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%I #33 Feb 10 2020 18:26:10
%S 7,11,23,47,127,139,211,223,251,331,367,379,383,463,487,499,607,619,
%T 691,727,739,743,811,823,863,887,967,971,983,1051,1063,1087,1171,1291,
%U 1303,1327,1367,1423,1447,1451,1459
%N Fixed points of a sequence h(n) defined by the minimum number of 6's in the relation n*[n,6,6,...,6,n] = [x,...,x] between simple continued fractions.
%C In a variant of A213891, multiply n by a number with simple continued fraction [n,6,6,...,6,n] and increase the number of 6'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, 6, 2] = [4, 3, 4],
%C 3 * [3, 6, 3] = [9, 2, 9],
%C 4 * [4, 6, 6, 6, 4] = [16, 1, 1, 1, 5, 1, 1, 1, 16],
%C 5 * [5, 6, 6, 6, 6, 5] = [25, 1, 4, 3, 3, 4, 1, 25],
%C 6 * [6, 6, 6] = [36, 1, 36],
%C 7 * [7, 6, 6, 6, 6, 6, 6, 6, 7] = [50, 7, 2, 1, 4, 4, 4, 1, 2, 7, 50].
%C The number of 6's needed defines the sequence h(n) = 1, 1, 3, 4, 1, 7, 7, 5, 9, ... (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 numbers is analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequences (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 generalized Fibonacci sequences satisfying f(n) = 6*f(n-1) + f(n-2), A005668, A015451, A179237, etc. This would mean that a prime is in the sequence if and only if it divides some term in each of the sequences satisfying f(n) = 6*f(n-1) + f(n-2).
%C The above sequence h() is recorded as A262216. - _M. F. Hasler_, Sep 15 2015
%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[6, #] == # &] (* _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,6), 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. A000057, A213891 - A213894, A213896 - A213899, A261311; A213358.
%Y Cf. A213648, A262212 - A262220, A213900, A262211.
%K nonn
%O 1,1
%A _Art DuPre_, Jun 23 2012
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