

A003558


Least number m > 0 such that 2^m == + 1 (mod 2n + 1).


32



1, 1, 2, 3, 3, 5, 6, 4, 4, 9, 6, 11, 10, 9, 14, 5, 5, 12, 18, 12, 10, 7, 12, 23, 21, 8, 26, 20, 9, 29, 30, 6, 6, 33, 22, 35, 9, 20, 30, 39, 27, 41, 8, 28, 11, 12, 10, 36, 24, 15, 50, 51, 12, 53, 18, 36, 14, 44, 12, 24, 55, 20, 50, 7, 7, 65, 18, 36, 34, 69, 46
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OFFSET

0,3


COMMENTS

Multiplicative suborder of 2 (mod 2n+1) (or sord(2, 2n+1)).
This is called quasiorder in the Hilton/Pederson reference.
For the complexity of computing this, see A002326.
Also, the order of the socalled "milk shuffle" of a deck of n cards, which maps cards (1,2,...,n) to (1,n,2,n1,3,n2,...). See the paper of Lévy.  Jeffrey Shallit, Jun 09 2019
It appears that under iteration of the basen Kaprekar map, for even n > 2 (A165012, A165051, A165090, A151949 in bases 4, 6, 8, 10), almost all cycles are of length a(n/2  1); proved under the additional constraint that the cycle contains at least one element satisfying "number of digits (n1)  number of digits 0 = o(total number of digits)".  Joseph Myers, Sep 05 2009
From Gary W. Adamson, Sep 20 2011: (Start)
a(n) can be determined by the cycle lengths of iterates using x^2  2, seed 2*cos(2*Pi/N); as shown in the A065941 comment of Sep 06 2011. The iterative map of the logistic equation 4x*(1x) is likewise chaotic with the same cycle lengths but initiating the trajectory with sin^2 2*Pi/N, N = 2n+1 [Kappraff & Adamson, 2004]. Chaotic terms with the identical cycle lengths can be obtained by applying Newton's method to i = sqrt(1) [Strang, also Kappraff and Adamson, 2003], resulting in the morphism for the cot 2Pi/N trajectory: (x^21)/2x. (End)
From Gary W. Adamson, Sep 11 2019: (Start)
Using x^2  2 with seed 2*cos(Pi/7), we obtain the periodthree trajectory 1.8019377...> 1.24697...> .445041... For an odd prime N, the trajectory terms represent diagonal lengths of regular star 2Ngons, with edge the shortest value (.445...in this case.) (Cf. "Polygons and Chaos", p. 9 Fig 4.) We can normalize such lengths by dividing through with the lowest value, giving 3 diagonals of the 14gon: (1, 2.801937..., 4.048917...). Label the terms ranked in magnitude with odd integers (1, 3, 5), and we find that the diagonal lengths are in agreement with the diagonal formula (sin(j*Pi)/14)/(sin(Pi/14)), with j = (1,3,5). (End)
Roots of signed nth row A054142 polynomials are chaotic with respect to the operation (2, x^2), with cycle lengths a(n). Example: starting with a root to x^3  5x^2 + 6x  1 = 0; (2 + 2*cos(2*Pi/N) = 3.24697...); we obtain the trajectory (3.24697...> 1.55495...> .198062...); the roots to the polynomial with cycle length 3 matching a(3) = 3.  Gary W. Adamson, Sep 21 2011
Also a(n1) = card {cos((2^k)*Pi/(2*n1)): k in N} for n >= 1 (see A216066, an essentially identical sequence, for more information).  Roman Witula, Sep 01 2012
From Juhani Heino, Oct 26 2015: (Start)
Start a sequence with numbers 1 and n. For next numbers, add previous numbers going backwards until the sum is even. Then the new number is sum/2. I conjecture that the sequence returns to 1,n and a(n) is the cycle length.
For example: 1,7,4,2,1,7,... so a(7) = 4.
1,6,3,5,4,2,1,6,... so a(6) = 6. (End)
