OFFSET 0,2
COMMENT In other words, least m 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) by 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 divisible by a(n). We conjecture that, conversely, if 2n is divisible by a(n) then 2n+1 is 1 or a prime. - Vladimir Shevelev (shevelev(AT)bgu.ac.il), Apr 29 2008
REFERENCES E. Bach and J. O. Shallit, Algorithmic Number Theory, I.
A. J. C. Cunningham, On Binal Fractions, Math. Gaz., 4 (1908), circa p. 266.
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.
S. W. Golomb, Permutations by cutting and shuffling, SIAM Rev., 3 (1961), 293-297.
V. I. Levenshtein, Conflict-avoiding codes and cyclic triple systems [in Russian], Problemy Peredachi Informatsii, 43 (No. 3, 2007), 39-53.
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
M. J. Gardner and C. A. McMahan, Riffling casino checks, Math. Mag., 50 (1977), 38-41.
LINKS T. D. Noe, Table of n, a(n) for n = 0..10000
Eric Weissteins World of Mathematics, Link to a section of The World of Mathematics.
Eric Weissteins World of Mathematics, Riffle Shuffle
Eric Weissteins World of Mathematics, In-Shuffle
Eric Weissteins World of Mathematics, Out-Shuffle
Eric Weissteins World of Mathematics, Multiplicative Order
MAPLE with(numtheory): f := n->order(2, 2*n+1);
PROGRAM (PARI) a(n)=if(n<0, 0, znorder(Mod(2, 2*n+1))) /* Michael Somos Mar 31 2005 */
CROSSREFS Cf. A070667-A070675, A070676, A053447, A070677, A070681, A070678, A053451, A070679, A070682, A070680, A070683.
Cf. A024222, A006694 (number of cyclotomic cosets), A014664 (order of 2 mod n-th prime).
Adjacent sequences: A002323 A002324 A002325 this_sequence A002327 A002328 A002329
Sequence in context: A083673 A131388 A131393 this_sequence A064273 A134561 A120947
KEYWORD nonn,easy,nice
AUTHOR njas
EXTENSIONS More terms from David W. Wilson (davidwwilson(AT)comcast.net), Jan 13, 2000.
More terms from Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 11 2003
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 (名单; 图表; 听) 换句话说,垂距0,2评论最少m这样2n+1划分2^m-1。 2n+2卡片riffle拖曳的数字要求的退回甲板到初始状态。 riffle拖曳被s (1), s ((i/2) +1), s (2), s替换一张名单s (1), s (2),…, s (m) ((i/2) +2),… a (1) = 2,因为riffle拖曳[1, 2, 3, 4]要求2叠代[1, 2, 3, 4] -> [1, 3, 2, 4] -> [1, 2, 3, 4]恢复原始的秩序。 关于计算这个序列的复杂,看见例如Bach和Shallit, p。 115,锻炼8。 证明是不难的,如果2n+1是最初然后2n由a (n)是可分的。 我们臆想那,相反地,如果2n由a (n)然后2n+1是可分的是1或最初。 - Vladimir Shevelev (shevelev (在) bgu.ac.il), 2008年4月29日 参考E。 Bach和J。 O. Shallit,算法数论, I。 A. J. C. Cunningham,在Binal分数,算术。 Gaz。, 4 (1908),大约p。 266. T. Folger, “拖曳入Hyperspace”,发现, 1991年(第12卷, no1),页66-67。 M. Gardner, “卡片拖曳”,数学狂欢节第10章,页123-138。 纽约: 葡萄酒Books 1977年。 S. W. Golomb,变更通过切开和拖曳,泰国3 (1961), 293-297 Rev.。 v. i. Levenshtein,相冲突避免代码和循环三倍系统[用俄语], Problemy Peredachi Informatsii, 43 (没有。 3, 2007), 39-53。 L. Lunelli和M。 Lunelli, Tavola di congruenza a^n每a=2,5,10, Atti Sem的== 1 mod K。 席子。 Fis。 大学. 摩德纳10 (1960/61), 219-236 (1961)。 J. H. Silverman,数论,第3编辑友好的介绍。, Pearson教育,公司2006年, p。 146,锻炼。 21.3 M. J. Gardner和C。 A. McMahan, Riffling赌博娱乐场检查,算术。 Mag., 50 (1977), 38-41。 链接T。 D. Noe,表n, a (n)为n = 0..10000 数学,数学世界的部分的链接埃里克Weissteins世界。 数学, Riffle拖曳埃里克Weissteins世界 数学埃里克Weissteins世界,在拖曳 数学埃里克Weissteins世界,拖曳 数学,乘命令埃里克Weissteins世界 槭树与(numtheory) : f := n->顺序(2, 2*n+1); 编程(PARI) a (n) =if (n< 0, 0, znorder (Mod (2, 2*n+1))) /*迈克尔Somos 2005年3月31日* CROSSREFS Cf。 A070667-A070675, A070676, A053447, A070677, A070681, A070678, A053451, A070679, A070682, A070680, A070683。 Cf. A024222, A006694 (割圆的cosets的数字), A014664 (2 mod第n最初命令)。 毗邻序列: A002323 A002324 A002325 this_sequence A002327 A002328 A002329 序列在上下文: A083673 A131388 A131393 this_sequence A064273 A134561 A120947 主题词nonn,容易,好 作者njas 引伸更多期限从大卫W。 威尔逊(davidwwilson (在) comcast.net), 2000年1月13日。 更多期限从Benoit Cloitre (benoit7848c (在) orange.fr), 2003年4月11日