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%I #13 Jan 27 2024 14:27:51
%S 3,51,9558,98612460,5887957252067828,
%T 499453888097584722752603691410216,
%U 64374739267553439324757181002125046128361976093811234838816138018
%N a(n) is defined by the condition that the decimal expansion of the Sum_{n>=1} 1/(Sum_{k=1..n} a(k)) = 1/a(1) + 1/(a(1)+a(2)) + 1/(a(1)+a(2)+a(3)) + ... begins with the concatenation of these numbers; also a(1) = 3 and a(n) > a(n-1).
%C At any step only the least value greater than a(n) is taken into consideration. As a(2) we could choose 51, 360, 3363, 33363, ..., 3...363.
%C Next term has 131 digits. - _Giovanni Resta_, Mar 20 2019
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/EgyptianFraction.html">Egyptian fraction</a>
%e 1/3 = 0.3333...
%e 1/3 + 1/(3+51) = 0.351851...
%e 1/3 + 1/(3+51) + 1/(3+51+9558) = 0.3519558884...
%e The sum is 0.3 51 9558 ...
%p P:=proc(q,h) local a,b,d,n,t,z; a:=1/h; b:=length(h);
%p d:=h; print(d); t:=h;for n from 1 to q do
%p z:=evalf(evalf(a+1/(t+n),100)*10^(b+ilog10(n)+1),100);
%p z:=trunc(z-frac(z)); if z=d*10^(ilog10(n)+1)+n then b:=b+ilog10(n)+1;
%p d:=d*10^(ilog10(n)+1)+n; t:=t+n; a:=a+1/(t); print(n);
%p fi; od; end: P(10^20,3)
%Y Cf. A304288, A304289, A305661, A305662, A305663, A305664, A305665, A305666, A305667, A305668, A307007, A307020, A307021, A320023, A320284, A320306, A320307, A320308, A320309, A320335, A320336, A324222, A324223.
%K base,nonn,more
%O 1,1
%A _Paolo P. Lava_, Mar 20 2019
%E a(4)-a(7) from _Giovanni Resta_, Mar 20 2019
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