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 A159200 Decimal expansion of Sum(k=1;inf)(1/(10^(4*k+2)-1))-(1/(10^(2*k+1)-1)). 0
 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, 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, 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 (list; constant; graph; refs; listen; history; text; internal format)
 OFFSET 0,9 COMMENTS = Sum(k=1;inf) 1/((2^(4*k+2)*5^(4*k+2))-1) - 1/((2^(2*k+1)*5^(2*k+1))-1). And Sum(k=1;inf) (1/(10^(k)-1)) / ((1/(10^(4*k+2)-1))-(1/(10^(2*k+1)-1))) = Sum(k=1;inf)A073668 / ((1/(10^(4*k+2)-1))-(1/(10^(2*k+1)-1))) = -1.111... My idea for this decimal expansion come from the Engel Expansion of(e) A000027 and the Engel Expansion of(e^-1) A059193 where a(n)=2(2n+1)(n-1), and i have transformed it into (2n+1)^2-(6n+3). It appears that Engel Expansion of 1/e works like a Sundaram sieve. LINKS EXAMPLE -0.00101010201010301010301020301010303010301010501020301030301... CROSSREFS Sequence in context: A067432 A192174 A129308 * A219201 A033764 A033784 Adjacent sequences:  A159197 A159198 A159199 * A159201 A159202 A159203 KEYWORD cons,nonn AUTHOR Eric Desbiaux, Apr 06 2009 STATUS approved

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