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Multiplicative suborder of 2 (mod 2n+1) (or sord(2, 2n+1)).

This is called quasi-order of 2 mod b, with b = 2*n+1, for n >= 1, in the Hilton/Pederson reference.

For the complexity of computing this, see A002326.

Also, the order of the so-called "milk shuffle" of a deck of n cards, which maps cards (1,2,...,n) to (1,n,2,n-1,3,n-2,...). See the paper of Lévy. - Jeffrey Shallit, Jun 09 2019

It appears that under iteration of the base-n 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 (n-1) - 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*(1-x) 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(2*Pi/N) trajectory: (x^2-1)/2x. (End)

From Gary W. Adamson, Sep 11 2019: (Start)

Using x^2 - 2 with seed 2*cos(Pi/7), we obtain the period-three trajectory 1.8019377...-> 1.24697...-> -0.445041... For an odd prime N, the trajectory terms represent diagonal lengths of regular star 2N-gons, with edge the shortest value (0.445... 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 14-gon: (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 n-th 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...-> 0.198062...); the roots to the polynomial with cycle length 3 matching a(3) = 3. - Gary W. Adamson, Sep 21 2011

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 well-defined 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 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: 0.021021021.... In base 9 the expansion of 1/7 is 0.125125125... Check: The first few terms are 1/9 + 2/81 + 5/729 = 104/279 = 0.1426611... (close to 1/7 = 0.142857...). (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 (N-1)^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) = 0.090909 where 1/11 = 0.09090909...). (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

The terms in the 2*cos(Pi/N) trajectory (roots to the minimal polynomials in A187360 and (Lang)), are quickly obtained from the doubling trajectory (mod N) by using the operation L(m) 2*cos(x)--> 2*cos(m*x), where L(2), the second degree Lucas polynomial (A034807) is x^2 - 2. Relating to the heptagon and using seed 2*cos(Pi/7), we obtain the trajectory 1.8019..., 1.24697..., and 0.445041....; cyclic with period 3. All such roots can be derived from the N-th roots of Unity and can be mapped on the Vesica Piscis. Given the roots of Unity (Polar 1Angle(k*2*Pi/N), k = 1, 2, ..., (N-1)/2) the Vesica Piscis maps these points on the left (L) circle to the (R) circle by adding 1A(0) or (a + b*I) = (1 + 0i). But this operation is the same as vector addition in which the resultant vector is 1 + 1A(k*(2*Pi/N)). Example: given the radius at 2*Pi/7 on the left circle, this maps to (1 + 1A(2*Pi/7)) on the right circle; or 1A(2*Pi/7) --> (1.8019377...A(Pi/7). Similarly, 1 + 1A((2)*2*Pi/7)) maps to (1.24697...A (2*Pi/7); and 1 + 1A(3*2*Pi/7) maps to (.0445041...A(3*Pi/7). - Gary W. Adamson, Oct 23 2019

From Gary W. Adamson, Dec 01 2021: (Start)

As to segregating the two sets: (A014659 terms are those N = (2*n+1), N divides (2^m - 1), and (A014657 terms are those N that divide (2^m + 1)); it appears that the following criteria apply: Given IcoS(N, 1) (cf. Lang link "On the Equivalence...", p. 16, Definition 20), if the number of odd terms is odd, then N belongs to A014659, otherwise A014657. In IcoS(11, 1): (1, 2, 4, 3, 5), three odd terms indicate that 11 is a term in A014657. IcoS(15, 1) has the orbit (1, 2, 4, 7) with two odd terms indicating that 15 is a term in A014659.

