A213900 The minimum number of 11's in the relation n*[n,11,11,...,11,n] = [x,...,x] between simple terminating continued fractions.

2, 3, 5, 4, 11, 7, 5, 11, 14, 1, 11, 6, 23, 19, 11, 8, 11, 17, 29, 7, 5, 23, 11, 24, 20, 35, 23, 13, 59, 5, 23, 3, 8, 39, 11, 18, 17, 27, 29, 3, 23, 43, 5, 59, 23, 15, 11, 55, 74, 35, 41, 26, 35, 9, 23, 35, 41, 57, 59, 2, 5, 23, 47, 34, 11, 67, 17, 23, 119, 13

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

Links

Table of n, a(n) for n=2..71.

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

Cf. A000057, A213358, A213891 - A213899.

Cf. A213648, A262212 - A262220, A213900.

Sequence in context: A112060 A084933 A355065 * A213648 A302849 A193971

Adjacent sequences: A213897 A213898 A213899 * A213901 A213902 A213903

Keyword

nonn

Author

Art DuPre, Jun 24 2012

Status

approved