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coincidentally, its double the fourth root of the tenth consecutive y-value in that equation (rounded off)

x^2 + 7 = 2^y

(1,3) (3,4) (5,5) (7,6)

1+3+3+4+5+5+2 = 23

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Your formula can be simplified to:
(x4-x1) + (y4-y1)

6 + 3 + 2 = 11

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From the above pattern of coordinates, it can be assumed that:
the nth x = 2n - 1
the nth y = n + 2

Therefore, the 10th y would be 12.
The 4th root of 12 is 1.86, which rounds to 2.

2 * 4root(y10) = 2 * 2 = 4

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So you are either 23, 11, or 4.

(7,6) is wrong, nemo
7^2 + 7 = 56 =/= 64
The next pair after (5,5) is actually (11,7):
11^2 + 7 = 121 + 7 = 128 = 2^7

So you get x4=11, x1=1, y4=7, y1=3
x4 - x1 + y4 - y1
= 11 - 1 + 7 - 3
=14
2 more is 16

And according to your other statement y10 is 4096? So 2^4096=x^2+7 for an integer x... I don't really want to test that

4276

its for some math problem i have currently sitting on my desk ---- instant inspiration

Oh wow ... after the first three coords were so neat, I assumed the rest would follow the pattern. (I didn't actually test the fourth.) My mistake.

In that case, I don't know how you managed to find a formula for that sequence, in order to find (4276,4096) .. .. at 16, what math level are you taking ?

no, its (x, 4276)

x would get the message "your message has to long words"

i had to scroll sideways for a couple of seconds when i was using a web calculator to see it all

and its for a thing called olymon (olympic monthly)

me and a friend have done a couple problems from there

Maths freak? olymon sounds a bit like that^^ Reminds me of other mathematics olympics

*** my *** on a **** dammit sorry guys the tenth term is not 4thousand whatever i ***-ed up

That's okay .. after all, none of the rest of us has even attempted to figure it out.