It appears that if sin(2^m * Pi/N) has a negative sign, then N is in A014659; otherwise N is in A014657. With N = 15, m is 4 and sin(16 * Pi/15) is -0.2079116... If N is 11, m is 5 and sin(32 * Pi/11) is 0.2817325. (End)

On the iterative map using x^2 - 2, (Devaney, p. 126) states that we must find the function that takes 2*cos(Pi) -> 2*cos(2*Pi). "However, we may write 2*cos(2*Pi) = 2*(2*cos^2(Pi) - 1) = (2*cos(Pi))^2 - 2. So the required function is x^2 - 2." On the period 3 implies chaos theorem of James Yorke and T.Y. Li, proved in 1975; Devaney (p. 133) states that if F is continuous and we find a cycle of period 3, there are infinitely many other cycles for this map with every possible period. Check: The x^2 - 2 orbit for 7 has a period of 3, so this entry has periodic points of all other periods. - Gary W. Adamson, Jan 04 2023

It appears that a(n) is the length of the cycle starting at 2/(2*n+1) for the map x->1 - abs(2*x-1). - Michel Marcus, Jul 16 2025

Since for some a < n, 2^a == 1 (mod n) (a consequence of Euler's Theorem), searching up to k=n is sufficient to determine whether an integer is in the sequence. - Michael B. Porter, Dec 06 2009

A195470(a(n)) > 0; A195610(n) gives the smallest k such that a(n) divides 2^k + 1. - Reinhard Zumkeller, Sep 21 2011

This sequence is the subset of odd integers > 1 as (2*n - 1) in A179480, such that the corresponding entry in A179480 is odd. Example: A179480(14) = 5, odd, with (2*14 - 1) = 27; and 5 is a term of this sequence. A014659 (odd and does not divide (2^k + 1) for any k >= 1) represents the subset of odd terms >1 corresponding to A179480 entries that are even. - Gary W. Adamson, Aug 20 2012

All prime factors of a(n) are in A091317. Sequence has asymptotic density 0. - Robert Israel, Aug 12 2014

This sequence, for m>2, is those m for which, for some e, (m-1)(2^e-1)/m is a term of A253608. Moreover, e(n) is 2*A195610(n) when m is a(n). - Donald M Davis, Jan 12 2018

From Wolfdieter Lang, Aug 22 2020: (Start)

Without a(2) = 2 this is the complement of A014659 relative to the odd positive integers A005408.

For the least nonnegative integer k(n) with 2^k(n) + 1 == d(n)*a(n), for n >= 1, see k(n) = A195610(n) and d(n) = A337220(n).

Starting with a(3) = 3 these numbers are the odd moduli, named 2*n+1 in the definition of A003558, for which the minus signs applies (see A332433(m) for the signs applying for A003558(m)). (End)

The length of row n is deg(n) + 1, n >= 1, with the degree deg(1) = deg(2) = 1, and deg(n) = phi(n)/2 = A023022(n) for n >= 3. That is: 2, 2, 2, 2, 3, 2, 4, 3, 4, 3, 6, 3, 7, 4, 5, 5, 9, 4, 10, 5, ...

2*cos(2*Pi/n) = R(2, rho(n)) = -2 + rho(n)^2, with rho(n) = 2*cos(Pi/n) and the monic Chebyshev T-polynomials R(n, x), n>=1, with coefficient table A127672. For even n 2*cos(2*Pi/n) becomes rho(n/2). Therefore, 2*cos(2*Pi/n) is an integer in the algebraic number field Q(rho(n/2)) or Q(rho(n)) if n is even or odd, respectively. The degree deg(n) of the minimal polynomials, call them MPR2(n, x), is delta(n/2) or delta(n) for even or odd n, respectively, with delta(n) = A055034(n). This becomes deg(n) as given above.

These minimal polynomials are C(n/2, x) if n is even, with C(k, x) the minimal polynomials of rho(k) given in A187360.

