## amistre64 3 years ago Think of 3 rules, such that they generate 2,3,6 as the first 3 terms of a sequence.

1. amistre64

1/2 3/2 -4 i came up with this one:$\frac{1}{2}\left(\frac{n-1.5}{|n-1.5|}+\frac{3n-7.5}{|n-2.5|}-\frac{8n-280}{|n-3.5|}\right)$

2. amistre64

lol, i forgot to delete my reference scribbles

3. amistre64

i was also considering coming up with something such that:$\int_{0}^{n}f(x)dx=2,3,6~for~n=1,2,3$

4. amistre64

the first one i thought of was$a_n=n^2-2n+3$

5. amistre64

|dw:1346159138909:dw| hmm; such that the first area is 2, and the second is 1, and the third is 3

6. amistre64

if i could determine the slopes in that figure, i could integrate the function to get what im thinking of :)

7. amistre64

|dw:1346159507196:dw|b-a=1$\int_{a}^{b}k~dx=F(b)-F(a)=1$F(b)=4, F(a)=-3

8. amistre64

err, F(a) = 3

9. amistre64

|dw:1346159801695:dw| 0 to b , b<1 ... its like right on the tip of my brain

10. amistre64

|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

11. amistre64

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 :)

12. amistre64

|dw:1346161272353:dw| hmmm

13. amistre64

not B=8, 4+4=8/2=4; that triangle area there is B=4

14. amistre64

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

15. amistre64

|dw:1346162349632:dw|

16. amistre64

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

17. amistre64

int mx-2 , [0,1] m/2 - 2 = 3 m/2 = 5 m=10

18. amistre64

|dw:1346163082261:dw|

19. amistre64

-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 $\int_{0}^{n}f(x) = -5|x-1|+8|x-2|-7|x-3|+10~dx=\{2,3,6,...\}~:~n=1,2,3,...$

20. amistre64

$\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

21. amistre64

ima goofy goober!!! lol

22. satellite73

wow!

23. amistre64

i know right, i should be soo ashamed

24. satellite73

i guess my idea of a second degree polynomial is too prosaic. you probably did that first anyway

25. satellite73

maybe change it to "find the most arcane sequences whose first three terms are ..."

26. amistre64

i did .... and my second original idea was:$a_{n+1}=a{n}*a_{n-1}~:~a_1=2, a_2=3$

27. satellite73

that is a good one

28. amistre64

if by arcane you mean deviously clever ... i agree

29. satellite73

i really meant "off the wall" not arcane

30. amistre64

i wonder what the math teachers gonna give me as a grade on the homework this year; or if shell even read it

31. satellite73

you will get a gold star for cleverness

32. amistre64

i gotta hobble over to physics and 2 other classes for awhile; yall try not to break stuff while im gone

33. UnkleRhaukus

$\large{s_n=2n-\frac n4\left(1+(-1)^n\right)}$ $s_1=2-\frac 14\left(1+(-1)^1\right)=2-0=2$ $s_2=2\times2-\frac 24\left(1+(-1)^2\right)=4-\frac12(2)=3$ $s_3=2\times3-\frac 34\left(1+(-1)^3\right)=6-0=6$

34. satellite73

$a_n=(-1)^{n}n^2\cos(\pi n)+(-1)^{n+1}2\cos(\pi n)+(-1)^n3\cos(\pi n)$

35. eliassaab

Why is interesting? I do not know. Here is my simple contribution. $a_1=2\\ a_2=3\\ a_n=a_{n-1}a_{n-2},\quad n>2$ $a_1=2\\ a_2=3\\ a_n=a_{n-1}+a_{n-2}+1,\quad n>2$ $a_1=2\\ a_2=3\\ a_n=a_{n-1}-a_{n-2}+5,\quad n>2$

36. eliassaab

Here are two more. You can have infinite such rules. $a_1=2\\ a_2=3\\ a_n=2 a_{n-1} ,\quad n>2$ $a_1=2\\ a_2=3\\ a_n=3 a_{n-2} ,\quad n>2$

37. amistre64

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.

38. amistre64

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 :)

39. Herp_Derp

$a_n=\Pi (n)-\Gamma (n) +2$ $$\Pi (n)=n!$$ and $$\Gamma (n) =(n-1)!$$ but it just looks so much cooler that way.

40. amistre64

lol, awesome

41. Herp_Derp

This one is just... idk $\large a_n=\left\lfloor\sqrt[n]{n+\Big\lfloor e^{\frac{1}{2}n^2}\Big\rfloor}\right\rfloor +\left\lfloor\int_0^n \frac{\frac{1}{2}xe^{-\sqrt{x}}}{\sin (x)}\,\text{d}x\right\rfloor$