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?


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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$$F_2k+1=2^k+2-3,qquad F_2k=2^k+2-6.$$


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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:

$x_1=1$

$x_3=2x_1+3=2+3$

$x_5=2x_3+3=2left(2+3 ight)+3=2^2+2.3+3$

$x_7=2x_5+3=2left(2^2+2.3+3 ight)+3=2^3+2^2.3+2.3+3$

et cetera.

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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.


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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.