login
Sequence gives the denominators, in increasing values from the second term on, of Egyptian fractions whose sum has the concatenation of these denominators as decimal part. Case a(1) = 5.
16

%I #17 Aug 05 2019 05:34:57

%S 5,3,524,414362,964900433080,568112044247363806135385,

%T 532690079132413282557643073103806291708903760733

%N Sequence gives the denominators, in increasing values from the second term on, of Egyptian fractions whose sum has the concatenation of these denominators as decimal part. Case a(1) = 5.

%C Next term has 97 digits. - _Giovanni Resta_, Jun 06 2018

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/EgyptianFraction.html">Egyptian fraction</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TrottConstants.html">Trott constants (similar but with continued fractions)</a>

%e We start from 5: 1/5 = 0.2. At the beginning we have 2 instead of 5 as first decimal digit but the second term will fix it.

%e In fact the next integer is 3 and 1/5 + 1/3 = 0.5333... and so on.

%e The sum is 0.5 3 524 414362 964900433080 ...

%p P:=proc(q) local a,b,d,n; a:=1/5; b:=ilog10(1/a)+1; d:=1/a; print(d);

%p for n from 1 to q do if trunc(evalf(a+1/n,100)*10^(b+ilog10(n)+1))=d*10^(ilog10(n)+1)+n then b:=b+ilog10(n)+1; d:=d*10^(ilog10(n)+1)+n; a:=a+1/n; print(n); fi; od; end: P(10^20);

%Y Cf. A302932, A302933, A303388, A304285, A304286.

%K nonn,base

%O 1,1

%A _Paolo P. Lava_, Jun 06 2018

%E a(4)-a(7) from _Giovanni Resta_, Jun 06 2018