A213892 - OEIS

%I #41 Feb 10 2020 18:07:30

%S 2,7,19,31,47,67,71,83,151,163,167,223,227,271,307,331,359,379,431,

%T 463,479,487,499,631,643,683,691,743,787,811,839,863,947,967,1019,

%U 1051,1087,1103,1123,1163,1259,1279,1307,1319,1399,1423,1451,1471

%N Fixed points of a sequence h(n) defined by the minimum number of 3's in the relation n*[n,3,3,...,3,n] = [x,...,x] between simple continued fractions.

%C In a variant of A213891, multiply n by a number with simple continued fraction [n,3,3,...,3,n] and increase the number of 3'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, 3, 3, 2] = [4, 1, 1, 1, 1, 4],

%C 3 * [3, 3, 3] = [9, 1, 9],

%C 4 * [4, 3, 3, 3, 3, 3, 4] = [17, 4, 1, 2, 1, 4, 17],

%C 5 * [5, 3, 3, 5] = [26, 1, 1, 26],

%C 6 * [6, 3, 3, 3, 3, 3, 6] = [37, 1, 4, 2, 4, 1, 37],

%C 7 * [7, 3, 3, 3, 3, 3, 3, 3, 7] = [51, 8, 2, 1, 2, 8, 51].

%C The number of 3's needed defines the sequence h(n) = 2, 1, 5, 2, 5, 7, 5, 9, 2, ... (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) = 3*f(n-1) + f(n-2), A006190, A003688, A052924, 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) = 3*f(n-1) + f(n-2).

%C The above sequence h() is recorded as A262213. - _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[3, #] == # &] (* _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,3), 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, A213893 - A213899, A261311; A213358.

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

%K nonn

%O 1,1

%A _Art DuPre_, Jun 23 2012