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COMMENTS
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In a variant of A213891, multiply n by a number with simple continued fraction [n,7,7,..,7,n] and increase the number of 7's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 7, 7, 2] = [4, 3, 1, 1, 3, 4],
3 * [3, 7, 7, 7, 3] = [9, 2, 2, 1, 1, 1, 2, 2, 9] ,
4 * [4, 7, 7, 7, 7, 7, 4] = [16, 1, 1, 3, 1, 1, 1, 6, 1, 1, 1, 3, 1, 1, 16],
5 * [5, 7, 7, 5] = [25, 1, 2, 2, 1, 25] ,
6 * [6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6] = [36, 1, 5, 3, 1, 4, 10, 1, 2, 2, 4, 2, 2, 1, 10, 4, 1, 3, 5, 1, 36],
7 * [7, 7, 7] = [49, 1, 49] .
The number of 7's needed defines the sequence h(n) = 2, 3, 5, 2, 11, 1, 5, 11, 2,... (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 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)=7*f(n-1)+f(n-2), A054413, A015453, etc. This would mean that a prime is in the sequence A213896 if and only if it divides some term in each of the sequences satisfying f(n)=7*f(n-1)+f(n-2).
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PROG
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(PARI)
{a(n) = local(t, m=1); if( n<2, 0, while( 1,
t = contfracpnqn( concat([n, vector(m, i, 7), n]));
t = contfrac(n*t[1, 1]/t[2, 1]);
if(t[1]<n^2 || t[#t]<n^2, m++, break));
m)};
for(k=1, 1500, if(k==a(k), print1(a(k), ", ")));
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