A364872 Lexicographically earliest continued fraction which is its own unit fraction series.

2, 1, 2, 2, 5, 95, 137447, 19092121105, 1456654254113777258001, 8728918703159963392919895262580124849062181

Offset

0,1

Comments

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 unit fraction series - they are dense on the interval (2,oo).

Links

Table of n, a(n) for n=0..9.

Example

The partial continued fraction must always be strictly larger than the partial unit fractions:
[1]      cannot be  since            1 = 1.
[2]      can be     since            2 > 1/2.
[2,1]    can be     since        2+1/1 > 1/2+1/1.
[2,1,1]  cannot be  since  2+1/(1+1/1) = 1/2+1/1+1/1.
[2,1,2]  can be     since  2+1/(1+1/2) > 1/2+1/1+1/2.
...
sum(1/a[n]) = 2.71053359137351078733864566... (A364873).

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};

Crossrefs

Cf. A364873.

Sequence in context: A340059 A095291 A285003 * A284002 A093659 A306714

Adjacent sequences: A364869 A364870 A364871 * A364873 A364874 A364875

Keyword

nonn,cofr

Author

Rok Cestnik, Aug 11 2023

Status

approved