A364872 - OEIS

%I #17 Aug 13 2023 03:33:30

%S 2,1,2,2,5,95,137447,19092121105,1456654254113777258001,

%T 8728918703159963392919895262580124849062181

%N Lexicographically earliest continued fraction which is its own unit fraction series.

%C Earliest infinite sequence {a0,a1,a2,a3,...} such that: a0+1/(a1+1/(a2+1/(a3+...))) = 1/a0 + 1/a1 + 1/a2 + 1/a3 + ...

%C There are infinitely many real numbers whose continued fraction is also their unit fraction series - they are dense on the interval (2,oo).

%e The partial continued fraction must always be strictly larger than the partial unit fractions:

%e [1]      cannot be  since            1 = 1.

%e [2]      can be     since            2 > 1/2.

%e [2,1]    can be     since        2+1/1 > 1/2+1/1.

%e [2,1,1]  cannot be  since  2+1/(1+1/1) = 1/2+1/1+1/1.

%e [2,1,2]  can be     since  2+1/(1+1/2) > 1/2+1/1+1/2.

%e ...

%e sum(1/a[n]) = 2.71053359137351078733864566... (A364873).

%o (PARI)

%o cf(a) = my(m=contfracpnqn(a)); m[1,1]/m[2,1];

%o uf(a) = sum(i=1, #a, 1/a[i]);

%o 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};

%Y Cf. A364873.

%K nonn,cofr

%O 0,1

%A _Rok Cestnik_, Aug 11 2023