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For these odd primes delta(p) = A055034(n) = (p-1)/2 is squarefree, and therefore the (Abelian) multiplicative group Modd p (see a comment on A203571 for Modd n, not to be confused with mod n) is guaranteed to be cyclic. This is because the number of Abelian groups of order n (A000688) is 1 precisely for the squarefree numbers A005117. See also A210845. One can in fact prove that the multiplicative group Modd p is cyclic for all primes (the case p=2 is trivial). - Wolfdieter Lang, Sep 24 2012

The trivial solutions of the congruence x^2 == 1 (mod 3*prime(n+2)), n>=1, with the primes prime(n+2) = A000040(n+2) have positive representatives 1 and 3*prime(n+2)-1. There are all-together four incongruent solutions due to a general theorem (see, e.g., the Hardy-Wright reference, Theorem 122, p. 96, and also A060594) and the fact that the number of incongruent solutions of this congruence with odd prime modulus p is two, namely with positive representative p and p-1 (see, e.g., Hardy-Wright, Theorem 109, p. 85). a(n) is the smallest positive odd representative >1 which solves this congruence. The other nontrivial even representative solving this congruence is 3*prime(n+2) - a(n), i.e. 4, 8, 10, 14, 16, 20, ... See 2*A207336.

a(n) solves also the congruence x^2 == 1 (Modd A001748(n+2)), n>=1. For Modd n (not to be confused with mod n) see a comment on A203571. This follows from floor(a(n)^2/3*prime(n+2)) being even, in fact it is 8*A024699(n) (see a comment there), hence a(n)^2 (Modd 3*prime(n+2)) = a(n)^2 (mod 3*prime(n+2)) = 1. For those multiplicative groups Modd 3*p with p an odd prime which are cyclic (this is not possible in the mod case, see A033949), a(n) is the representative of the only other nontrivial solution of this congruence. The representative of the trivial solution is 1 (-1 belongs to the same Modd class). (The conjecture stated here earlier is wrong, that is, the multiplicative group Modd (91=7*13) is non-cyclic. It may still be true for 3*p. - Wolfdieter Lang, Mar 15 2012)

Equivalently, primes of the form (K^2 + (K+1)^2)/13. The

connection to the primes of the form (m^2+1)/26 is given by m=2*K+1 (m is necessarily odd).

The corresponding m=m(n) values are given in A208293(n).

Equivalently, primes of the form (4*T(K)+1)/13, with the

corresponding triangular numbers T(K):=A000217(K), for

K=K(n)=(m(n)-1)/2, given in A208294(n).

For n>=2 the smallest positive representative of the class of

nontrivial solutions of the congruence x^2==1 (Modd a(n)) is

x=m(n). The trivial solution is the class with representative x=1, which also includes -1. For the prime

a(1)=17 the nontrivial solution is 13 (see A002733(2)). Unique nontrivial smallest positive representatives exist for the solutions for any prime of the form 4*k+1, given in A002144. Here the subset with k=k(n)=(a(n)-1)/4 appears, namely 4,9,114,150,175,219,.... For Modd n see a comment on A203571.

These primes with corresponding m values are such that floor(m(n)^2/p(n)) = 5^2, n>=1.

The corresponding primes (n^2+1)/26 are given in A208292(n).

a(n) is the smallest positive representative of the class of

nontrivial solutions of the congruence x^2==1 (Modd A208292(n)), if n>=2. The trivial solution is the class with representative x=1, which also includes -1. For Modd n see a comment on A203571. For n=1: a(1) = 21 == 13 (Modd 17), and 13 is the smallest positive solution >1.

The unique class of nontrivial solutions of the congruence x^2==1 (Modd p), with p an odd prime, exists for any p of the form 4*k+1, given in A002144. Here a subset of these primes is covered, the ones for k=k(n)=(a(n)^2-25)/(4*26). These values are 4, 9, 114, 150, 175, 219, ...

This sequence can be continued periodically for negative values of n.

This is the sixth sequence of a k-family of sequences P_k, k>=1, which starts with A000007(n+1), n>=0, (the 0-sequence), A000035, A193680, A193682, A203571 for k=1,...,5, respectively.

See a comment on A203571 for the general case of the P_k sequences. For a(n)=P_6(n) the nonnegative members of the equivalence classes [0], [1],...,[5], defined by p==q iff P_6(p)=P_6(q), are found in the array A092260 if there class [6], starting with 6, is replaced by 0,6,12,..., which is class [0] (nonnegative part).

For general Modd n (not to be confused with mod n) see a comment on A203571. The present sequence gives the residues Modd 11 for the positive odd numbers not divisible by 11, which are given in A204454.

The underlying period length 22 sequence with offset 0 is P_11, also called Modd11, periodic([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1]).

For general Modd n (not to be confused with mod n) see a comment on A203571. The present sequence gives the residues Modd 15 of the positive odd numbers relatively prime to 15 (the positive odd numbers from all reduced residue classes mod 15), shown in A007775. The underlying periodic sequence with period length 30 is [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,0,14,13,12,11,10,9,8,7,6,5,4,3,2,1], called, with offset 0, P_15 or Modd15.

For the double factorials (2*n-1)!!, for n >= 1, see A001147, and n!! = A006882(n).

For a(n) Modd prime(n) see a comment on A209389 stating the analog of Wilson's theorem for Modd prime(n). For Modd n, (not to be confused with mod n) see a comment on A203571. - Wolfdieter Lang, Mar 28 2012

a(n)^2 == A212159(n) (mod prime(n)), n >= 2. See also the W. Holsztyński link given there. - Wolfdieter Lang, May 07 2012

This is the product over the smallest positive representatives of the odd reduced residue class Modd n. For Modd n (not to be confused with mod n) see a comment on A203571. This reduced residue class has delta(n)=A055034(n) members.

The Moddn values of this sequence are given in A209339.

A055034(n) is the degree delta(n) of the minimal polynomial of the algebraic number rho(n):=2*cos(pi/n), n>=1, whose coefficients are shown in A187360. It is also the order of multiplicative abelian group Modd n (for multiplication Modd n see a comment on A203571). This is the Galois group Gal(Q(rho(n))/Q). If the number of abelian groups of order delta(n) is 1 then this group is necessarily cyclic.

Because A000688 is 1 exactly for the squarefree numbers A005117, the set of a(n) values of the present sequence is a (proper) subset of A206551. Hence it is immediately clear that the multiplicative group Modd a(n) is cyclic, but there are other cyclic Modd n groups, e.g., for n = 8, 10, 15, 16, 17, 19, 26, 27, 32, 34, 35, 37, 38, 39, 41,...