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Sequence gives the denominators, in increasing values, of Egyptian fractions whose sum has the concatenation of these denominators as decimal part. Case a(1) = 4.
20

%I #22 Aug 05 2019 05:34:42

%S 4,5,316,617610,803725588973,456253083482572713037551,

%T 9436780443304881627624731251391047815103579902912

%N Sequence gives the denominators, in increasing values, of Egyptian fractions whose sum has the concatenation of these denominators as decimal part. Case a(1) = 4.

%C There are only three possible sequences of this kind: one starting from 3 (A302932), another from 4 (this sequence) and another from 10 (A302933).

%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 4: 1/4 = 0.25. At the beginning we have 2 instead of 4 as first decimal digit but the second term will fix it.

%e In fact the next integer is 5 and 1/4 + 1/5 = 0.45 and so on.

%e The sum is 0.4 5 316 617610 803725588973 ...

%p P:=proc(q) local a,b,d,n; a:=1/4; 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.

%K nonn,base

%O 1,1

%A _Paolo P. Lava_, Jun 06 2018

%E a(5)-a(7) from _Giovanni Resta_, Jun 06 2018