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A213894
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Fixed points of a sequence h(n) defined by the minimum number of 5's in the relation n*[n,5,5,...,5,n] = [x,...,x] between simple continued fractions.
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3
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2, 3, 11, 19, 31, 43, 47, 79, 127, 131, 163, 211, 251, 271, 307, 311, 331, 367, 379, 443, 503, 563, 599, 607, 659, 743, 751, 823, 839, 859, 887, 907, 911, 947, 967, 1063, 1087, 1091, 1123, 1163, 1171, 1187, 1259, 1279, 1291, 1303, 1307, 1319, 1423, 1447, 1471, 1487
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OFFSET
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1,1
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COMMENTS
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In a variant of A213891, multiply n by a number with simple continued fraction [n,5,5,...,5,n] and increase the number of 5's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 5, 5, 2] = [4, 2, 1, 1, 2, 4],
3 * [3, 5, 5, 5, 3] = [9, 1, 1, 2, 1, 2, 1, 1, 9],
4 * [4, 5, 5, 5, 5, 5, 4] = [16, 1, 3, 2, 1, 4, 1, 2, 3, 1, 16] ,
5 * [5, 5, 5] = [25, 1, 25].
The number of 5's needed defines the sequence h(n) = 2, 3, 5, 1, 11, 5, 5, 3, 5, 11, 11, ... (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 generalized Fibonacci sequences satisfying f(n) = 5*f(n-1) + f(n-2), A052918, A015449, A164581, 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) = 5*f(n-1) + f(n-2).
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LINKS
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MATHEMATICA
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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[5, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)
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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, 5), 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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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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