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A214030 The multiplication continued fraction sequence for 2,3,2,3...

%I A214030 #20 Jul 30 2012 14:05:05
%S A214030 0,1,5,3,4,5,2,3,5,9,4,11,13,5,29,3,15,5,19,19,5,9,23,11,13,17,11,14,
%T A214030 29,7,7,29,15,14,11,37,19,41,19,41,5,13,19,29,23,47,11,20,49,47,27,25,
%U A214030 17,4,11,59,29,28,59,11,7,5,15,69,29,65,15,23,29,9,11,36,37,149
%N A214030 The multiplication continued fraction sequence for 2,3,2,3...
%C A214030 This is the first sequence which comes from continued fractions which cannot, so far, be looked at as an apparition sequence for a linearly recursive sequence.
%C A214030 The first term a(1), is always 0. We will always refer to the continued fraction [a,b,c] for short, when we really mean the continued fraction with quotients [a,b,c].
%C A214030 To calculate a(2), we consider [2,2], and take a(2) to be the least number of terms of the sequence 2,3,2,3,... which will have the property that 2[2,2,3,2,3,...,2] = [2^2+k,...,2^2+l], where k,l>=0. We find that 2[2,2,2]=[4,1,4], and both the first term 4 and the last term, 4 are >= 2^2, so 1 is the correct number of terms from 2,3,2,3,.... Thus we have shown that a(2)=1.
%C A214030 Next, to calculate a(3), we consider the sequence [3,2,3,...,3] and notice that if we put only one term from the sequence 2,3,2,3,... between the 3s at the ends, we get 3[3,2,3]=[3^2+1,3,2] and the last term 2 is not >= 3^2, so we consider 3[3,2,3,3]=[3^2+1,3,3,2] and 2 is not >= 3^2, so we look at 3[3,2,3,2,3]=[3^2+1,3,4,4], and 4, the last term, is still not >= 3^2. Next, we consider 3[3,2,3,2,3,3] = [3^2+1,3,4,3,4] and we see that the last term 4 is still not >= 3^2. Finally when we put 5 terms of the sequence 2,3,2,3,.. between, the 3s, we get 3[3,2,3,2,3,2,3] = [10,3,4,3,10], and now the first element 10 is >=3^2, and the last number 10 is also >=3^2, and this proves that a(3)=5. By using the PARI program below, we get the sequence a(n) above, shown in the data portion.
%o A214030 (PARI)
%o A214030 {b23(n)=local(t,m=1,s=[n]); if (n<2,0,while(1,
%o A214030 if(m%2,s=concat(s,2),s=concat(s,3));
%o A214030 t=contfracpnqn(concat(s,n));
%o A214030 t=contfrac(n*t[1,1]/t[2,1]);
%o A214030 if(t[1]<n^2||t[#t]<n^2,m++,break));m)};
%o A214030 /* To print the sequence to the screen,use */
%o A214030 for(i=1,500,print1(b23(i),", "));
%Y A214030 Cf. A213891, A213892.
%K A214030 nonn
%O A214030 1,3
%A A214030 _Art DuPre_, Jul 12 2012