%I #40 Feb 22 2021 13:27:37
%S 3,52,58130,684605953,18209086488275508678,
%T 62003660502140726224833872663126972818,
%U 8808009238542973573505163998155451603728137116913210531450093169728092387825
%N Sequence gives the denominators, in increasing values, of Egyptian fractions such that their sum has the concatenation of these denominators as decimal part. Case a(1) = 3.
%C There are only three possible sequences of this kind: one starting from 3 (this sequence), another from 4 (A304286) and another from 10 (A302933).
%C Next term a(8) has 152 digits (see b-file). - _Giovanni Resta_, Apr 16 2018
%H Giovanni Resta, <a href="/A302932/b302932.txt">Table of n, a(n) for n = 1..8</a>
%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 3 because 1/3 = 0.3333...
%e Then the next integer is 52 because 1/3 + 1/52 = 0.352564... and so on.
%e The sum is 0.3 52 58130 684605953 18209086488275508678 ...
%p P:=proc(q) local a,b,d,n; a:=1/3; b:=1; d:=3; 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;
%p od; end: P(10^20);
%Y Cf. A302933, A303388, A304286.
%K nonn,base
%O 1,1
%A _Paolo P. Lava_, Apr 16 2018
%E a(4)-a(7) from _Giovanni Resta_, Apr 16 2018