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Conjecture: For every n>=0 the sequence {f^(m)(n)} is eventually periodic. Further calculations may show if there exist terms different from 1 and 3.

A subsequence of A077816.

Odd numbers n > 1 such that A237292((n-1)/2) = 1.

Indices k such that A077816(k) is not in this sequence are 2, 13, 14, 16, 22, 24, 25, 27, 28, 30, ... - Altug Alkan, Dec 11 2015

There are no other terms, unless there are other Wieferich primes (A001220) besides 1093 and 3511. - Max Alekseyev, Dec 11 2015

It seems that a(n) = (A115591(n)-1)/2. Indeed, it follows from the definition of A115591 that if a prime p is listed in A115591, then (p-1)/2 is also listed in this sequence.

The related case of A002326(k) = 2k is true if (and conjecturally only if) 2k+1 is a prime with primitive root 2, see A001122.

Sequence is believed to be infinite.

Joseph Silverman showed that the abc-conjecture implies that there are infinitely many primes which are not in the sequence. - Benoit Cloitre, Jan 09 2003

Graves and Murty (2013) improved Silverman's result by showing that for any fixed k > 1, the abc-conjecture implies that there are infinitely many primes == 1 (mod k) which are not in the sequence. - Jonathan Sondow, Jan 21 2013

The squares of these numbers are Fermat pseudoprimes to base 2 (A001567) and Catalan pseudoprimes (A163209). - T. D. Noe, May 22 2003

Primes p that divide the numerator of the harmonic number H((p-1)/2); that is, p divides A001008((p-1)/2). - T. D. Noe, Mar 31 2004

In a 1977 paper, Wells Johnson, citing a suggestion from Lawrence Washington, pointed out the repetitions in the binary representations of the numbers which are one less than the two known Wieferich primes; i.e., 1092 = 10001000100 (base 2); 3510 = 110110110110 (base 2). It is perhaps worth remarking that 1092 = 444 (base 16) and 3510 = 6666 (base 8), so that these numbers are small multiples of repunits in the respective bases. Whether this is mathematically significant does not appear to be known. - John Blythe Dobson, Sep 29 2007

A002326((a(n)^2 - 1)/2) = A002326((a(n)-1)/2). - Vladimir Shevelev, Jul 09 2008, Aug 24 2008

It is believed that p^2 does not divide 3^(p-1) - 1 if p = a(n). This is true for n = 1 and 2. See A178815, A178844, A178900, and Ostafe-Shparlinski (2010) Section 1.1. - Jonathan Sondow, Jun 29 2010

These primes also divide the numerator of the harmonic number H(floor((p-1)/4)). - H. Eskandari (hamid.r.eskandari(AT)gmail.com), Sep 28 2010

1093 and 3511 are prime numbers p satisfying congruence 429327^(p-1) == 1 (mod p^2). Why? - Arkadiusz Wesolowski, Apr 07 2011. Such bases are listed in A247208. - Max Alekseyev, Nov 25 2014. See A269798 for all such bases, prime and composite, that are not powers of 2. - Felix Fröhlich, Apr 07 2018

A196202(A049084(a(1))) = A196202(A049084(a(2))) = 1. - Reinhard Zumkeller, Sep 29 2011

If q is prime and q^2 divides a prime-exponent Mersenne number, then q must be a Wieferich prime. Neither of the two known Wieferich primes divide Mersenne numbers. See Will Edgington's Mersenne page in the links below. - Daran Gill, Apr 04 2013

There are no other terms below 4.97*10^17 as established by PrimeGrid (see link below). - Max Alekseyev, Nov 20 2015. The search was done via PrimeGrid's PRPNet and the results were not double-checked. Because of the unreliability of the testing, the search was suspended in May 2017 (cf. Goetz, 2017). - Felix Fröhlich, Apr 01 2018. On Nov 28 2020, PrimeGrid has resumed the search (cf. Reggie, 2020). - Felix Fröhlich, Nov 29 2020. As of Dec 29 2022, PrimeGrid has completed the search to 2^64 (about 1.8 * 10^19) and has no plans to continue further. - Charles R Greathouse IV, Sep 24 2024

Are there other primes q >= p such that q^2 divides 2^(p-1)-1, where p is a prime? - Thomas Ordowski, Nov 22 2014. Any such q must be a Wieferich prime. - Max Alekseyev, Nov 25 2014

