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A159200 Decimal expansion of Sum_{k >= 1} (1/(10^(4*k + 2) - 1)) - (1/(10^(2*k + 1) - 1)), negated.

%I A159200 #48 Feb 16 2025 08:33:10
%S A159200 0,0,1,0,1,0,1,0,2,0,1,0,1,0,3,0,1,0,1,0,3,0,1,0,2,0,3,0,1,0,1,0,3,0,
%T A159200 3,0,1,0,3,0,1,0,1,0,5,0,1,0,2,0,3,0,1,0,3,0,3,0,1,0,1,0,5,0,3,0,1,0,
%U A159200 3,0,1,0,1,0,5,0,3,0,1,0,4,0,1,0,3,0,3,0,1,0,3,0,3,0,3,0,1,0,5,0
%N A159200 Decimal expansion of Sum_{k >= 1} (1/(10^(4*k + 2) - 1)) - (1/(10^(2*k + 1) - 1)), negated.
%C A159200 It equals Sum_{k >= 1} 1/((2^(4*k + 2)*5^(4*k + 2)) - 1) - 1/((2^(2*k + 1)*5^(2*k + 1)) - 1).
%C A159200 Note that Sum_{k >= 1} (1/(10^k - 1)) / Sum_{k >= 1} ((1/(10^(4*k + 2) - 1)) -(1/(10^(2*k + 1) - 1))) = A073668 / Sum_{k >= 1} ((1/(10^(4*k + 2) - 1)) - (1/(10^(2*k + 1) - 1))) = -121.100.
%C A159200 My idea for this decimal expansion came from the Engel expansion of e - 1, i.e., A000027(n) = n, and the Engel expansion of e^(-1), i.e., A059193(n) = 2*(2*n + 1)*(n - 1), which I have transformed into (2*n + 1)^2 - (6*n + 3) (since 2*(2*n + 1)*(n - 1) = (2*n + 1)^2 - (6*n + 3)). It appears that the Engel expansion of 1/e works like a Sundaram sieve.
%C A159200 Decimal expansion of Sum_{n>=0} (d(2*n+1) - 1)/(10^(2*n+1) - 1), where d = A000005. - _Jianing Song_, Apr 12 2021
%H A159200 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/EngelExpansion.html">Engel expansion</a>.
%H A159200 English Wikipedia, <a href="https://en.wikipedia.org/wiki/Sieve_of_Sundaram">Sieve of Sundaram</a>.
%H A159200 French Wikipedia, <a href="http://fr.wikipedia.org/wiki/Crible_de_Sundaram">Crible de Sundaram</a>.
%e A159200 -0.00101010201010301010301020301010303010301010501020301030301...
%o A159200 (PARI) suminf(k=1, 1/(10^(4*k + 2) - 1) - 1/(10^(2*k + 1) - 1)) \\ _Michel Marcus_, Jun 25 2019
%K A159200 cons,nonn
%O A159200 0,9
%A A159200 _Eric Desbiaux_, Apr 06 2009
%E A159200 Comments edited by _Petros Hadjicostas_, Jun 19 2019