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Sequence gives the denominators, in increasing values, of Egyptian fractions such that their alternating sum has the concatenation of these denominators as decimal part. a(1) = 3.
21

%I #21 Apr 25 2018 11:50:01

%S 3,299,98957,118885566690,7993597110943689863079

%N Sequence gives the denominators, in increasing values, of Egyptian fractions such that their alternating sum has the concatenation of these denominators as decimal part. a(1) = 3.

%C Similar to A302932.

%C a(6) > 10^40. - _Giovanni Resta_, Apr 23 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 3 because 1/3 = 0.3333...

%e Then the next integer is 299 because 1/3 - 1/299 = 0.32998885.

%e Next term is 98957 because 1/3 - 1/299 + 1/98957 = 0.3299989571272... and so on.

%e The alternating sum is 0.3 299 98957 118885566690 ...

%p with(numtheory): P:=proc(q) local a,b,d,n,t; a:=1/3; b:=1; d:=3; print(d); t:=1;

%p for n from 1 to q do if trunc(evalf(a+(-1)^t/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)^t/n; t:=t+1; print(n); fi; od; end: P(10^20);

%Y Cf. A302932, A302933.

%K nonn,base,more

%O 1,1

%A _Paolo P. Lava_, Apr 23 2018

%E a(4)-a(5) from _Giovanni Resta_, Apr 23 2018