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Denominators a(n) of the fractions Sum_{n>=1} {n/a(n)} = 1/a(1) + 2/a(2) + 3/a(3) + ... such that the sum has the concatenation of these denominators as decimal part. Case a(1) = 11.
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%I #15 Jun 10 2018 22:36:29

%S 11,75,795297,93992106533,45114643856137363285734,

%T 5244406139948692731854965782168773850666243474

%N Denominators a(n) of the fractions Sum_{n>=1} {n/a(n)} = 1/a(1) + 2/a(2) + 3/a(3) + ... such that the sum has the concatenation of these denominators as decimal part. Case a(1) = 11.

%C It appears that fractions of this kind exist only for a(1) equal to 3 (A304288), 10 (A304289), 11 (this sequence), 14 (A305662), and 31 (A304663).

%C a(7) has 92 digits. - _Giovanni Resta_, Jun 08 2018

%e 1/11 = 0.090909... At the beginning instead of 11 we have 09 as first decimal digits. Adding the second term this is fixed.

%e 1/11 + 2/75 = 0.117575...

%e 1/11 + 2/75 + 3/795297 = 0.1175795297514...

%e The sum is 0.11 75 795297 93992106533 ...

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

%p t:=2; for n from 1 to q do if trunc(evalf(a+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+t/n; t:=t+1; print(n); fi; od; end: P(10^20,11);

%Y Cf. A302932, A302933, A303388, A304285, A304286, A304287, A304288, A304289, A305662, A305663, A305664, A305665, A305666.

%K nonn,base

%O 1,1

%A _Paolo P. Lava_, Jun 08 2018

%E a(3)-a(6) from _Giovanni Resta_, Jun 08 2018