A364873 Decimal expansion of the lexicographically earliest continued fraction which equals its own sum of reciprocals.

2, 7, 1, 0, 5, 3, 3, 5, 9, 1, 3, 7, 3, 5, 1, 0, 7, 8, 7, 3, 3, 8, 6, 4, 5, 6, 6, 2, 0, 4, 8, 1, 7, 0, 1, 1, 1, 5, 1, 8, 3, 3, 4, 9, 9, 3, 0, 7, 0, 4, 4, 7, 6, 3, 7, 9, 4, 3, 4, 3, 9, 0, 9, 5, 0, 8, 3, 0, 4, 7, 0, 0, 0, 8, 2, 0, 7, 6, 8, 6, 1, 8, 7, 3, 1, 3, 1, 8, 2, 2, 1, 9, 6, 8, 7, 2, 2

Offset

1,1

Comments

This continued fraction (A364872) is the earliest infinite sequence {a0,a1,a2,a3,...} such that: a0+1/(a1+1/(a2+1/(a3+...))) = 1/a0 + 1/a1 + 1/a2 + 1/a3 + ....

There are infinitely many real numbers whose continued fraction is also their sum of reciprocals - they are dense on the interval (2,oo).

Links

Table of n, a(n) for n=1..97.

Example

2.71053359137351078733864566...

Prog

(PARI)
cf(a) = my(m=contfracpnqn(a)); m[1, 1]/m[2, 1];
uf(a) = sum(i=1, #a, 1/a[i]);
A364872(N) = {a=[2]; for(i=2, N, a=concat(a, if(cf(a)==uf(a), a[i-1], ceil(1/(cf(a)-uf(a))))); while(cf(a)<=uf(a), a[i]++)); a};
A364873(N) = {t=2; while(floor(10^N*cf(A364872(t))) != floor(10^N*cf(A364872(t+1))), t++); digits(floor(10^(N-1)*cf(A364872(t))))};

Crossrefs

Cf. A364872.

Sequence in context: A113651 A021373 A199950 * A011047 A347589 A345930

Adjacent sequences: A364870 A364871 A364872 * A364874 A364875 A364876

Keyword

nonn,cons

Author

Rok Cestnik, Aug 11 2023

Status

approved