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amistre64
Think of 3 rules, such that they generate 2,3,6 as the first 3 terms of a sequence.
1/2 3/2 -4 i came up with this one:\[\frac{1}{2}\left(\frac{n-1.5}{|n-1.5|}+\frac{3n-7.5}{|n-2.5|}-\frac{8n-280}{|n-3.5|}\right)\]
lol, i forgot to delete my reference scribbles
i was also considering coming up with something such that:\[\int_{0}^{n}f(x)dx=2,3,6~for~n=1,2,3\]
the first one i thought of was\[a_n=n^2-2n+3\]
|dw:1346159138909:dw| hmm; such that the first area is 2, and the second is 1, and the third is 3
if i could determine the slopes in that figure, i could integrate the function to get what im thinking of :)
|dw:1346159507196:dw|b-a=1\[\int_{a}^{b}k~dx=F(b)-F(a)=1\]F(b)=4, F(a)=-3
|dw:1346159801695:dw| 0 to b , b<1 ... its like right on the tip of my brain
|dw:1346160036988:dw| at a, A=2, at (b-a)/2, A=0 so there has to be a sweet spot between a and the midpoint that satisfies my quandry
if b= 3/4; and a = 1/4 \[\frac{4}{3/4}x=\frac{16}{3}x; \int\to\ \frac83 \left((\frac34)^2-(\frac14)^2\right)\]\[\frac83 \left(\frac{9-1}{16}\right)=1\frac13\] so its between 1/4 and 1/2 :)
|dw:1346161272353:dw| hmmm
not B=8, 4+4=8/2=4; that triangle area there is B=4
if i guess that: (4+d)/2 = 3; d=2 mx-d 4+2 = 6, 6 in 1 = 6x, 6x-2 3x^2-2x, (0,1) = 3-2=1
|dw:1346162349632:dw|
2,3,6 is my goal; and i wonder if alternating signs, and doubling would give me my needed slopes, making the next one 12 12x-2; 6x^2-2x = 4, not 3 it was a nice guess
int mx-2 , [0,1] m/2 - 2 = 3 m/2 = 5 m=10
|dw:1346163082261:dw|
-a-b-c = 4 a-b-c=-6 a+b-c=10 rref{{-1,-1,-1,4},{1,-1,-1,-6},{1,1,-1,10}} -5,8,-7 f(x) = -5|x-1|+8|x-2|-7|x-3| and since n=0 = -10; lets add 10 to it to move it to the proper spot \[\int_{0}^{n}f(x) = -5|x-1|+8|x-2|-7|x-3|+10~dx=\{2,3,6,...\}~:~n=1,2,3,...\]
\[\int_{0}^{n}f(x)~dx\]\[\int_{0}^{n} (-5|x-1|+8|x-2|-7|x-3|+10)~dx=\{2,3,6,...\}~:~n=1,2,3,...\] that fixes the typos
ima goofy goober!!! lol
i know right, i should be soo ashamed
i guess my idea of a second degree polynomial is too prosaic. you probably did that first anyway
maybe change it to "find the most arcane sequences whose first three terms are ..."
i did .... and my second original idea was:\[a_{n+1}=a{n}*a_{n-1}~:~a_1=2, a_2=3\]
if by arcane you mean deviously clever ... i agree
i really meant "off the wall" not arcane
i wonder what the math teachers gonna give me as a grade on the homework this year; or if shell even read it
you will get a gold star for cleverness
i gotta hobble over to physics and 2 other classes for awhile; yall try not to break stuff while im gone
\[\large{s_n=2n-\frac n4\left(1+(-1)^n\right)}\] \[s_1=2-\frac 14\left(1+(-1)^1\right)=2-0=2\] \[s_2=2\times2-\frac 24\left(1+(-1)^2\right)=4-\frac12(2)=3\] \[s_3=2\times3-\frac 34\left(1+(-1)^3\right)=6-0=6\]
\[a_n=(-1)^{n}n^2\cos(\pi n)+(-1)^{n+1}2\cos(\pi n)+(-1)^n3\cos(\pi n)\]
Why is interesting? I do not know. Here is my simple contribution. \[ a_1=2\\ a_2=3\\ a_n=a_{n-1}a_{n-2},\quad n>2 \] \[ a_1=2\\ a_2=3\\ a_n=a_{n-1}+a_{n-2}+1,\quad n>2 \] \[ a_1=2\\ a_2=3\\ a_n=a_{n-1}-a_{n-2}+5,\quad n>2 \]
Here are two more. You can have infinite such rules. \[ a_1=2\\ a_2=3\\ a_n=2 a_{n-1} ,\quad n>2 \] \[ a_1=2\\ a_2=3\\ a_n=3 a_{n-2} ,\quad n>2 \]
the question was so bland that i figured id see if there were more interesting ways to develop a sequence that began with the given digits.
if my teacher insists that i have to do the homework for a grade .... im at least going to try to get some fun out of it :)
\[a_n=\Pi (n)-\Gamma (n) +2\] \(\Pi (n)=n!\) and \(\Gamma (n) =(n-1)!\) but it just looks so much cooler that way.
This one is just... idk \[\large a_n=\left\lfloor\sqrt[n]{n+\Big\lfloor e^{\frac{1}{2}n^2}\Big\rfloor}\right\rfloor +\left\lfloor\int_0^n \frac{\frac{1}{2}xe^{-\sqrt{x}}}{\sin (x)}\,\text{d}x\right\rfloor\]