Puzzle time - slicing the pie
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OK, here's what all other sequences that start with 1,2,4,6,10,14,19 stored at that nice website say about it's 10th/11th element:
A023115 - 45
A076101 - 52
A094281 - 51
A323623 - 43
A023536 - 46
A047808 - 44
A076268 - 42
A024536 - 49
A117237 - 52It must be one of those

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OK, here's what all other sequences that start with 1,2,4,6,10,14,19 stored at that nice website say about it's 10th/11th element:
A023115 - 45
A076101 - 52
A094281 - 51
A323623 - 43
A023536 - 46
A047808 - 44
A076268 - 42
A024536 - 49
A117237 - 52It must be one of those

@Klaus said in Puzzle time - slicing the pie:
OK, here's what all other sequences that start with 1,2,4,6,10,14,19 stored at that nice website say about it's 10th/11th element:
A023115 - 45
A076101 - 52
A094281 - 51
A323623 - 43
A023536 - 46
A047808 - 44
A076268 - 42
A024536 - 49
A117237 - 52It must be one of those

And it’s none of those!
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Hint:
||Thinking of the cuts as lines on a circle, what’s the largest number of lines the nth cut can intersect?||
@jon-nyc said in Puzzle time - slicing the pie:
Hint:
||Thinking of the cuts as lines on a circle, what’s the largest number of lines the nth cut can intersect?||
Hey, I'm in a Zoom meeting right now. I can only do this as a low-priority background task that doesn't require me to think

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Oh, I see, my problem was that I screwed up on the start of the sequence. It's 1,2,4,7,11,16.The first answer I get from OEIS is A000124, which has the promising subtitle "maximal number of pieces formed when slicing a pancake with n cuts.".
So the answer seems to be 56.
I refer to the references given at OEIS for the rationale

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Cheater.
||How about noticing, based on my hint, that the nth cut can, at most, intersect the other n-1 cuts just once. That forms n new line segments (considering the edges of the pizza as end points as well). Each new line segment corresponds to a new slice of pie.
So f(n)=f(n-1)+n
And we know f(0) is 1So:
f(1) = 2
f(2) = 4
f(3) = 7
f(4) = 11
f(5) = 16
f(6) = 22
f(7) = 29
f(8) = 37
f(9) = 46
f(10) = 56So to get the nth number:
1+1+2+3+4+5+6..+n
Note that's just 1 plus the sum of the natural numbers to n.
So 1 + n(n+1)/2 or
(n^2 + n + 2)/2||
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@jon-nyc said in Puzzle time - slicing the pie:
How about noticing, based on my hint, that the nth cut can, at most, intersect the other n-1 cuts just once.
That makes sense, but how do you know that this is not merely an upper bound but that you can always find a line that intersects n-1 cuts?
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