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

1. amistre64 Group Title

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 Group Title

lol, i forgot to delete my reference scribbles

3. amistre64 Group Title

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 Group Title

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

5. amistre64 Group Title

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

6. amistre64 Group Title

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

7. amistre64 Group Title

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

8. amistre64 Group Title

err, F(a) = 3

9. amistre64 Group Title

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

10. amistre64 Group Title

|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 Group Title

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 Group Title

|dw:1346161272353:dw| hmmm

13. amistre64 Group Title

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

14. amistre64 Group Title

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 Group Title

|dw:1346162349632:dw|

16. amistre64 Group Title

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 Group Title

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

18. amistre64 Group Title

|dw:1346163082261:dw|

19. amistre64 Group Title

-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 Group Title

$\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 Group Title

ima goofy goober!!! lol

22. satellite73 Group Title

wow!

23. amistre64 Group Title

i know right, i should be soo ashamed

24. satellite73 Group Title

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

25. satellite73 Group Title

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

26. amistre64 Group Title

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

27. satellite73 Group Title

that is a good one

28. amistre64 Group Title

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

29. satellite73 Group Title

i really meant "off the wall" not arcane

30. amistre64 Group Title

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 Group Title

you will get a gold star for cleverness

32. amistre64 Group Title

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

33. UnkleRhaukus Group Title

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

34. satellite73 Group Title

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

35. eliassaab Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

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

40. amistre64 Group Title

lol, awesome

41. Herp_Derp Group Title

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