Skip to main content
MathOverflow

Return to Answer

edited body
Source Link
Barry Cipra
Barry Cipra
  • 5.4k
  • 1
  • 33
  • 29

If I haven't made a mistake then the first few values of your sequence look like this:

$$ (a_{m,n})=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 1 & 2 & 3 & 5 & 8 & ... \newline 1 & 1 & 3 & 9 & 21 & 57 & \newline 1 & 1 & 4 & 21 & 91 & ... & \newline 1 & 1 & 5 & 41 & 329 & ... & \newline 1 & 1 & \vdots & \vdots & \vdots & & \newline \vdots & \vdots & & & & & \end{pmatrix} $$$$ (a_{m,n})=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 1 & 2 & 3 & 5 & 8 & ... \newline 1 & 1 & 3 & 9 & 21 & 59 & \newline 1 & 1 & 4 & 21 & 91 & ... & \newline 1 & 1 & 5 & 41 & 329 & ... & \newline 1 & 1 & \vdots & \vdots & \vdots & & \newline \vdots & \vdots & & & & & \end{pmatrix} $$

Maybe the proposed induction gets easier if you look at the columns: It holds that $a(m,2)=a(m-1,2)+1$ for $m\geq 1$ and $a(0,2)=1$, thus $a(m,2)=m+1$, which is technically easier to deal with then Fibonacci, I guess.

If I haven't made a mistake then the first few values of your sequence look like this:

$$ (a_{m,n})=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 1 & 2 & 3 & 5 & 8 & ... \newline 1 & 1 & 3 & 9 & 21 & 57 & \newline 1 & 1 & 4 & 21 & 91 & ... & \newline 1 & 1 & 5 & 41 & 329 & ... & \newline 1 & 1 & \vdots & \vdots & \vdots & & \newline \vdots & \vdots & & & & & \end{pmatrix} $$

Maybe the proposed induction gets easier if you look at the columns: It holds that $a(m,2)=a(m-1,2)+1$ for $m\geq 1$ and $a(0,2)=1$, thus $a(m,2)=m+1$, which is technically easier to deal with then Fibonacci, I guess.

If I haven't made a mistake then the first few values of your sequence look like this:

$$ (a_{m,n})=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 1 & 2 & 3 & 5 & 8 & ... \newline 1 & 1 & 3 & 9 & 21 & 59 & \newline 1 & 1 & 4 & 21 & 91 & ... & \newline 1 & 1 & 5 & 41 & 329 & ... & \newline 1 & 1 & \vdots & \vdots & \vdots & & \newline \vdots & \vdots & & & & & \end{pmatrix} $$

Maybe the proposed induction gets easier if you look at the columns: It holds that $a(m,2)=a(m-1,2)+1$ for $m\geq 1$ and $a(0,2)=1$, thus $a(m,2)=m+1$, which is technically easier to deal with then Fibonacci, I guess.

corrected an entry in the matrix
Source Link
Barry Cipra
Barry Cipra
  • 5.4k
  • 1
  • 33
  • 29

If I haven't made a mistake then the first few values of your sequence look like this:

$$ (a_{m,n})=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 1 & 2 & 3 & 5 & 8 & ... \newline 1 & 1 & 3 & 9 & 21 & 91 & \newline 1 & 1 & 4 & 21 & 91 & ... & \newline 1 & 1 & 5 & 41 & 329 & ... & \newline 1 & 1 & \vdots & \vdots & \vdots & & \newline \vdots & \vdots & & & & & \end{pmatrix} $$$$ (a_{m,n})=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 1 & 2 & 3 & 5 & 8 & ... \newline 1 & 1 & 3 & 9 & 21 & 57 & \newline 1 & 1 & 4 & 21 & 91 & ... & \newline 1 & 1 & 5 & 41 & 329 & ... & \newline 1 & 1 & \vdots & \vdots & \vdots & & \newline \vdots & \vdots & & & & & \end{pmatrix} $$

Maybe the proposed induction gets easier if you look at the columns: It holds that $a(m,2)=a(m-1,2)+1$ for $m\geq 1$ and $a(0,2)=1$, thus $a(m,2)=m+1$, which is technically easier to deal with then Fibonacci, I guess.

