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

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

Table of n, a(n) for n=0..99.

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

Adjacent sequences: A159197 A159198 A159199 * A159201 A159202 A159203

Keyword

cons,nonn

Author

Eric Desbiaux, Apr 06 2009

Extensions

Comments edited by Petros Hadjicostas, Jun 19 2019

Status

approved