Primes p such that p^2 divides 2^r - 1 for some r, 0 < r < p. - Thomas Ordowski, Nov 28 2014, corrected by Max Alekseyev, Nov 28 2014

For some reason, both p=a(1) and p=a(2) also have more bases b with 1 < b < p that make b^(p-1) == 1 (mod p^2) than any smaller prime p; in other words, a(1) and a(2) belong to A248865. - Jeppe Stig Nielsen, Jul 28 2015

Let r_1, r_2, r_3, ..., r_i be the set of roots of the polynomial X^((p-1)/2) - (p-3)! * X^((p-3)/2) - (p-5)! * X^((p-5)/2) - ... - 1. Then p is a Wieferich prime iff p divides sum{k=1, p}(r_k^((p-1)/2)) (see Example 2 in Jakubec, 1994). - Felix Fröhlich, May 27 2016

Arthur Wieferich showed that if p is not a term of this sequence, then the First Case of Fermat's Last Theorem has no solution in x, y and z for prime exponent p (cf. Wieferich, 1909). - Felix Fröhlich, May 27 2016

Let U_n(P, Q) be a Lucas sequence of the first kind, let e be the Legendre symbol (D/p) and let p be a prime not dividing 2QD, where D = P^2 - 4*Q. Then a prime p such that U_(p-e) == 0 (mod p^2) is called a "Lucas-Wieferich prime associated to the pair (P, Q)". Wieferich primes are those Lucas-Wieferich primes that are associated to the pair (3, 2) (cf. McIntosh, Roettger, 2007, p. 2088). - Felix Fröhlich, May 27 2016

Any repeated prime factor of a term of A000215 is a term of this sequence. Thus, if there exist infinitely many Fermat numbers that are not squarefree, then this sequence is infinite, since no two Fermat numbers share a common factor. - Felix Fröhlich, May 27 2016

If the Diophantine equation p^x - 2^y = d has more than one solution in positive integers (x, y), with (p, d) not being one of the pairs (3, 1), (3, -5), (3, -13) or (5, -3), then p is a term of this sequence (cf. Scott, Styer, 2004, Corollary to Theorem 2). - Felix Fröhlich, Jun 18 2016

Odd primes p such that Chi_(D_0)(p) != 1 and Lambda_p(Q(sqrt(D_0))) != 1, where D_0 < 0 is the fundamental discriminant of the imaginary quadratic field Q(sqrt(1-p^2)) and Chi and Lambda are Iwasawa invariants (cf. Byeon, 2006, Proposition 1 (i)). - Felix Fröhlich, Jun 25 2016

If q is an odd prime, k, p are primes with p = 2*k+1, k == 3 (mod 4), p == -1 (mod q) and p =/= -1 (mod q^3) (Jakubec, 1998, Corollary 2 gives p == -5 (mod q) and p =/= -5 (mod q^3)) with the multiplicative order of q modulo k = (k-1)/2 and q dividing the class number of the real cyclotomic field Q(Zeta_p + (Zeta_p)^(-1)), then q is a term of this sequence (cf. Jakubec, 1995, Theorem 1). - Felix Fröhlich, Jun 25 2016

From Felix Fröhlich, Aug 06 2016: (Start)

Primes p such that p-1 is in A240719.

Prime terms of A077816 (cf. Agoh, Dilcher, Skula, 1997, Corollary 5.9).

p = prime(n) is in the sequence iff T(2, n) > 1, where T = A258045.

p = prime(n) is in the sequence iff an integer k exists such that T(n, k) = 2, where T = A258787. (End)

Conjecture: an integer n > 1 such that n^2 divides 2^(n-1)-1 must be a Wieferich prime. - Thomas Ordowski, Dec 21 2016

