cp's OEIS Frontend

In the formula, the relation a(p^e) = p^(e-1)*a(p) is called Wall's conjecture, which has been verified for primes up to 10^14. See A060305. Primes for which this relation fails are called Wall-Sun-Sun primes. - T. D. Noe, Mar 03 2009

All solutions to F_m == 0 (mod n) are given by m == 0 (mod a(n)). For a proof see, e.g., Vajda, p. 73. [Old comment changed by Wolfdieter Lang, Jan 19 2015]

If p is a prime of the form 10n +- 1 then a(p) is a divisor of p-1. If q is a prime of the form 10n +- 3 then a(q) is a divisor of q+1. - Robert G. Wilson v, Jul 07 2007

Definition 1 in Riasat (2011) calls this k(n), or sometimes just k. Corollary 1 in the same paper, "every positive integer divides infinitely many Fibonacci numbers," demonstrates that this sequence is infinite. - Alonso del Arte, Jul 27 2013

If p is a prime then a(p) <= p+1. This is because if p is a prime then exactly one of the following Fibonacci numbers is a multiple of p: F(p-1), F(p) or F(p+1). - Dmitry Kamenetsky, Jul 23 2015

From Renault 1996:

1. a(lcm(n,m)) = lcm(a(n), a(m)).

2. if n|m then a(n)|a(m).

3. if m has prime factorization m=p1^e1 * p2^e2 * ... * pn^en then a(m) = lcm(a(p1^e1), a(p2^e2), ..., a(pn^en)). - Dmitry Kamenetsky, Jul 23 2015

a(n)=n if and only if n=5^k or n=12*5^k for some k >= 0 (see Marques 2012). - Dmitry Kamenetsky, Aug 08 2015

Every positive integer (except 2) eventually appears in this sequence. This is because every Fibonacci number bigger than 1 (except Fibonacci(6)=8 and Fibonacci(12)=144) has at least one prime factor that is not a factor of any earlier Fibonacci number (see Knott reference). Let f(n) be such a prime factor for Fibonacci(n); then a(f(n))=n. - Dmitry Kamenetsky, Aug 08 2015

We can reconstruct the Fibonacci numbers from this sequence using the formula Fibonacci(n+2) = 1 + Sum_{i: a(i) <= n} phi(i)*floor(n/a(i)), where phi(n) is Euler's totient function A000010 (see the Stroinski link). For example F(6) = 1 + phi(1)*floor(4/a(1)) + phi(2)*floor(4/a(2)) + phi(3)*floor(4/a(4)) = 1 + 1*4 + 1*1 + 2*1 = 8. - Peter Bala, Sep 10 2015

Conjecture: Sum_{d|n} phi(d)*a(d) = A232656(n). - Logan J. Kleinwaks, Oct 28 2017

a(F_m) = m for all m > 1. Indeed, let (b(j)) be defined by b(1)=b(2)=1, and b(j+2) = (b(j) + b(j+1)) mod n. Then a(n) equals the index of the first occurrence of 0 in (b(j)). Example: if n=4 then b = A079343 = 1,1,2,3,1,0,1,1,..., so a(4)=6. If n is a Fibonacci number n=F_m, then obviously a(n)=m. Note that this gives a simple proof of the fact that all integers larger than 2 occur in (a(n)). - Michel Dekking, Nov 10 2017

Here a Fibonacci sequence is a sequence which begins with any two integers and continues using the rule s(n+2) = s(n+1) + s(n). These primes divide at least one number in each such sequence. - Don Reble, Dec 15 2006

Primes p such that the smallest positive m for which Fibonacci(m) == 0 (mod p) is m = p + 1. In other words, the n-th prime p is in this sequence iff A001602(n) = p + 1. - Max Alekseyev, Nov 23 2007

Cubre and Rouse comment that this sequence is not known to be infinite. - Charles R Greathouse IV, Jan 02 2013

Number of terms up to 10^n: 3, 7, 38, 249, 1894, 15456, 130824, 1134404, 10007875, 89562047, .... - Charles R Greathouse IV, Nov 19 2014

These are also the fixed points of sequence A213648 which gives the minimal number of 1's such that n*[n; 1,..., 1, n] = [x; ..., x], where [...] denotes simple continued fractions. - M. F. Hasler, Sep 15 2015

It appears that for n >= 2, all first differences are congruent to 0 (mod 4). - Christopher Hohl, Dec 28 2018

The comment above is equivalent to a(n) == 3 (mod 4) for n >= 2. This is indeed correct. Actually it can be proved that a(n) == 3, 7 (mod 20) for n >= 2. Let p != 2, 5 be a prime, then: A001175(p) divides (p - 1)/2 if p == 1, 9 (mod 20); p - 1 if p == 11, 19 (mod 20); (p + 1)/2 if p == 13, 17 (mod 20). So the remaining cases are p == 3, 7 (mod 20). - Jianing Song, Dec 29 2018

It has long been a problem to find "natural" functions which will produce only primes. The sequence here apparently does just that, and it may well be the most natural function yet doing just that. There is apparently no reason why these sequences should produce only primes.

Let [a,b,...,c] = a+1/(b+(1/...+1/c)) represent a simple continued fraction.

Consider for n=2 the continued fraction [2,1,2] = 8/3. If we multiply 8/3 by 2, we get 16/3. If we write 16/3 as a continued fraction, we get [5,3]. Since the first entry 5 of this sequence is not equal to the last, 3, we insert another 1 in [2,1,2] between n and n to get [n,1,1,n] = 13/5. If we multiply 13/5 by 2, we get 26/5. If we write 26/5 as a continued fraction, we get [5,5] and now the first entry 5, of [5,5] is the same as the last entry 5 of [5,5]. Therefore 2 is the first number of 1s that we had to insert between the 1s in order for twice the resulting continued fraction to have equal first and last entries. Therefore, we define g(2)=2.

