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A159200
Decimal expansion of Sum_{k >= 1} (1/(10^(4*k + 2) - 1)) - (1/(10^(2*k + 1) - 1)), negated.
1
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
OFFSET
0,9
COMMENTS
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).
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.
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.
Decimal expansion of Sum_{n>=0} (d(2*n+1) - 1)/(10^(2*n+1) - 1), where d = A000005. - Jianing Song, Apr 12 2021
LINKS
Eric Weisstein's World of Mathematics, Engel expansion.
English Wikipedia, Sieve of Sundaram.
French Wikipedia, Crible de Sundaram.
EXAMPLE
-0.00101010201010301010301020301010303010301010501020301030301...
PROG
(PARI) suminf(k=1, 1/(10^(4*k + 2) - 1) - 1/(10^(2*k + 1) - 1)) \\ Michel Marcus, Jun 25 2019
CROSSREFS
Sequence in context: A284413 A323879 A129308 * A338021 A318721 A219201
KEYWORD
cons,nonn
AUTHOR
Eric Desbiaux, Apr 06 2009
EXTENSIONS
Comments edited by Petros Hadjicostas, Jun 19 2019
STATUS
approved