20 web 2.0 Backlinks DA 30 Plus social Bookmarking Tiers for $5

20 web 2.0 Backlinks DA 30 Plus social Bookmarking Tiers

20 web 2.0 Backlinks DA 30 Plus + Social Bookmarking Tiers

I will create 20 web 2.0 Backlinks and tier them with social bookmarking backlinks as shown in the image.

All these web 2.0s have a domain autority of 30 Plus (100%)

What i need from you :

  1. 1 url
  2. 3 -5 keywords

I scrape articles related to your keywords and spin them and insert 2 links, 1 keyword and a generic anchor (click here, read more…etc) per url.
These web 2.0s will boost your targeted keywords to rank higher in the search engines. don’t expect to get lots of traffic from these blogs, they are purely created to boost rankings.

To your succes,

Michel (Jammerjoh)

.(tagsToTranslate)web2.0(t)backlink(t)ranking(t)booster(t)links(t)seo

Algorithm Analysis – Evidence of the Master Thorem case with tiers and b = 2

This question refers to my previous one.

The following repetition relationship is given: $ T (n) = aT (floor / floor) + f (n) $ from where $ a geq 1, b> 1 $ and $ f (n) = theta (n ^ { log_ba}) $, Then $ T (n) = n ^ { log_ba} + sum_ {j = 0} ^ { lfloor log_bn rfloor – 1} a ^ {j} f (n_j) = n ^ { log_ba} + g ( n) $, It should be proved $ g (n) = omega (n ^ { log_ba} logn) $,

It does not seem to be a difficult problem, as there is evidence of a similar (for the upper bound and a presence of the ceiling operation on recursive calls) in the Introduction to the algorithms of Cormen, Leiserson, Rivest, Stein, But I've found it harder to handle $ b = 2 $:

$ n_j = begin {cases}
n, & text {if j = 0} \
lfloor n_ {j-1} / b rfloor, & text {if j> 0}
end {cases} $

$ n_j geq n / b ^ {j} – sum_ {i = 0} ^ {j-1} 1 / b ^ {i} geq n / b ^ {j} – sum_ {i = 0} ^ { infty} 1 / b ^ {i} = n / b ^ {j} – b / (b-1) $

Under the condition there is such a constant $ c $ The

$ g (n) geq $

$ c sum_ {j = 0} ^ { lfloor log_bn rfloor – 1} a ^ {j} ( frac {n} {b ^ {j}} – frac {b} {b-1}) ^ {log_ba} = $

$ cn ^ {log_ba} sum_ {j = 0} ^ { lfloor log_b n rfloor – 1} (1- frac {b ^ {j}} {n} frac {b} {b-1} ) ^ {log_ba} geq $ //from where $ frac {b ^ {j}} {n} le 1 $

$ cn ^ {log_ba} sum_ {j = 0} ^ { lfloor log_b n rfloor – 1} (1- frac {b ^ { lfloor log_bn lfloor -1}} {n} frac {b } {b-1}) ^ {log_ba} = $

$ cn ^ {log_ba} sum_ {j = 0} ^ { lfloor log_b n rfloor – 1} (1- frac {b ^ { lfloor log_bn lfloor}} {n} frac {1} { b-1}) ^ {log_ba} geq $

$ c n ^ {log_ba} sum_ {j = 0} ^ { lfloor log_b n rfloor – 1} (1- frac {1} {b-1}) ^ {log_ba} $

Note iff $ b = 2 $ when $ (1- frac {1} {b-1}) ^ {log_ba} = 0 $ and can not be given as a permissible constant with respect to $ Omega $ (it has to be a positive one).

I tried to prove this special case in other ways, though $ n_j geq frac {n} {2 ^ {j}} – (2 – frac {1} {2 ^ {j-1}}) $ (It is the sum of the geometric progression without limit), but ultimately failed on the same problem.