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

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, 29, 7, 7, 29, 15, 14, 11, 37, 19, 41, 19, 41, 5, 13, 19, 29, 23, 47, 11, 20, 49, 47, 27, 25, 17, 4, 11, 59, 29, 28, 59, 11, 7, 5, 15, 69, 29, 65, 15, 23, 29, 9, 11, 36, 37, 149

Offset: 1

Author

Art DuPre, Jul 12 2012

Keywords

• nonn

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.

Crossrefs

Cf. A213891, A213892.

Programs

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]