For odd n the known zeros of C(n, x) are rho(n) and its conjugates, call them rho(n;j), j=1, 2, ..., delta(n), with rho(n;1) = rho(n). These conjugates can be written in the power basis of Q(rho(2*l+1)), l >= 1. See the link to the Q(2cos(Pi/n)) paper in A187360, and there Table 4. Then the (monic) minimal polynomial MPR2(2*l+1, x) = Product_{j=1..delta(2*l+1)} (x - (-2 + rho(2*l+1;j)^2)), l >= 0. After expansion all powers of rho(2*l+1) not smaller than delta(2*l+1) are reduced with the help of C(2*l+1,rho(2*l+1)) = 0, leading automatically to integer coefficients (without using the trigonometric version of rho(2*l+1)).

Compare the present minimal polynomials with the (non-monic) minimal polynomials of cos(2*Pi/n) given in an Artur Jasinski comment from Oct 28 2008 on A023022.

The present monic integer minimal polynomials of 2*cos(2*Pi/n), called MPR2(n, x), are related to the non-monic integer minimal polynomials of 2*cos(2*Pi/n) of A181877, called there psi(n, x) by MPR2(n, x) = psi(n, x/2). See Table 5 of the Wolfdieter Lang link given there. - Wolfdieter Lang, Nov 29 2013

The present minimal polynomials MPR2(n, x) are C(n/2, x) if n is even (see above) and (-1)^degC(n)*C(n, -x) if n is odd, with the C polynomials from A187360 of degree degC(n) = A055034(n). Note that degC(2*k+1) = deg(2*k+1) = A023022(2*k+1), k >= 0. - Wolfdieter Lang, Apr 12 2018

Let {U(n, x)} be defined as: U(0, x) = 0, U(1, x) = 1, U(n, x) = x*U(n-1, x) - U(n-2, x) for n >= 2, then U(n, x) = Product_{k|2n, k>=3} MPR2(k, x) for n > 0, because U(n, x) = Product_{m=1..n-1} (x - 2*cos(Pi*m/n)) for n > 0. - Jianing Song, Jul 08 2019

Conjecture: For odd n > 1, the term of the highest degree of (MPR2(2n, x) - MPR2(n, x))/2 is (-1)^omega(n) * x^(phi(n)/2-n/rad(n)) = A076479(n) * x^(A023022(n)-A003557(n)). For example, for n = 15, (MPR2(30, x) - MPR2(15, x))/2 = x^3 - 4x; for n = 105, (MPR2(210, x) - MPR2(105, x))/2 = -x^23 + ...; for n = 225, (MPR2(450, x) - MPR2(225, x))/2 = x^45 + ... If this is true, then for odd n > 1, a(n,A023022(n)-k) = a(2n,A023022(n)-k) = 0 for k = 1, 3, ..., A003557(n)-2; a(n,A023022(n)-A003557(n)) = -A076479(n) and a(2n,A023022(n)-A003557(n)) = A076479(n). - Jianing Song, Jul 11 2019

Conjecture: Let MPR2(n, x) equal the odd indexed (n) monic polynomial. If the number of roots with negative signs is even, then n is a term in A014659. Example: n = 7 for x^3 + x^2 - 2x - 1, having two negative roots, (-445041..., and -1.801937...). Two is even so the integer 7 is in A014659. n = 9 for the polynomial x^3 - 3x + 1, with one negative root, (-1.87938). The term 9 is in A014657. - Gary W. Adamson, Oct 20 2021

From Gary W. Adamson, Nov 30 2021 (Start)

Given the first (phi(n))/2 terms for odd n, the number of even terms in the set is equal to the number of positive roots in MPR2(n, x). The number of odd terms is equal to the number of negative roots in MPR2(n, x). For n = 11, (phi(11))/2 = 5, and the set is (1, 2, 3, 4, 5); having two even and three odd terms.

Given MPR2(11, x) = x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1, there are two roots with positive signs: 1.682508..., and .830830...; and three roots with negative signs: -1.918985..., -1.309921..., and -.284629....Using the Descartes' rule for signs, MPR2(11, x) has coefficients signed (+ + - - + +); having two sign changes indicating two positive roots. With all real roots there are three (= 5 - 2) roots signed negative. (End)