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A213898
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Fixed points of a sequence h(n) defined by the minimum number of 9's in the relation n*[n,9,9,...,9,n] = [x,...,x] between simple continued fractions.
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4
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2, 11, 31, 43, 47, 67, 79, 103, 127, 199, 211, 223, 263, 307, 311, 383, 431, 439, 463, 467, 499, 523, 563, 571, 587, 691, 719, 751, 811, 839, 863, 883, 911, 967, 991, 1051, 1063, 1087, 1091, 1123, 1151, 1231, 1307, 1327, 1399, 1447, 1451, 1459, 1483, 1499
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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,9,9,..,9,n] and increase the number of 9's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 9, 9, 2] = [4, 4, 1, 1, 4, 4],
3 * [3, 9, 3] = [9, 3, 9],
4 * [4, 9, 9, 9, 9, 9, 4] = [16, 2, 3, 1, 1, 1, 1, 8, 1, 1, 1, 1, 3, 2, 16] ,
5 * [5, 9, 9, 9, 9, 5] = [25, 1, 1, 4, 1, 1, 1, 1, 1, 1, 4, 1, 1, 25],
6 * [6, 9, 9, 9, 9, 9, 6] = [36, 1, 1, 1, 13, 6, 13, 1, 1, 1, 36],
7 * [7, 9, 9, 9, 9, 9, 7] = [49, 1, 3, 3, 6, 1, 6, 3, 3, 1, 49].
The number of 9's needed defines the sequence h(n) = 2, 1,5, 4, 5, 5, 5, 1, 14,... (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 sequence (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)=9*f(n-1)+f(n-2) like A099371, A015455 etc. This would mean that a prime is in the sequence A213898 if and only if it divides some term in each of the sequences satisfying f(n)=9*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[9, #] == # &] (* 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, 9), 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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