A081088 Smallest partial quotients of a series of continued fractions that have a sum of unity, where the n-th continued fraction term is [0; a(n),a(n-1),...,a(1)], so that sum(n>0, [0; a(n),a(n-1),...,a(1)] ) = 1, with a(1)=2.

2, 2, 10, 260, 703300, 128651592765800, 11640481755119007104771565805489000, 17432395734015190050301181934013684788461125501100342391858949624062957005321114000

Offset

1,1

Comments

a(n+1) appears to be divisible by a(n) for n>0; a(n+1)/a(n) = A081089(n). Also log(a(n+1))/log(a(n)) -> 1+sqrt(2). The 8th term has 79 digits, while the 9th term has 199 digits.

The convergents of the continued fraction series equal the ratios of the terms of A081090: [0; a(n),a(n-1),...,a(1)] = A081090(n)/A080190(n+1) for n>1; thus a(n) = ( A081090(n+1) - A081090(n-1) )/A081090(n) for n>1. - Paul D. Hanna, Mar 05 2003

Links

Table of n, a(n) for n=1..8.

Formula

a(n) = A081090(n)*A081090(n-1) for n>2, where A081090(n)^2 + 1 = a(n+1)/a(n).

Example

1 = [0;2] + [0;2,2] + [0;10,2,2] + [0;260,10,2,2] + [0;703300,260,10,2,2] + [0;128651592765800,703300,260,10,2,2] +... = .5 + .4 + .0961538461 + .0038447319 + .0000014218 + ...
At n=4, [0;260,10,2,2] = A081090(4)/A081090(5) = 52/13525. - Paul D. Hanna, Mar 05 2003

Crossrefs

Cf. A081086, A081089, A081090.

Sequence in context: A011248 A069240 A000371 * A236369 A001038 A283454

Adjacent sequences: A081085 A081086 A081087 * A081089 A081090 A081091

Keyword

cofr,nonn

Author

Hans Havermann and Paul D. Hanna, Mar 05 2003

Status

approved