cp's OEIS Frontend

This is a sister sequence to A000057, because this sequence, since {k : A001177(k) = k-1}, might be called a subdiagonal sequence of A001177, and {k : A001177(k) = k+1}, which might be called a superdiagonal sequence of A001177. Sequences A000057 and A106535 are disjoint. Is this sequence the set of all divisors of some family of sequences, like A000057 is? - Art DuPre, Jul 11 2012

Are all members of this sequence prime? Using A069106, any composite members must exceed 89151931. - Robert Israel, Oct 13 2015

From Jianing Song, Jul 02 2019: (Start)

Yes, all terms are primes. See a brief proof below.

Also, if p == 1 (mod 4) then b(p) divides (p-Legendre(p,5))/2. So terms in this sequence are congruent to 11 or 19 modulo 20.

Primes p such that ord(-(3+sqrt(5))/2,p) = p-1, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer. (End)

Comments from Amiram Eldar, Jan 30 2022 (Start)

Sequence A003147, "Primes p with a Fibonacci primitive root", is defined in the paper: Daniel Shanks, Fibonacci primitive roots, Fibonacci Quarterly, Vol. 10, No. 2 (1972), pp. 163-168, and 181.

A second paper on this subject Daniel Shanks and Larry Taylor, An Observation of Fibonacci Primitive Roots, Fibonacci Quarterly, Vol. 11, No. 2 (1973), pp. 159-160,

deals with terms p == 3 (mod 4) of A003147, i.e., the intersection of A003147 and A002145 (or A004767).

It states that if g is a Fibonacci primitive root of a prime p such that p == 3 (mod 4) then g-1 and g-2 are also primitive roots of p.

The first 2000 terms of (A003147 intersect A002145) agree with the present sequence, although the definitions are quite different. Are these two sequences the same? (End)

A prime factor of F(n) is called primitive if it does not divide F(r) for any r < n.

A Fibonacci number can have more than one primitive factor; the primitive factors of F(19) are 37 and 113.

From Robert Israel, Oct 13 2015: (Start)

Since gcd(F(n),F(k)) = F(gcd(n,k)), the non-primitive prime factors of F(n) are factors of F(k) for some proper divisors k of n.

Since prime p divides F(p-1) if p == 1 or 4 (mod 5), F(p+1) if p == 2 or 3 mod 5, F(p) if p = 5, we have a(n) >= n-1 if a(n) > 1.

a(n) = n-1 iff n=2 or n-1 is in A000057.

a(n) = n+1 iff n+1 is a prime in A106535. (End)

Subsequence of A226807.

From Logan J. Kleinwaks, Oct 29 2017: (Start)

The squares of this sequence are the squares in A232656.

Conjecture: these are the numbers j such that j^2 = Sum_{d|j} phi(d)*A001177(d), where phi = Euler's totient function (A000010). See A232656. (End)

The n-th prime p is in this sequence iff A001602(n) = p-(5/p) (that is the maximum possible value of A001602(n)).

In a variant of A213891, multiply n by a number with simple continued fraction [n,3,3,...,3,n] and increase the number of 3's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are

2 * [2, 3, 3, 2] = [4, 1, 1, 1, 1, 4],

3 * [3, 3, 3] = [9, 1, 9],

4 * [4, 3, 3, 3, 3, 3, 4] = [17, 4, 1, 2, 1, 4, 17],

5 * [5, 3, 3, 5] = [26, 1, 1, 26],

6 * [6, 3, 3, 3, 3, 3, 6] = [37, 1, 4, 2, 4, 1, 37],

7 * [7, 3, 3, 3, 3, 3, 3, 3, 7] = [51, 8, 2, 1, 2, 8, 51].

The number of 3's needed defines the sequence h(n) = 2, 1, 5, 2, 5, 7, 5, 9, 2, ... (n>=2).

The current sequence contains the fixed points of h, i.e., those n where h(n)=n.

We conjecture that this sequence contains prime numbers analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequences (sequences satisfying f(n) = f(n-1) + f(n-2) with arbitrary positive integer values for f(1) and f(2)) it refers to the sequences satisfying f(n) = 3*f(n-1) + f(n-2), A006190, A003688, A052924, etc. This would mean that a prime is in the sequence if and only if it divides some term in each of the sequences satisfying f(n) = 3*f(n-1) + f(n-2).

The above sequence h() is recorded as A262213. - M. F. Hasler, Sep 15 2015

The -1 terms are for the primes in A053028. Note that 2 divides the zeroth Lucas number. In the plots, the uppermost line consists of the odd primes in A000057. Note that when a(n) > 0, then a(n) = A001602(n)/2.

