OFFSET
2,1
COMMENTS
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
From M. Jeremie Lafitte (Levitas), Aug 19 2026: (Start)
The conjectures in the preceding comments are true. The fixed points of this sequence are exactly the terms of A000057, and these are precisely the primes p that divide some term of every integer sequence satisfying f(k+2) = 11*f(k+1)+f(k).
By the Formula section and Lehmer's maximal-rank theorem, every fixed point is indeed prime; see A214028. To identify them, put T_a(x) = a+1/x, M_a = {{a,1},{1,0}}, C = {{5,0},{3,1}}. The identity M_1^5*C = C*M_11 = {{55,5},{34,3}} shows that T_11 is conjugate to T_1^5 on P^1(F_p) for every prime p!=5.
For an odd prime p!=5, a complete Fibonacci orbit implies (5/p)=-1, hence gcd(5,p+1)=1; therefore T_1^5 is a cycle of length p+1. Conversely, transitivity of T_1^5 implies transitivity of T_1, so T_1 is itself a cycle of length p+1. For p=2 or p=5, T_11=T_1 over F_p.
Thus T_1 and T_11 have complete projective orbits for exactly the same primes, namely the terms of A000057. The complete-cycle interpretation gives the universal-divisibility characterization above. (End)
FORMULA
For n>=2, let U(0)=0, U(1)=1, and U(k+2) = 11*U(k+1)+U(k). The canonical-endpoint equivalence in A213648, with 11 in place of 1, gives a(n) = min{m>=1: n | U(m+1)} = z_U(n)-1, where z_U(n) = min{k>=1: n | U(k)}. Consequently, n is a fixed point of this sequence if and only if z_U(n) = n+1. - M. Jeremie Lafitte (Levitas), Aug 19 2026
MATHEMATICA
f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; f[11, #] & /@ Range[2, 120] (* Michael De Vlieger, Sep 16 2015 *)
PROG
(PARI) \\ This PARI program will generate sequence A000057
{a(n) = local(t, m=1); if( n<2, 0, while( 1,
t = contfracpnqn( concat([n, vector(m, i, 11), n]));
t = contfrac(n*t[1, 1]/t[2, 1]);
if(t[1]<n^2 || t[#t]<n^2, m++, break));
m)};
for(k=1, 1500, if(k==a(k), print1(a(k), ", ")));
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Art DuPre, Jun 24 2012
STATUS
approved