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A002326 Multiplicative order of 2 mod 2n+1.
(Formerly M0936 N0350)
197
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
In other words, least m > 0 such that 2n+1 divides 2^m-1.
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^(2t-1))=4t. Thus if n=2^(2t-1), where, for any m > 0, t=2^(m-1) 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^3-1)/2) = p * a((p^2-1)/2). Because otherwise a((p^3-1)/2) < p * a((p^2-1)/2) iff a((p^3-1)/2) = a((p-1)/2) for a prime p. Equivalently p^3 divides 2^(p-1)-1, but no such prime p is known. - Thomas Ordowski, Feb 10 2014
A generalization of the previous conjecture: For each k>=2, if p is an odd prime then a(((p^(k+1))-1)/2) = p * a((p^k-1)/2). Computer testing of this generalized conjecture shows that there is no counterexample for k and p both up to 1000. - Ahmad J. Masad, Oct 17 2020
a(n) = a((N-1)/2), with odd N = 2*n+1 >= 3 (n >= 1), is also the primitive period length of (1/N) in binary notation: (1/N)_2 = 0.repeat(a[1]a[2]...a[P(N)]), and P(N) = a((N-1)/2). E.g., N = 11 (n = 5), (1/11)_2 = 0.repeat(0001011101), with P(11) = 10 = a(5). Proof: Use a cyclic shift operation sigma (1 step to the left) on the cycle: sigma((1/N)_2) = .repeat(a[2]...a[P(N)]a[1]). Then one can prove for the composition sigma^[k] (k=0 is the identity map) written back in decimal notation the result (sigma^[k]((1/N)_2))_10 = (1/N)*2^k (mod N). E.g. N = 11, sigma^[2]((1/11)_2) = .repeat(0101110100), written in base 10 as 4/11, etc. Hence P(N) and the order of 2 modulo N coincide. - Gary W. Adamson and Wolfdieter Lang, Oct 14 2020
REFERENCES
E. Bach and Jeffrey Shallit, Algorithmic Number Theory, I.
T. Folger, "Shuffling Into Hyperspace," Discover, 1991 (vol 12, no 1), pages 66-67.
M. Gardner, "Card Shuffles," Mathematical Carnival chapter 10, pages 123-138. New York: Vintage Books, 1977.
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), 219-236 (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
M. Baake, U. Grimm and J. Nilsson, Scaling of the Thue-Morse diffraction measure, arXiv preprint arXiv:1311.4371 [math-ph], 2013.
D. Bayer and P. Diaconis, Trailing the dovetail shuffle to its lair, Ann. Appl. Prob. 2 (2) (1992) 294-313.
Matthew Brand, Choosing 1 of N with and without lucky numbers, arXiv:1808.07994 [math.NT], 2018.
J. Brillhart, J. S. Lomont and P. Morton, Cyclotomic properties of the Rudin-Shapiro polynomials, J. Reine Angew. Math.288 (1976), 37--65. 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): 542-556.
A. J. C. Cunningham, On Binal Fractions, Math. Gaz., 4 (71) (1908), circa p. 266.
P. Diaconis, R. L. Graham, and W. M. Kantor, The mathematics of perfect shuffles, Adv. Appl. Math. 4 (2) (1983) 175-196
M. J. Gardner and C. A. McMahan, Riffling casino checks, Math. Mag., 50 (1) (1977), 38-41.
S. W. Golomb, Permutations by cutting and shuffling, SIAM Rev., 3 (1961), 293-297.
Jonas Kaiser, On the relationship between the Collatz conjecture and Mersenne prime numbers, arXiv preprint arXiv:1608.00862 [math.GM], 2016.
Torleiv Klove, On covering sets for limited-magnitude errors, Cryptogr. Commun. 8 (3) (2016) 415-433
V. I. Levenshtein, Conflict-avoiding codes and cyclic triple systems [in Russian], Problemy Peredachi Informatsii, 43 (No. 3, 2007), 39-53.
V. I. Levenshtein, Conflict-avoiding codes and cyclic triple systems, Problems of Information Transmission, September 2007, Volume 43, Issue 3, pp 199-212 (translated from Russian)
Yuan-Hsun Lo, Kenneth W. Shum, Wing Shing Wong and Yijin Zhang, Multichannel Conflict-Avoiding Codes of Weights Three and Four, arXiv:2009.11754 [cs.IT], 2020.
Jarkko Peltomäki and Aleksi Saarela, Standard words and solutions of the word equation X_1^2 ... X_n^2 = (X_1 ... X_n)^2, Journal of Combinatorial Theory, Series A (2021) Vol. 178, 105340. See also arXiv:2004.14657 [cs.FL], 2020.
Vladimir Shevelev, Gilberto Garcia-Pulgarin, Juan Miguel Velasquez-Soto and John H. Castillo, Overpseudoprimes, and Mersenne and Fermat numbers as primover numbers, arXiv preprint arXiv:1206:0606 [math.NT], 2012.
Vladimir Shevelev, G. Garcia-Pulgarin, 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, In-Shuffle
Eric Weisstein's World of Mathematics, Out-Shuffle
Eric Weisstein's World of Mathematics, Multiplicative Order
Wikipedia, Riffle Shuffle
FORMULA
a((3^n-1)/2) = A025192(n). - Vladimir Shevelev, May 09 2008
Bisection of A007733: a(n) = A007733(2*n+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^n-1) = 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*n-1]. It is easy to prove that the first 1 appears not later than in the n-th 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)
(GAP) List([0..100], n->OrderMod(2, 2*n+1)); # Muniru A Asiru, Feb 01 2019
(Python)
from sympy import n_order
[n_order(2, 2*n+1) for n in range(73)] # Hermann Stamm-Wilbrandt, Jul 27 2021
CROSSREFS
Cf. A024222, A006694 (number of cyclotomic cosets).
Cf. A014664 (order of 2 mod n-th prime).
Cf. A001122 (primes for which 2 is a primitive root).
Cf. A216838 (primes for which 2 is not a primitive root).
Bisections give A274298, A274299.
Partial sums: A359147.
Sequence in context: A131388 A131393 A216476 * A285493 A064273 A257986
KEYWORD
nonn,easy,nice
AUTHOR
EXTENSIONS
More terms from David W. Wilson, Jan 13 2000
More terms from Benoit Cloitre, Apr 11 2003
STATUS
approved

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