From Juhani Heino, Nov 06 2015: (Start)
Proof of the above conjecture: Let n = 1/2; thus 2n + 1 = 0, so operations are performed mod (2n + 1). When the member is even, it is divided by 2. When it is odd, multiply by n, so effectively divide by 2. This is all welldefined in the sense that new members m are 1 <= m <= n. Now see what happens starting from an odd member m. The next member is m/2. As long as there are even members, divide by 2 and end up with an odd m/(2^k). Now add all the members starting with m. The sum is m/(2^k). It's divided by 2, so the next member is m/(2^(k+1)). That is the same as (m/(2^k))/(2), as with the definition.
So actually start from 1 and always divide by 2, although the sign sometimes changes. Eventually 1 is reached again. The chain can be traversed backwards and then 2^(cycle length) == + 1 (mod 2n + 1).
To conclude, we take care of a(0): sequence 1,0 continues with zeros and never returns to 1. So let us declare that cycle length 0 means unavailable. (End)
From Gary W. Adamson, Aug 20 2019: (Start)
Terms in the sequence can be obtained by applying the doubling sequence mod (2n + 1), then counting the numbers of terms until the next term is == +1 (mod 2n + 1). Example: given 25, the trajectory is (1, 2, 4, 8, 16, 7, 14, 3, 6, 12).
The cycle ends since the next term is 24 == 1 (mod 25) and has a period of 10. (End)
From Gary W. Adamson, Sep 04 2019: (Start)
Conjecture of Kappraff and Adamson in "Polygons and Chaos", p. 13 Section 7, "Chaos and Number": Given the cycle length for N = 2n + 1, the same cycle length is present in bases 4, 9, 16, 25, ..., m^2, for the expansion of 1/N.
Examples: The cycle length for 7 is 3, likewise for 1/7 in base 4: .021021021.... In base 9 the expansion of 1/7 is .125125125... Check: The first few terms are 1/9 + 2/81 + 5/729 = 104/279 = .1426611... (close to 1/7 = .142857). (End)
From Gary W. Adamson, Sep 24 2019: (Start) An exception to the rule
for 1/N in bases m^2: (when N divides m^2 as in 1/7 in base 49, = 7/49, rational). When all terms in the cycle are the same, the identity reduces to
1/N in (some bases) = .a, a, a, .... The minimal values of "a" for 1/N are provided as examples, with the generalization 1/N in base (N1)^2 = .a, a, a,... for N odd:
1/3 in base 4 = .1, 1, 1,...
1/5 in base 16 = .3, 3, 3,...
1/7 in base 36 = .5, 5, 5,...
1/9 in base 64 = .7, 7, 7,...
1/11 in base 100 = .9, 9, 9,... (Check: the first three terms are 9/100 +
9/(100^2) + 9/(100^3) = .090909 where 1/11 = .09090909...). (End)
For N = 2n+1, the corresponding entry is equal to the degree of the polynomial for N shown in (Lang, Table 2, p. 46). As shown, x^3  3x  1 is the minimal polynomial for N = 9, with roots (1.87938..., 1.53208..., 0.347296...); matching the (abs) values of the 2*cos(Pi/9) trajectory using x^2  2. Thus, a(4) = 3. If N is prime, the polynomials shown in Table 2 are the same as those for the same N in A065941. If different, the minimal polynomials shown in Table 2 are factors of those in A065941.  Gary W. Adamson, Oct 01 2019


REFERENCES

Peter Hilton and Jean Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, 2010, pp. 261264.
Carl Schick, Trigonometrie und unterhaltsame Zahlentheorie, Zurich, 2003.