If I haven't made a mistake then the first few values of your sequence look like this:

$$ (a_{m,n})=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 1 & 2 & 3 & 5 & 8 & ... \newline 1 & 1 & 3 & 9 & 21 & 91 & \newline 1 & 1 & 4 & 21 & 91 & ... & \newline 1 & 1 & 5 & 41 & 329 & ... & \newline 1 & 1 & \vdots & \vdots & \vdots & & \newline \vdots & \vdots & & & & & \end{pmatrix} $$

Maybe the proposed induction gets easier if you look at the columns: It holds that $a(m,2)=a(m-1,2)+1$ for $m\geq 1$ and $a(0,2)=1$, thus $a(m,2)=m+1$, which is technically easier to deal with then Fibonacci, I guess.

If I haven't made a mistake then the first few values of your sequence look like this:

$$ (a_{m,n})=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 1 & 2 & 3 & 5 & 8 & ... \newline 1 & 1 & 3 & 9 & 21 & 57 & \newline 1 & 1 & 4 & 21 & 91 & ... & \newline 1 & 1 & 5 & 41 & 329 & ... & \newline 1 & 1 & \vdots & \vdots & \vdots & & \newline \vdots & \vdots & & & & & \end{pmatrix} $$

Maybe the proposed induction gets easier if you look at the columns: It holds that $a(m,2)=a(m-1,2)+1$ for $m\geq 1$ and $a(0,2)=1$, thus $a(m,2)=m+1$, which is technically easier to deal with then Fibonacci, I guess.

corrected calculation errors in the previous version.
Source Link
MHMertens
MHMertens
  • 105
  • 6

If I haven't made a mistake then the first few values of your sequence look like this:

$$ (a_{m,n})=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 2 & 3 & 4 & 5 & ... & \newline 1 & 3 & 9 & 21 & ... & & \newline 1 & 4 & 39/2 & ... & & & \newline 1 & 5 & \vdots & & & & \newline \vdots & \vdots & & & & & \end{pmatrix} $$$$ (a_{m,n})=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 1 & 2 & 3 & 5 & 8 & ... \newline 1 & 1 & 3 & 9 & 21 & 91 & \newline 1 & 1 & 4 & 21 & 91 & ... & \newline 1 & 1 & 5 & 41 & 329 & ... & \newline 1 & 1 & \vdots & \vdots & \vdots & & \newline \vdots & \vdots & & & & & \end{pmatrix} $$

In other wordsMaybe the proposed induction gets easier if you look at the columns: It holds that $a(m,2)=a(m-1,2)+1$ for $m\geq 1$ and $a(0,2)=1$, thus (again provided that I didn't miscalculate anything) your assertion$a(m,2)=m+1$, which is unfortunately not correcttechnically easier to deal with then Fibonacci, I guess.

If I haven't made a mistake then the first few values of your sequence look like this:

$$ (a_{m,n})=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 2 & 3 & 4 & 5 & ... & \newline 1 & 3 & 9 & 21 & ... & & \newline 1 & 4 & 39/2 & ... & & & \newline 1 & 5 & \vdots & & & & \newline \vdots & \vdots & & & & & \end{pmatrix} $$

In other words, (again provided that I didn't miscalculate anything) your assertion is unfortunately not correct.

If I haven't made a mistake then the first few values of your sequence look like this:

$$ (a_{m,n})=\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & ... \newline 1 & 1 & 2 & 3 & 5 & 8 & ... \newline 1 & 1 & 3 & 9 & 21 & 91 & \newline 1 & 1 & 4 & 21 & 91 & ... & \newline 1 & 1 & 5 & 41 & 329 & ... & \newline 1 & 1 & \vdots & \vdots & \vdots & & \newline \vdots & \vdots & & & & & \end{pmatrix} $$

Maybe the proposed induction gets easier if you look at the columns: It holds that $a(m,2)=a(m-1,2)+1$ for $m\geq 1$ and $a(0,2)=1$, thus $a(m,2)=m+1$, which is technically easier to deal with then Fibonacci, I guess.

Source Link
MHMertens
MHMertens
  • 105
  • 6