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).
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
AUTHOR
Art DuPre, Jun 24 2012
STATUS
approved