%I #33 Feb 10 2020 18:26:16
%S 3,7,31,43,47,71,107,151,167,179,211,223,239,251,271,283,419,431,463,
%T 467,487,491,523,547,563,571,631,839,859,883,907,967,971,1087,1103,
%U 1171,1187,1279,1283,1291,1367,1399,1423,1459,1471,1483,1487,1499
%N Fixed points of a sequence h(n) defined by the minimum number of 10's in the relation n*[n,10,10,...,10,n] = [x,...,x] between simple continued fractions.
%C In a variant of A213891, multiply n by a number with simple continued fraction [n,10,10,...,10,n] and increase the number of 10'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, 10, 2] = [4, 5, 4],
%C 3 * [3, 10, 10, 10, 3] = [9, 3, 2, 1, 2, 1, 2, 3, 9],
%C 4 * [4, 10, 10, 10, 4] = [16, 2, 1, 1, 9, 1, 1, 2, 16],
%C 5 * [5, 10, 5] = [25, 2, 25],
%C 6 * [6, 10, 10, 10, 6] = [36, 1, 1, 2, 6, 2, 1, 1, 36],
%C 7 * [7, 10, 10, 10, 10, 10, 10, 10, 7] = [49, 1, 2, 3, 1, 6, 2, 1, 2, 2, 2, 1, 2, 6, 1, 3, 2, 1, 49].
%C The number of 10's needed defines the sequence h(n) = 1, 3, 3, 1, 3, 7, 7, 11, 1, ... (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 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 sequences satisfying f(n) = 10*f(n-1) + f(n-2), A041041, A015456, etc. This would mean that a prime is in the sequence A213899 if and only if it divides some term in each of the sequences satisfying f(n) = 10*f(n-1) + f(n-2).
%C The sequence h() is given in A262220. - _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[10, #] == # &] (* _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,10), 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 - A213898, A261311.
%Y Cf. A213648, A262212 - A262220, A213900, A262211.
%K nonn
%O 1,1
%A _Art DuPre_, Jun 24 2012
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