Solve by induction.The recurrence

$\displaystyle S(n) = 1$, for $\displaystyle n=1$

$\displaystyle S(n) = [S(\frac{n}{2})]+[S(\frac{n}{2})]+1$, for $\displaystyle n \geq 2$

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- May 6th 2009, 03:29 PMApprentice123Mathematical inductionSolve by induction.The recurrence

$\displaystyle S(n) = 1$, for $\displaystyle n=1$

$\displaystyle S(n) = [S(\frac{n}{2})]+[S(\frac{n}{2})]+1$, for $\displaystyle n \geq 2$ - May 6th 2009, 04:16 PMProve It
- May 6th 2009, 05:32 PMSoroban
Hello, Apprentice123!

I believe you have misplaced the brackets . . .

Quote:

Solve by induction the recurrence:

$\displaystyle S(1) = 1$

$\displaystyle S(n) \:=\:S\bigg(\!\!\left[\tfrac{n}{2}\right]\!\!\bigg) + S\bigg(\!\!\left[\tfrac{n}{2}\right]\!\!\bigg) +1$ . for $\displaystyle n \geq 2$

. . $\displaystyle \begin{array}{ccc} \text{Group} & & \text{Value} \\ \hline S(1) &=& 1 \\ S(2),S(3) &=& 3 \\ S(4),S(5),S(6),S(7) &=& 7 \\ S(8),S(9),\;\hdots\; S(15) &=& 15 \\ S(16),S(17),\hdots S(31) &=& 31 \\ \vdots & & \vdots \end{array}$

Each group doubles in length.

The $\displaystyle n^{th}$ group has the value: .$\displaystyle 2^n-1$

Good luck with the inductive proof . . .