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 hardlungemine.com
inquiry Jul 20 "14 in ~ 11:13 GuestGuest
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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 start wearing purpleStart wearing violet
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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.

You are watching: 1-2-5-10-13-26-29-48

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 drhabdrhab
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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.