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![[parent]](http://images.planetmath.org:8080/images/uparrow.png)
sequences  and  are divisible by 
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(Derivation)
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"sequences  and  are divisible by  " is owned by perucho.
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(view preamble | get metadata)
Cross-references: divisible, clear, base, length, palindromic number, even, sequences, application, integer, series, finite, alternating
This is version 3 of sequences  and  are divisible by  , born on 2006-09-11, modified 2006-09-14.
Object id is 8340, canonical name is SequencesB2n1AndB2n11AreDivisibleByB1.
Accessed 850 times total.
Classification:
| AMS MSC: | 11A63 (Number theory :: Elementary number theory :: Radix representation; digital problems) |
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Pending Errata and Addenda
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![$\displaystyle S_{m+1}(\mu)=\frac{(-1)^\mu[1-(-1)^{m+1}b^{m+1}]}{b+1}= \sum_{i=0}^m(-1)^{i+\mu}b^i.$](http://images.planetmath.org:8080/cache/objects/8340/js/img3.png "$\displaystyle S_{m+1}(\mu)=\frac{(-1)^\mu[1-(-1)^{m+1}b^{m+1}]}{b+1}= \sum_{i=0}^m(-1)^{i+\mu}b^i.$")
![$\displaystyle (EPN)_n=\sum_{k=0}^{n-1}b_k(b^{2n-1-k}+b^k)= \sum_{k=0}^{n-1}\frac{b_k}{b^k}[(b^{2n-1}+1)+(b^{2k}-1)],$](http://images.planetmath.org:8080/cache/objects/8340/js/img6.png "$\displaystyle (EPN)_n=\sum_{k=0}^{n-1}b_k(b^{2n-1-k}+b^k)= \sum_{k=0}^{n-1}\frac{b_k}{b^k}[(b^{2n-1}+1)+(b^{2k}-1)],$")