The above conjecture is equivalent to the statement that no "Wieferich pseudoprimes" (WPSPs) exist. While base-b WPSPs are known to exist for several bases b > 1 other than 2 (see for example A244752), no base-2 WPSPs are known. Since two necessary conditions for a composite to be a base-2 WPSP are that, both, it is a base-2 Fermat pseudoprime (A001567) and all its prime factors are Wieferich primes (cf. A270833), as shown in the comments in A240719, it seems that the first base-2 WPSP, if it exists, is probably very large. This appears to be supported by the guess that the properties of a composite to be a term of A001567 and of A270833 are "independent" of each other and by the observation that the scatterplot of A256517 seems to become "less dense" at the x-axis parallel line y = 2 for increasing n. It has been suggested in the literature that there could be asymptotically about log(log(x)) Wieferich primes below some number x, which is a function that grows to infinity, but does so very slowly. Considering the above constraints, the number of WPSPs may grow even more slowly, suggesting any such number, should it exist, probably lies far beyond any bound a brute-force search could reach in the forseeable future. Therefore I guess that the conjecture may be false, but a disproof or the discovery of a counterexample are probably extraordinarily difficult problems. - Felix Fröhlich, Jan 18 2019

Named after the German mathematician Arthur Josef Alwin Wieferich (1884-1954). a(1) = 1093 was found by Waldemar Meissner in 1913. a(2) = 3511 was found by N. G. W. H. Beeger in 1922. - Amiram Eldar, Jun 05 2021

From Jianing Song, Jun 21 2025: (Start)

The ring of integers of Q(2^(1/k)) is Z[2^(1/k)] if and only if k does not have a prime factor in this sequence (k is in A342390). See Theorem 5.3 of the paper of Keith Conrad. For example, we have:

(1 + 2^(364/1093) + 2^(2*364/1093) + ... + 2^(1092*364/1093))/1093 is an algebraic integer, but it is not in Z[2^(1/1093)];

(1 + 2^(1755/3511) + 2^(2*1755/3511) + ... + 2^(3510*1755/3511))/3511 is an algebraic integer, but it is not in Z[2^(1/3511)]. (End)

Artin conjectured that this sequence is infinite.

Conjecture: sequence contains infinitely many pairs of twin primes. - Benoit Cloitre, May 08 2003

Pieter Moree writes (Oct 20 2004): Assuming the Generalized Riemann Hypothesis, it can be shown that the density of primes p such that a prescribed integer g has order (p-1)/t, with t fixed, exists and, moreover, it can be computed. This density will be a rational number times the so-called Artin constant. For 2 and 10 the density of primitive roots is A, the Artin constant itself.

It seems that this sequence consists of A050229 \ {1,2}.

Primes p such that 1/p, when written in base 2, has period p-1, which is the greatest period possible for any integer.

Positive integer 2*m-1 is in the sequence iff A179382(m)=m-1. - Vladimir Shevelev, Jul 14 2010

These are the odd primes p for which the polynomial 1+x+x^2+...+x^(p-1) is irreducible over GF(2). - V. Raman, Sep 17 2012 [Corrected by N. J. A. Sloane, Oct 17 2012]

Prime(n) is in the sequence if (and conjecturally only if) A133954(n) = prime(n). - Vladimir Shevelev, Aug 30 2013

Pollack shows that, on the GRH, that there is some C such that a(n+1) - a(n) < C infinitely often (in fact, 1 can be replaced by any positive integer). Further, for any m, a(n), a(n+1), ..., a(n+m) are consecutive primes infinitely often. - Charles R Greathouse IV, Jan 05 2015

From Jianing Song, Apr 27 2019: (Start)

All terms are congruent to 3 or 5 modulo 8. If we define

Pi(N,b) = # {p prime, p <= N, p == b (mod 8)};

Q(N) = # {p prime, p <= N, p in this sequence},

then by Artin's conjecture, Q(N) ~ C*N/log(N) ~ 2*C*(Pi(N,3) + Pi(N,5)), where C = A005596 is Artin's constant.

Conjecture: if we further define

Q(N,b) = # {p prime, p <= N, p == b (mod 8), p in this sequence},

then we have:

Q(N,3) ~ (1/2)*Q(N) ~ C*Pi(N,3);

Q(N,5) ~ (1/2)*Q(N) ~ C*Pi(N,5). (End)

Conjecture: for a prime p > 5, p has primitive root 2 iff p == +-3 (mod 8) divides 2^k + 3 for some k < p - 1 and divides 2^m + 5 for some m < p - 1. It seems that all primes of the form 2^k + 3 for k <> 2 (A057732) have primitive root 2. - Thomas Ordowski, Nov 27 2023

Warning: there is considerable overlap between this entry and the essentially identical A008776.

