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A213897
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Fixed points of a sequence h(n) defined by the minimum number of 8's in the relation n*[n,8,8,...,8,n] = [x,...,x] between simple continued fractions.
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4
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3, 7, 23, 31, 71, 107, 131, 139, 163, 199, 211, 227, 283, 347, 367, 379, 419, 431, 439, 487, 499, 503, 547, 571, 607, 619, 643, 691, 719, 751, 787, 811, 823, 827, 907, 911, 983, 991, 1031, 1051, 1091, 1151, 1163, 1231, 1303, 1319, 1367, 1399, 1423, 1439, 1459, 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,8,8,..,8,n] and increase the number of 8's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 8, 2] = [4, 4, 4],
3 * [3, 8, 8, 8, 3] = [9, 2, 1, 2, 2, 2, 1, 2, 9],
4 * [4, 8, 4] = [16, 2, 16],
5 * [5, 8, 8, 5] = [25, 1, 1, 1, 1, 1, 1, 25],
6 * [6, 8, 8, 8, 6] = [36, 1, 2, 1, 4, 1, 2, 1, 36],
7 * [7, 8, 8, 8, 8, 8, 8, 8, 7] = [49, 1, 6, 4, 3, 2, 1, 2, 1, 2, 3, 4, 6, 1, 49].
The number of 8's needed defines the sequence h(n) = 1, 3, 1, 2, 3, 7, 1, 11, 5,.. (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)=8*f(n-1)+f(n-2), A041025, A015454, etc. This would mean that a prime is in the sequence A213897 if and only if it divides some term in each of the sequences satisfying f(n)=8*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[8, #] == # &] (* 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, 8), 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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