/a(n) polynomials of the same degree a(n) over GF(2).
If (x^(2n+1)+1)/(x+1) is irreducible over GF(2), then 2n+1 is prime, and 2 is a primitive root (mod 2n+1) (cf. A001122).
For all n > 0, a(n) is the degree of the largest irreducible polynomial factor for the polynomial (x^(2n+1)+1)/(x+1) over GF(2). (End)
a(n) is a factor of phi(2n+1) (A000010(2n+1)).  Douglas Boffey, Oct 21 2013
Conjecture: if p is an odd prime then a((p^31)/2) = p * a((p^21)/2). Because otherwise a((p^31)/2) < p * a((p^21)/2) iff a((p^31)/2) = a((p1)/2) for a prime p. Equivalently p^3 divides 2^(p1)1, but no such prime p is known.  Thomas Ordowski, Feb 10 2014


REFERENCES

E. Bach and Jeffrey Shallit, Algorithmic Number Theory, I.
T. Folger, "Shuffling Into Hyperspace," Discover, 1991 (vol 12, no 1), pages 6667.
M. Gardner, "Card Shuffles," Mathematical Carnival chapter 10, pages 123138. New York: Vintage Books, 1977.
V. I. Levenshtein, Conflictavoiding codes and cyclic triple systems [in Russian], Problemy Peredachi Informatsii, 43 (No. 3, 2007), 3953.
L. Lunelli and M. Lunelli, Tavola di congruenza a^n == 1 mod K per a=2,5,10, Atti Sem. Mat. Fis. Univ. Modena 10 (1960/61), 219236 (1961).
J. H. Silverman, A Friendly Introduction to Number Theory, 3rd ed., Pearson Education, Inc, 2006, p. 146, Exer. 21.3
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

T. D. Noe, Table of n, a(n) for n = 0..10000
M. Baake, U. Grimm, J. Nilsson, Scaling of the ThueMorse diffraction measure, arXiv preprint arXiv:1311.4371 [mathph], 2013.
D. Bayer, P. Diaconis, Trailing the dovetail shuffle to its lair, Ann. Appl. Prob. 2 (2) (1992) 294313
Brillhart, John; Lomont, J. S.; Morton, Patrick. Cyclotomic properties of the RudinShapiro polynomials, J. Reine Angew. Math.288 (1976), 3765. See Table 2. MR0498479 (58 #16589).
Steve Butler, Persi Diaconis and R. L. Graham, The mathematics of the flip and horseshoe shuffles, arXiv:1412.8533 [math.CO], 2014.
Steve Butler, Persi Diaconis and R. L. Graham, The mathematics of the flip and horseshoe shuffles, The American Mathematical Monthly 123.6 (2016): 542556.
A. J. C. Cunningham, On Binal Fractions, Math. Gaz., 4 (71) (1908), circa p. 266.
P. Diaconis, R. L. Graham, W. M. Kantor, The mathematics of perfect shuffles, Adv. Appl. Math. 4 (2) (1983) 175196
M. J. Gardner and C. A. McMahan, Riffling casino checks, Math. Mag., 50 (1) (1977), 3841.
S. W. Golomb, Permutations by cutting and shuffling, SIAM Rev., 3 (1961), 293297.
Jonas Kaiser, On the relationship between the Collatz conjecture and Mersenne prime numbers, arXiv preprint arXiv:1608.00862 [math.GM], 2016.
V. I. Levenshtein, Conflictavoiding codes and cyclic triple systems, Problems of Information Transmission, September 2007, Volume 43, Issue 3, pp 199212 (translated from Russian)
Vladimir Shevelev, Gilberto GarciaPulgarin, Juan Miguel VelasquezSoto and John H. Castillo, Overpseudoprimes, and Mersenne and Fermat numbers as primover numbers, arXiv preprint arXiv:1206:0606 [math.NT], 2012.
V. Shevelev, G. GarciaPulgarin, J. M. Velasquez and J. H. Castillo, Overpseudoprimes, and Mersenne and Fermat Numbers as Primover Numbers, J. Integer Seq. 15 (2012) Article 12.7.7
Eric Weisstein's World of Mathematics, Riffle Shuffle
Eric Weisstein's World of Mathematics, InShuffle
Eric Weisstein's World of Mathematics, OutShuffle
Eric Weisstein's World of Mathematics, Multiplicative Order
Wikipedia, Riffle Shuffle