Shifts one place left when plus-convolved (PLUSCONV) with itself. a(n) = 2*Sum_{i=0..n-1} a(i). - Antti Karttunen, May 15 2001

Let M = { 0, 1, ..., 2^n-1 } be the set of all n-bit numbers. Consider two operations on this set: "sum modulo 2^n" (+) and "bitwise exclusive or" (XOR). The results of these operations are correlated.

To give a numerical measure, consider the equations over M: u = x + y, v = x XOR y and ask for how many pairs (u,v) is there a solution? The answer is exactly a(n) = 2*3^(n-1) for n >= 1. The fraction a(n)/4^n of such pairs vanishes as n goes to infinity. - Max Alekseyev, Feb 26 2003

Number of (s(0), s(1), ..., s(2n+2)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+2, s(0) = 3, s(2n+2) = 3. - Herbert Kociemba, Jun 10 2004

Number of compositions of n into parts of two kinds. For a string of n objects, before the first, choose first kind or second kind; before each subsequent object, choose continue, first kind, or second kind. For example, compositions of 3 are 3; 2,1; 1,2; and 1,1,1. Using parts of two kinds, these produce respectively 2, 4, 4 and 8 compositions, 2+4+4+8 = 18. - Franklin T. Adams-Watters, Aug 18 2006

In the compositions the kinds of parts are ordered inside a run of identical parts, see example. Replacing "ordered" by "unordered" gives A052945. - Joerg Arndt, Apr 28 2013

Number of permutations of {1, 2, ..., n+1} such that no term is more than 2 larger than its predecessor. For example, a(3) = 18 because all permutations of {1, 2, 3, 4} are valid except 1423, 1432, 2143, 3142, 2314, 3214, in which 1 is followed by 4. Proof: removing (n + 1) gives a still-valid sequence. For n >= 2, can insert (n + 1) either at the beginning or immediately following n or immediately following (n - 1), but nowhere else. Thus the number of such permutations triples when we increase the sequence length by 1. - Joel B. Lewis, Nov 14 2006

Antidiagonal sums of square array A081277. - Philippe Deléham, Dec 04 2006

Equals row sums of triangle A160760. - Gary W. Adamson, May 25 2009

Let M = a triangle with (1, 2, 4, 8, ...) as the left border and all other columns = (0, 1, 2, 4, 8, ...). A025192 = lim_{n->oo} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010

Number of nonisomorphic graded posets with 0 and uniform hasse graph of rank n with no 3-element antichain. ("Uniform" used in the sense of Retakh, Serconek and Wilson. By "graded" we mean that all maximal chains have the same length n.) - David Nacin, Feb 13 2012

Equals partial sums of A003946 prefaced with a 1: (1, 1, 4, 12, 36, 108, ...). - Gary W. Adamson, Feb 15 2012

Number of vertices (or sides) of the (n-1)-th iteration of a Gosper island. - Arkadiusz Wesolowski, Feb 07 2013

Row sums of triangle in A035002. - Jon Perry, May 30 2013

a(n) counts walks (closed) on the graph G(1-vertex; 1-loop, 1-loop, 2-loop, 2-loop, 3-loop, 3-loop, ...). - David Neil McGrath, Jan 01 2015

From Tom Copeland, Dec 03 2015: (Start)

For n > 0, a(n) are the traces of the even powers of the adjacency matrix M of the simple Lie algebra B_3, tr(M^(2n)) where M = Matrix(row 1; row 2; row 3) = Matrix[0,1,0; 1,0,2; 0,1,0], same as the traces of Matrix[0,2,0; 1,0,1; 0,1,0] (cf. Damianou). The traces of the odd powers vanish.

The characteristic polynomial of M equals determinant(x*I - M) = x^3 - 3x = A127672(3,x), so 1 - 3*x^2 = det(I - x M) = exp(-Sum_{n>=1} tr(M^n) x^n / n), implying Sum_{n>=1} a(n+1) x^(2n) / (2n) = -log(1 - 3*x^2), giving a logarithmic generating function for the aerated sequence, excluding a(0) and a(1).

a(n+1) = tr(M^(2n)), where tr(M^n) = 3^(n/2) + (-1)^n * 3^(n/2) = 2^n*(cos(Pi/6)^n + cos(5*Pi/6)^n) = n-th power sum of the eigenvalues of M = n-th power sum of the zeros of the characteristic polynomial.

