7 10 13 18 25 36

On a quiz I saw today. I’m sure it’s really obvious, but I just can’t work it out.

I’ve got the answer… but still I can’t get it. Anyone help?

7 10 13 18 25 36

On a quiz I saw today. I’m sure it’s really obvious, but I just can’t work it out.

I’ve got the answer… but still I can’t get it. Anyone help?

Is the answer 49 or 53?

All I’ve got is that the intervals are all prime numbers:

7 + **3** = 10,

10 + **3** = 13,

13 + **5** = 18,

18 + **7** = 25,

25 + **11** = 36

It looks to me like:

n[sub]1[/sub]=7

n[sub]2[/sub]=10

for k >= 3:

n[sub]k[/sub]=n[sub]k-2[/sub]+n[sub]k-1[/sub]-(k+1)

IOW, each number is the sum of the previous two minus a value that increases by one for each term.

7 + 10 - 4 = 13

10 + 13 - 5 = 18

13 + 18 - 6 = 25

18 + 25 - 7 = 36

25 + 36 - 8 = 53

So… 53?

My guess is 51 – and that’s not arrived at by taking the mean of **ultrafilter**’s answers.

7 + 10 = 17 - 4 = 13

10 + 13 = 23 - 5 = 18

13 + 18 = 31 - 6 = 25

18 + 25 = 43 - 7 = 36

25 + 36 = 61 - 8 = 53

I should add that my justification was by taking first differences – 3,3,5,7,11 – then second differences – 0,2,2,4. The best way I saw to continue that pattern was a second 4, giving 15 as the next first difference, thus 36+15 = 51 as the next number. But with short samples like this you can usually generate multiple plausible answers.

I think **brad d** has it.

Yes! 53 was the answer given. Can see it now. I can sleep easily tonight.

Thanks all.

I think this is an alternative expression of **brad_d**’s solution:

n[sub]0[/sub] = 7

n[sub]k[/sub] = n[sub]k-1[/sub] + 2F[sub]k[/sub] + 1, where F[sub]n[/sub] is the nth Fibonacci number.

The Fibonacci pattern in the differences is what jumped out at me.

In fairness, the next number could be anything. Somebody thought this a cute way to create a sequence, and Ultrafilter and BradD both apparently thought of that cute way or its equivalent. But there must be infinite different ways of creating any given sequence of any length, albeit perhaps messy.

If nothing else, you can come up with a polynomial which will pass through any given finite set of points. I could construct a polynomial which passes through the points (1,7) (2,10) (3,13) (4,18) (5,25) (6,36) (7,53) , but then again, I could also construct a polynomial which passes through the points (1,7) (2,10) (3,13) (4,18) (5,25) (6,36) (7,6973246892)

This is always my problem with “lateral thinking” puzzles that I see. I read the answer, and I say, “yes, that’s an answer; but until you can prove to me that’s the *only* answer, then you haven’t helped me any.”

“There’s a guy in a field with two empty shotgun shells and a fifth of ripple. He’s covered in acne and wearing an ‘I’m with stupid’ t-shirt. Where did he go to college?” *Pfft*.

Michigan.

Frakkin’ show-off.

And then Carolina Law.

Explains the ripple.