If we do the same for n=3, [3,1,3], we see that 3 is the minimum number of 1s that we have to insert between the 3s in order that when we multiply the continued fraction [3,1,1,1,3] by 3, we get [10,1,10], so the first and last entries are the same, namely 10. Therefore we define g(3)=3.

If we do this for n=4, [4,1,4] we see that 5 is the minimum number of 1s we have to insert before the first and last entries of 4*[4,1,1,1,1,1,4] are the same, namely, we get [18,2,18]. If we had multiplied [4,1,4], [4,1,1,4], [4,1,1,1,4],[4,1,1,1,1,4] by 4 we get, respectively [19,5],[18,4,2],[18,1,1,3],[18,2,2,3], none of which has its first and last entries equal. Therefore we define g(4)=5.

It turns out, proceeding as we just have, we get g(5)=4, g(6)=11, g(7)=7, which is A213648. If we define a sequence b(n) to contain the fixed points for which g(n)=n, considering that the sequence A213648 starts with 2 as its second term, then we get A000057 connected with the prime divisors of all the Fibonacci sequences.

If we do the same for inserting 2s as we just described for 1s, we get this sequence here.

These primes arise by first looking at the sequence h(n), whose n-th term is the minimum number of twos in [n,2,2,....,2,n], so that the continued fraction of n times the fraction corresponding to the above quotients has its first and last term equal. Next we construct the sequence of fixed points where h(n)=n. This sequence consists of prime numbers (conjecture). We conjecture that this sequence of prime numbers is analogous to A000057, in the sense that, instead of referring to the Fibonacci sequences it refers to the generalized Fibonacci sequences satisfying f(n)=2*f(n-1)+f(n-2). This would mean that a prime is in this sequence here if and only if it divides some term in each of the sequences satisfying f(n)=2*f(n-1)+f(n-2).

In a variant of A213891, multiply n by a number with simple continued fraction [n,11,11,..,11,n] and increase the number of 11's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are

2 * [2, 11, 11, 2] = [4, 5, 1, 1, 5, 4],

3 * [3, 11, 11, 11, 3] = [9, 3, 1, 2, 3, 2, 1, 3, 9],

4 * [4, 11, 11, 11, 11, 11, 4] = [16, 2, 1, 3, 2, 1, 1, 10, 1, 1, 2, 3, 1, 2, 16],

5 * [5, 11, 11, 11, 11, 5] = [25, 2, 4, 1, 1, 2, 2, 1, 1, 4, 2, 25] ,

6 * [6, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 6] = [36, 1, 1, 5, 1, 1, 2, 7, 16, 1, 1, 1, 2, 1, 6, 1, 2, 1, 1, 1, 16, 7, 2, 1, 1, 5, 1, 1, 36].

The number of 11's needed defines the sequence a(n).

If we consider the fixed points such that a(n)=n, we conjecture to obtain the sequence A000057. This sequence consists of prime numbers. We conjecture that this sequence of prime numbers, in addition to its well-known relation to the collection of Fibonacci sequences (sequences satisfying f(n)=f(n-1)+f(n-2) with arbitrary positive integer values for f(1) and f(2)) it also refers to the sequences satisfying f(n)=11*f(n-1)+f(n-2), A049666, A015457, etc. This would mean that a prime is in the sequence A000057 if and only if it divides some term in each of the sequences satisfying f(n)=11*f(n-1)+f(n-2).

It is surprising that the fixed points of this sequence seem to be the same as for the variant A213648 where 11 is replaced by 1, while for the other variants A262212 - A262220 (where the repeated term is 2, ..., 10) the fixed points are different, see A213891 - A213899. - M. F. Hasler, Sep 15 2015

In a variant of A213891, multiply n by a number with simple continued fraction [n,10,10,...,10,n] and increase the number of 10's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are

2 * [2, 10, 2] = [4, 5, 4],

3 * [3, 10, 10, 10, 3] = [9, 3, 2, 1, 2, 1, 2, 3, 9],

4 * [4, 10, 10, 10, 4] = [16, 2, 1, 1, 9, 1, 1, 2, 16],

5 * [5, 10, 5] = [25, 2, 25],

6 * [6, 10, 10, 10, 6] = [36, 1, 1, 2, 6, 2, 1, 1, 36],

7 * [7, 10, 10, 10, 10, 10, 10, 10, 7] = [49, 1, 2, 3, 1, 6, 2, 1, 2, 2, 2, 1, 2, 6, 1, 3, 2, 1, 49].

The number of 10's needed defines the sequence h(n) = 1, 3, 3, 1, 3, 7, 7, 11, 1, ... (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) = 10*f(n-1) + f(n-2), A041041, A015456, etc. This would mean that a prime is in the sequence A213899 if and only if it divides some term in each of the sequences satisfying f(n) = 10*f(n-1) + f(n-2).

The sequence h() is given in A262220. - M. F. Hasler, Sep 15 2015

Sequence A213891 lists fixed points of this sequence.

Sequence A213899 lists fixed points of this sequence.

Sequence A261311 lists fixed points of this sequence.

It is surprising that the variant A213900 with 11 instead of 12 has the same fixed points A000057 as the variant A213648 with 1 instead of 12, but other variants (A262212 - A262220 and this one) have different sets of fixed points (A213891 - A213899 and A261311).

Surprisingly, the variant A213900 with 11 instead of 12 has the same fixed points A000057 as the variant with 1 instead of 12, but other variants (A262212 - A262220 and A262211) have different sets of fixed points (A213891 - A213899 and this).

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