LINKS

T. D. Noe, Table of n, a(n) for n = 0..1000
R. Bekes, J. Pedersen, B. Shao, Mad tea party cyclic partitions, Coll. Math. J. 43 (1) (2012) 2536, Jstor.
Daniel Gabric, Jeffrey Shallit, Borders, Palindrome Prefixes, and Square Prefixes, arXiv:1906.03689 [cs.DM], 2019.
P. Hilton, J. Pedersen, On Factoring 2^k+1, The Math. Educ. 5 (1) (1994) 2931.
Jay Kappraff and Gary W. Adamson, The Relationship of the Cotangent Function to Special Relativity Theory, Silver Means, pcycles, and Chaos Theory; FORMA, Vol. 18, No. 3, pp. 249262 (2003).
Jay Kappraff and Gary W. Adamson, Polygons and Chaos, Journal of Dynamical Systems and Geometric Theories, Vol 2 (2004), p 65.
Wolfdieter Lang, "The Field Q(2Cos Pi/N), its Galois group and length ratios in the regular ngon". https://arxiv.org/pdf/1210.1018.pdf
Paul Lévy, Sur quelques classes de permutations, Compositio Math. 8 (1951), 148.
H. J. Smith, XICalc  Extra Precision Integer Calculator. [broken link?]
Gilbert Strang, A Chaotic Search for i, College Mathematics Journal 22, 312, (1991) [JSTOR]
Eric Weisstein's World of Mathematics, Multiplicative Order.
Eric Weisstein's World of Mathematics, Suborder Function
S. Wolfram, Algebraic Properties of Cellular Automata (1984), Appendix B.


FORMULA

a(n) = log_2(A160657(n) + 2)  1.  Nathaniel Johnston, May 22 2009
a(n) <= n.  Charles R Greathouse IV, Sep 15 2012
a(n) = min{k > 0  q_k = q_0} where q_0 = 1 and q_k = 2*n+1  2*q_{k1} (cf. [Schick, p.4]; q_k=1 for n=1; q_k=A010684(k) for n=2; q_k=A130794(k) for n=3; q_k=A154870(k1) for n=4; q_k=A135449(k) for n=5.)  Jonathan Skowera, Jun 29 2013


EXAMPLE

a(3) = 3 since f(x), x^2  2 has a period of 3 using seed 2*cos(2*Pi/7), where 7 = 2*3 + 1.
a(15) = 5 since the iterative map of the logistic equation 4x*(1x) has a period 5 using seed sin^2(2*Pi)/N; N = 31 = 2*15 + 1.


MAPLE

A003558 := proc(n)
local m, mo ;
if n = 0 then
return 0 ;
end if;
for m from 1 do
mo := modp(2^m, 2*n+1) ;
if mo in {1, 2*n} then
return m;
end if;
end do:
end proc:
seq(A003558(n), n=0..20) ; # R. J. Mathar, Dec 01 2014
f:= proc(n) local t;
t:= numtheory:mlog(1, 2, n);
if t = FAIL then numtheory:order(2, n) else t fi
end proc:
0, seq(f(2*k+1), k=1..1000); # Robert Israel, Oct 26 2015


MATHEMATICA

Suborder[a_, n_] := If[n>1 && GCD[a, n]==1, Min[MultiplicativeOrder[a, n, {1, 1}]], 0];
Table[Suborder[2, 2n+1], {n, 0, 100}] (* T. D. Noe, Aug 02 2006 *)


PROG

(PARI) a(n) = {m=1; while(m, if( (2^m) % (2*n+1) == 1  (2^m) % (2*n+1) == 2*n, return(m)); m++)} \\ Altug Alkan, Nov 06 2015


CROSSREFS

Cf. A054142, A065941, A085478, A160657, A179480, A135303 (coach numbers), A216371 (odd primes with one coach), A000215 (Fermat numbers).
A216066 is an essentially identical sequence apart from the offset.
Sequence in context: A023160 A085312 A046530 * A216066 A234094 A301853
Adjacent sequences: A003555 A003556 A003557 * A003559 A003560 A003561


KEYWORD

nonn,changed


AUTHOR

N. J. A. Sloane


EXTENSIONS

More terms from Harry J. Smith, Feb 11 2005
Entry revised by N. J. A. Sloane, Aug 02 2006 and again Dec 10 2017


STATUS

approved