FORMULA

a((3^n1)/2) = A025192(n).  Vladimir Shevelev, May 09 2008
Bisection of A007733: a(n) = A007733(2n+1).  Max Alekseyev, Jun 11 2009
a((b(n)1)/2) = n for odd n and even n such that b(n/2) != b(n), where b(n) = A005420(n).  Thomas Ordowski, Jan 11 2014
Note that a(2^n1) = n+1 and a(2^n) = 2(n+1).  Thomas Ordowski, Jan 16 2014
a(n) = A056239(A292239(n)) = A048675(A292265(n)).  Antti Karttunen, Oct 04 2017


EXAMPLE

From Vladimir Shevelev, Oct 03 2017: (Start)
Our algorithm for the calculation of a(n) in the author's comment in A179680 (see also the Sage program below) could be represented in the form of a "finite continued fraction". For example let n = 8, 2*n+1 = 17. We have
1 + 17
 + 17
2
 + 17
2
 + 17
2
 = 1
32
Here the denominators are the A006519 of the numerators: A006519(1+17) = 2, A006519(9+17) = 2, A006519(13+17) = 2, A006519(15+17) = 32. Summing the exponents of these powers of 2, we obtain the required result: a(8) = 1 + 1 + 1 + 5 = 8. Indeed, we have (((1*32  17)*2  17)*2  17)*2  17 = 1. So 32*2*2*2  1 == 0 (mod 17), 2^8  1 == 0 (mod 17). In the general case, note that all "partial fractions" (which indeed are integers) are odd residues modulo 2*n+1 in the interval [1, 2*n1]. It is easy to prove that the first 1 appears not later than in the nth step. (End)


MAPLE

a := n > `if`(n=0, 1, numtheory:order(2, 2*n+1)):
seq(a(n), n=0..72);


MATHEMATICA

Table[MultiplicativeOrder[2, 2*n + 1], {n, 0, 100}] (* Robert G. Wilson v, Apr 05 2011 *)


PROG

(PARI) a(n)=if(n<0, 0, znorder(Mod(2, 2*n+1))) /* Michael Somos, Mar 31 2005 */
(MAGMA) [ 1 ] cat [ Modorder(2, 2*n+1): n in [1..72] ]; // Klaus Brockhaus, Dec 03 2008
(Haskell)
import Data.List (findIndex)
import Data.Maybe (fromJust)
a002326 n = (+ 1) $ fromJust $
findIndex ((== 0) . (`mod` (2 * n + 1))) $ tail a000225_list
 Reinhard Zumkeller, Apr 22 2013
(Sage)
# From Peter Luschny, Oct 06 2017: (Start)
[Mod(2, n).multiplicative_order() for n in (0..145) if gcd(n, 2) == 1]
# Algorithm from Vladimir Shevelev as described in A179680 and presented in Example.
def A002326VS(n):
s, m, N = 0, 1, 2*n + 1
while True:
k = N + m
v = valuation(k, 2)
s += v
m = k >> v
if m == 1: break
return s
[A002326VS(n) for n in (0..72)] # (End)


CROSSREFS

Cf. A003571, A003573, A217469, A070667A070683, A053447, A053451, A292239, A292265.
Cf. A024222, A006694 (number of cyclotomic cosets).
Cf. A014664 (order of 2 mod nth prime).
Cf. A001122 (primes for which 2 is a primitive root).
Cf. A216838 (primes for which 2 is not a primitive root).
Cf. A000225.
Bisections give A274298, A274299.
Sequence in context: A131388 A131393 A216476 * A285493 A064273 A257986
Adjacent sequences: A002323 A002324 A002325 * A002327 A002328 A002329


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane


EXTENSIONS

More terms from David W. Wilson, Jan 13 2000
More terms from Benoit Cloitre, Apr 11 2003


STATUS

approved



