A213898 Fixed points of a sequence h(n) defined by the minimum number of 9's in the relation n*[n,9,9,...,9,n] = [x,...,x] between simple continued fractions.

2, 11, 31, 43, 47, 67, 79, 103, 127, 199, 211, 223, 263, 307, 311, 383, 431, 439, 463, 467, 499, 523, 563, 571, 587, 691, 719, 751, 811, 839, 863, 883, 911, 967, 991, 1051, 1063, 1087, 1091, 1123, 1151, 1231, 1307, 1327, 1399, 1447, 1451, 1459, 1483, 1499

Offset: 1

Author

Art DuPre, Jun 24 2012

Keywords

• nonn

Comments

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).

Crossrefs

Cf. A213358; A000057, A213891 - A213897, A213899, A261311.

Cf. A213648, A262212 - A262220, A213900, A262211.

Programs

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]; Select[Range[2, 1000], f[9, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)

PARI

{a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,9), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]