I see the sequence has a team of 3 terms that repeats: twin first term come get 2nd term, include three to get third term, repeat.
What about: $2, 3, 6, 7, 14, 15, 30,... $?
Again the sequence has actually a group of 3 terms that repeats: add one to first term come get second term, then twin second term come get third term.
How perform you compute the $n$th term of together sequences directly, without iterating through all coming before terms?
edited Jul 21 "14 at 20:30
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inquiry Jul 20 "14 in ~ 11:13
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For example, consider the first sequence. One can write two recurrence relations,$$F_2k=2F_2k-1,qquad F_2k+1=F_2k+3,$$and use them to deduce a relation entailing only weird terms:$$F_2k+1=2F_2k-1+3.$$The basic solution of this is$$F_2k+1=alphacdot 2^k-3,$$and the worth of the continuous $alpha=2$ is fixed by $F_1=1$. Hence$$F_2k+1=2^k+2-3,qquad F_2k=2^k+2-6.$$
edited Jul 20 "14 in ~ 18:09
answered Jul 20 "14 in ~ 11:45
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Denote the $n$-th ax by $x_n$ and also have a look in ~ the state withodd index $n$. Based upon what friend remarked yourself:
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This starts to look like
$x_2n+1=2^n+left(2^n-1+2^n-2+cdots+1 ight)3=2^n+left(2^n-1 ight)3=2^n+2-3$
This deserve to be confirmed by induction and its easy currently to discover that $x_2n=x_2n+1-3=2^n+2-6$.
On a sortlike method you involved $y_2n+1=2^n+2-2$ and also $y_2n=2^n+1-1$ wherein $y_n$denotes the $n$-th term of the 2nd sequence.
edited Jul 20 "14 at 12:02
answer Jul 20 "14 at 11:49
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A an ext general approach to fixing for the $n$th term of such sequences uses matrix multiplication. Expect the even terms room nonzero constant $r eq 1$ times the coming before odd terms, when the odd state are continuous $d$ add to the preceding also terms. Us have:
$$ eginpmatrix 1 & 0 & d \ 1 & 0 & 0 \ 0 & 0 & 1 endpmatrix eginpmatrix a_2n \ a_2n-1 \ 1 endpmatrix = eginpmatrix a_2n+1 \ a_2n \ 1 endpmatrix $$
$$ eginpmatrix r & 0 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 endpmatrix eginpmatrix a_2n+1 \ a_2n \ 1 endpmatrix = eginpmatrix a_2n+2 \ a_2n+1 \ 1 endpmatrix $$
Combining this by procession multiplication offers the dual step:
$$ eginpmatrix r & 0 & rd \ 1 & 0 & d \ 0 & 0 & 1 endpmatrix eginpmatrix a_2n \ a_2n-1 \ 1 endpmatrix = eginpmatrix a_2n+2 \ a_2n+1 \ 1 endpmatrix $$
The difficulty is then diminished to finding a closed kind for herbal powers the the matrix:
$$ A = eginpmatrix r & 0 & rd \ 1 & 0 & d \ 0 & 0 & 1 endpmatrix $$
which have the right to be done by diagonalization, because $A$ has three distinctive eigenvalues $0,1,r$.
Represent $A$ with respect come the matching basis that eigenvectors:
$$ left eginpmatrix 0 \ 1 \ 0 endpmatrix, eginpmatrix -rd \ - d \ r-1 endpmatrix, eginpmatrix r \ 1 \ 0 endpmatrix ight $$
and the result similarity transformation diagonalizes $A$, say $A = S D S^-1$ where $D= operatornamediag(0,1,r)$. Thus, presume an initial value $a_1$ and also $a_2 = ra_1$:
$$ A^n eginpmatrix a_2 \ a_1 \ 1 endpmatrix = S D^n S^-1 eginpmatrix ra_1 \ a_1 \ 1 endpmatrix =eginpmatrix a_2n+2 \ a_2n+1 \ 1 endpmatrix $$
The strength of $D$ are clearly $D^n = operatornamediag(0,1,r^n)$, therefore this gives a straight expression for any kind of terms in the sequence beginning from $a_1$:
$$ a_2n+1 = r^n a_1 + fracr^n -1r-1 d $$
$$ a_2n+2 = r a_2n+1 = r^n+1 a_1 + fracr^n -1r-1 rd $$
taking benefit of the calculate DanielV carried out in the Comment below.
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This procession multiplication method can it is in modified come handle much more general mixture of arithmetic and geometric rules.