%I
%S 2,11,31,43,47,67,79,103,127,199,211,223,263,307,311,383,431,439,463,
%T 467,499,523,563,571,587,691,719,751,811,839,863,883,911,967,991,1051,
%U 1063,1087,1091,1123,1151,1231,1307,1327,1399,1447,1451,1459,1483,1499
%N 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.
%C 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
%C 2 * [2, 9, 9, 2] = [4, 4, 1, 1, 4, 4],
%C 3 * [3, 9, 3] = [9, 3, 9],
%C 4 * [4, 9, 9, 9, 9, 9, 4] = [16, 2, 3, 1, 1, 1, 1, 8, 1, 1, 1, 1, 3, 2, 16] ,
%C 5 * [5, 9, 9, 9, 9, 5] = [25, 1, 1, 4, 1, 1, 1, 1, 1, 1, 4, 1, 1, 25],
%C 6 * [6, 9, 9, 9, 9, 9, 6] = [36, 1, 1, 1, 13, 6, 13, 1, 1, 1, 36],
%C 7 * [7, 9, 9, 9, 9, 9, 7] = [49, 1, 3, 3, 6, 1, 6, 3, 3, 1, 49].
%C The number of 9's needed defines the sequence h(n) = 2, 1,5, 4, 5, 5, 5, 1, 14,... (n>=2).
%C The current sequence contains the fixed points of h, i. e., those n where h(n)=n.
%C 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(n1)+f(n2) with arbitrary positive integer values for f(1) and f(2)) it refers to the sequences satisfying f(n)=9*f(n1)+f(n2) 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(n1)+f(n2).
%t 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 *)
%o (PARI)
%o {a(n) = local(t, m=1); if( n<2, 0, while( 1,
%o t = contfracpnqn( concat([n, vector(m,i,9), n]));
%o t = contfrac(n*t[1,1]/t[2,1]);
%o if(t[1]<n^2  t[#t]<n^2, m++, break));
%o m)};
%o for(k=1,1500,if(k==a(k),print1(a(k),", ")));
%Y Cf. A213358; A000057, A213891  A213897, A213899, A261311.
%Y Cf. A213648, A262212  A262220, A213900, A262211.
%K nonn
%O 1,1
%A _Art DuPre_, Jun 24 2012