The relation det(I - x M) = exp(-Sum_{n>=1} tr(M^n) x^n / n) = Sum_{n>=0} P_n(-tr(M), -tr(M^2), ..., -tr(M^n)) x^n/n! = exp(P.(-tr(M), -tr(M^2), ...)x), where P_n(x(1), ..., x(n)) are the partition polynomials of A036039 implies that with x(2n) = -tr(M^(2n)) = -a(n+1) for n > 0 and x(n) = 0 otherwise, the partition polynomials evaluate to zero except for P_2(x(1), x(2)) = P_2(0,-6) = -6.

Because of the inverse relation between the partition polynomials of A036039 and the Faber polynomials F_k(b1,b2,...,bk) of A263916, F_k(0,-3,0,0,...) = tr(M^k) gives aerated a(n), excluding n=0,1. E.g., F_2(0,-3) = -2(-3) = 6, F_4(0,-3,0,0) = 2 (-3)^2 = 18, and F_6(0,-3,0,0,0,0) = -2(-3)^3 = 54. (Cf. A265185.)

(End)

Number of permutations of length n > 0 avoiding the partially ordered pattern (POP) {1>2, 1>3, 1>4} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is the largest. - Sergey Kitaev, Dec 08 2020

For n > 0, a(n) is the number of 3-colorings of the grid graph P_2 X P_(n-1). More generally, for q > 1, the number of q-colorings of the grid graph P_2 X P_n is given by q*(q - 1)*((q - 1)*(q - 2) + 1)^(n - 1). - Sela Fried, Sep 25 2023

For n > 1, a(n) is the largest solution to the equation phi(x) = a(n-1). - M. Farrokhi D. G., Oct 25 2023

Number of dotted compositions of degree n. - Diego Arcis, Feb 01 2024

In other words, a(n), n >= 2, is the least k such that prime(n) divides 2^k-1.

Concerning the complexity of computing this sequence, see for example Bach and Shallit, p. 115, exercise 8.

Also A002326((p_n-1)/2). Conjecture: If p_n is not a Wieferich prime (1093, 3511, ...) then A002326(((p_n)^k-1)/2) = a(n)*(p_n)^(k-1). - Vladimir Shevelev, May 26 2008

If for distinct i,j,...,k we have a(i)=a(j)=...=a(k) then the number N = p_i*p_j*...*p_k is in A001262 and moreover A137576((N-1)/2) = N. For example, a(16)=a(37)=a(255)=52. Therefore we could take N = p_16*p_37*p_255 = 53*157*1613 = 13421773. - Vladimir Shevelev, Jun 14 2008

Also degree of the irreducible polynomial factors for the polynomial (x^p+1)/(x+1) over GF(2), where p is the n-th prime. - V. Raman, Oct 04 2012

Is this the same as the smallest k > 1 not already in the sequence such that p = prime(n) is a factor of 2^k-1 (A270600)? If the answer is yes, is the sequence a permutation of the positive integers > 1? - Felix Fröhlich, Feb 21 2016. Answer: No, it is easy to prove that 6 is missing and obviously 11 appears twice. - N. J. A. Sloane, Feb 21 2016

pi(A112927(m)) is the index at which a given number m first appears in this sequence. - M. F. Hasler, Feb 21 2016

Also sequence of period lengths for n's when you do primality testing and calculate "2^k mod n" from k = 0..n. - Gottfried Helms, Oct 05 2000

Fractal sequence related to A002326: the even terms of this sequence are this sequence itself, constructed on A002326, whose terms are the odd terms of this sequence. - Alexandre Wajnberg, Apr 27 2005

It seems that a(n) is also the sum of the terms in one period of the base-2 MR-expansion of 1/n (see A136042 for definition). - John W. Layman, Jan 22 2009

Indices n such that a(n) divides n are listed in A068563. - Max Alekseyev, Aug 25 2013

a(n) is the smallest k such that x^n - 1 factors into n linear polynomials over GF(2^k). For example, a(12) = 2, and x^12 - 1 = (x - 1)^4*(x - w)^4*(x - (w + 1))^4 in GF(4), where w^2 + w + 1 = 0. - Jianing Song, Jan 20 2019

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