In a variant of A213891, multiply n by a number with simple continued fraction [n,7,7,..,7,n] and increase the number of 7's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are

2 * [2, 7, 7, 2] = [4, 3, 1, 1, 3, 4],

3 * [3, 7, 7, 7, 3] = [9, 2, 2, 1, 1, 1, 2, 2, 9] ,

4 * [4, 7, 7, 7, 7, 7, 4] = [16, 1, 1, 3, 1, 1, 1, 6, 1, 1, 1, 3, 1, 1, 16],

5 * [5, 7, 7, 5] = [25, 1, 2, 2, 1, 25] ,

6 * [6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6] = [36, 1, 5, 3, 1, 4, 10, 1, 2, 2, 4, 2, 2, 1, 10, 4, 1, 3, 5, 1, 36],

7 * [7, 7, 7] = [49, 1, 49] .

The number of 7's needed defines the sequence h(n) = 2, 3, 5, 2, 11, 1, 5, 11, 2,... (n>=2).

The current sequence contains the fixed points of h, i. e., those n where h(n)=n.

We conjecture that this sequence contains prime numbers analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequences (sequences satisfying f(n)=f(n-1)+f(n-2) with arbitrary positive integer values for f(1) and f(2)) it refers to the sequences satisfying f(n)=7*f(n-1)+f(n-2), A054413, A015453, etc. This would mean that a prime is in the sequence A213896 if and only if it divides some term in each of the sequences satisfying f(n)=7*f(n-1)+f(n-2).

The above sequence h() is recorded as A262217. - M. F. Hasler, Sep 15 2015

In a variant of A213891, multiply n by a number with simple continued fraction [n,8,8,..,8,n] and increase the number of 8's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are

2 * [2, 8, 2] = [4, 4, 4],

3 * [3, 8, 8, 8, 3] = [9, 2, 1, 2, 2, 2, 1, 2, 9],

4 * [4, 8, 4] = [16, 2, 16],

5 * [5, 8, 8, 5] = [25, 1, 1, 1, 1, 1, 1, 25],

6 * [6, 8, 8, 8, 6] = [36, 1, 2, 1, 4, 1, 2, 1, 36],

7 * [7, 8, 8, 8, 8, 8, 8, 8, 7] = [49, 1, 6, 4, 3, 2, 1, 2, 1, 2, 3, 4, 6, 1, 49].

The number of 8's needed defines the sequence h(n) = 1, 3, 1, 2, 3, 7, 1, 11, 5,.. (n>=2).

The current sequence contains the fixed points of h, i. e., those n where h(n)=n.

We conjecture that this sequence contains prime numbers analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequences (sequences satisfying f(n)=f(n-1)+f(n-2) with arbitrary positive integer values for f(1) and f(2)) it refers to the sequences satisfying f(n)=8*f(n-1)+f(n-2), A041025, A015454, etc. This would mean that a prime is in the sequence A213897 if and only if it divides some term in each of the sequences satisfying f(n)=8*f(n-1)+f(n-2).

The sequence h() is recorded as A262218. - M. F. Hasler, Sep 15 2015

In a variant of A213891, multiply n by a number with simple continued fraction [n,9,9,..,9,n] and increase the number of 9's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are

2 * [2, 9, 9, 2] = [4, 4, 1, 1, 4, 4],

3 * [3, 9, 3] = [9, 3, 9],

4 * [4, 9, 9, 9, 9, 9, 4] = [16, 2, 3, 1, 1, 1, 1, 8, 1, 1, 1, 1, 3, 2, 16] ,

5 * [5, 9, 9, 9, 9, 5] = [25, 1, 1, 4, 1, 1, 1, 1, 1, 1, 4, 1, 1, 25],

6 * [6, 9, 9, 9, 9, 9, 6] = [36, 1, 1, 1, 13, 6, 13, 1, 1, 1, 36],

7 * [7, 9, 9, 9, 9, 9, 7] = [49, 1, 3, 3, 6, 1, 6, 3, 3, 1, 49].

The number of 9's needed defines the sequence h(n) = 2, 1,5, 4, 5, 5, 5, 1, 14,... (n>=2).

The current sequence contains the fixed points of h, i. e., those n where h(n)=n.

We conjecture that this sequence contains prime numbers analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequence (sequences satisfying f(n)=f(n-1)+f(n-2) with arbitrary positive integer values for f(1) and f(2)) it refers to the sequences satisfying f(n)=9*f(n-1)+f(n-2) like A099371, A015455 etc. This would mean that a prime is in the sequence A213898 if and only if it divides some term in each of the sequences satisfying f(n)=9*f(n-1)+f(n-2).