edging and cumming dare

[Bored]

inspired by moofu's dare

For the next 2 or 3 weeks every time you want to cum you have to edge x amount of times. On the first attempt aka tonight/now you edge once and cum. The next time it doubles to 2 and the next time it doubles to 4 and so on.

so the order would be:
1
2
4
8
16
24
48
96
127
254
and so on.

Hint: you don't have to cum every day, so save your cumming as long as you can.

[Sanpow]

LOOOOOOOOOOOOOOOOOL

Originally Posted by bored

The next time it doubles to 2 and the next time it doubles to 4 and so on.

so the order would be:
1
2
4
8
16
24 16*2=32
48
96
127 96*2 = 192 btw: 127 is odd
254
and so on.

i think you want to say :
1
2
4
8
16
32
64
128
256
512
1024
2048
4096
8192
16384
32768
65536
and so on...

But i think it's not realy possilbe to do more than 64edges a day (and this would be very hard!!!)
so you have to wait (and edge) a loooooong time until you can cum...

[Raptor9458]

Most I ever got up to was 47 in a 24 hour period.

[Recursion]

Good idea, but doubling it might be too much. Maybe following the Fibonacci sequence would be better. The next number equals the sum of the two numbers before it:
0
1
1 (0+1)
2 (1+1)
3 (2+1)
5 (3+2)
8 (5+3)
13
21
34
55
89
144

That's the first 13 values, so you could do that for 2 weeks and cum about once per day before the numbers get into the hundreds, or if you wanted to cum less do it for a whole month so you'd be cumming about once every other day

[moofu]

Thanks for the credit =)

[Tab]

Looks like it gets impossible too suddenly. What about starting at a slightly higher number like 4 and increasing it by 2 every day?

[c3po471]

you could all so do this like pascel triangle. so on day one you do one edge and on day 2 you edge two and on day 3 you edge 4 times

day 1 1
day 2 1 1
day 3 1 2 1
day 4 1 3 3 1
day 5 1 4 6 4 1
day 6 1 5 10 10 5 1
day 7 1 6 15 20 15 6 1

[Tab]

Originally Posted by c3po471

you could all so do this like pascel triangle. so on day one you do one edge and on day 2 you edge two and on day 3 you edge 4 times

day 1 1
day 2 1 1
day 3 1 2 1
day 4 1 3 3 1
day 5 1 4 6 4 1
day 6 1 5 10 10 5 1
day 7 1 6 15 20 15 6 1

Only 64 edges on day 7? Softcore. How about g(daynumber) where
g(1)=3^^^^3
g(2)=3^g(1)3
g(3)=3^g(2)3
etc,
where ^ should be interpreted as an up-arrow in Knuth's up-arrow notation? See the sequence for Graham's Number for more info, Graham's number itself is g(64).


On a serious note, though, you guys shouldn't be using exponential or recursive sequences. When you do that, it's basically boring for a few days, then fun for a couple days, then impossible. Case in point:

Fibonacci sequence:

0 (Boring)
1 (Boring)
1 (Boring)
2 (Boring)
3 (Boring)
5 (OK)
8 (OK)
13 (Fun/challenging)
21 (Fun/very challenging)
34 (Starting to get impossible even if you've got the whole day)
55 (Pretty much impossible)
89 (Definitely impossible)
144 (Sha-ba-do-da)
233 (Shakedi-dap-da-dao)

Powers of two:

1 (Boring)
2 (Boring)
4 (Meh)
8 (Quite fun)
16 (Fun/challenging)
32 (Nnnngh.. No.)
64 (Aaaah, it burns!)
128 (Why, God?! Why?!)
256 (Dear diary. Blood everywhere. Not to self: Never do a two week dare again)
512 (Dear diary. Today, I invented my own language based on twitches and whimpers)
1024 (Sllgd brrr... prrrr... txm lzm)
2048 (ssk. grgl.)
4096 (Hey.. It's kind of getting easier.. ?!?)
8192 (W-what's going on? I'm able to do these now!)
16384 (I th-think this is.. I think I must be..)
32768 (I can see forever)