OFFSET
1,3
COMMENTS
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.
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].
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.
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.
PROG
(PARI)
{b23(n)=local(t, m=1, s=[n]); if (n<2, 0, while(1,
if(m%2, s=concat(s, 2), s=concat(s, 3));
t=contfracpnqn(concat(s, n));
t=contfrac(n*t[1, 1]/t[2, 1]);
if(t[1]<n^2||t[#t]<n^2, m++, break)); m)};
/* To print the sequence to the screen, use */
for(i=1, 500, print1(b23(i), ", "));
CROSSREFS
KEYWORD
nonn
AUTHOR
Art DuPre, Jul 12 2012
STATUS
approved