6 added 551 characters in body

# Haskell, 29 bytes (Cracked: Cracked1,2)

a n=n*ceiling(realToFrac n/2)


Try it online!

This is A093005: $$\ a(n)=n\lceil \frac{n}{2} \rceil\$$.

Test cases for $$\0 \leq n \leq 20 \$$, that is map a [0..20]:

[0,1,2,6,8,15,18,28,32,45,50,66,72,91,98,120,128,153,162,190,200]


Edit:

## Intended solution (20 bytes)

b n=sum$n<$show(3^n)


BMO found anTry it online! Differs at 22 byte crack by noticing that$$\ n=23 \$$, with ceiling(n/2) can be computed as$$\ a(23) = 276 \$$ and div(n+1)2$$\ b(23) = 253\$$. However, my intended crack

This function is only 20 bytes and accordingequivalent to $$\b(n) = n\ len(3^n) = n \lceil log_{10}(1+3^n)\rceil\$$. Thanks to the rulesceiling, both functions overlap for integer arguments in the robbers, this means BMO's crack can be cracked again.range from $$\0\$$ to $$\22\$$:

source

a n=n*ceiling(realToFrac n/2)


Try it online!

This is A093005: $$\ a(n)=n\lceil \frac{n}{2} \rceil\$$.

Test cases for $$\0 \leq n \leq 20 \$$, that is map a [0..20]:

[0,1,2,6,8,15,18,28,32,45,50,66,72,91,98,120,128,153,162,190,200]


Edit:

BMO found an 22 byte crack by noticing that ceiling(n/2) can be computed as div(n+1)2. However, my intended crack is only 20 bytes and according to the rules for the robbers, this means BMO's crack can be cracked again.

# Haskell, 29 bytes (Cracked: 1,2)

a n=n*ceiling(realToFrac n/2)


Try it online!

This is A093005: $$\ a(n)=n\lceil \frac{n}{2} \rceil\$$.

Test cases for $$\0 \leq n \leq 20 \$$, that is map a [0..20]:

[0,1,2,6,8,15,18,28,32,45,50,66,72,91,98,120,128,153,162,190,200]


## Intended solution (20 bytes)

b n=sum$n<$show(3^n)


Try it online! Differs at $$\ n=23 \$$, with $$\ a(23) = 276 \$$ and $$\ b(23) = 253\$$.

This function is equivalent to $$\b(n) = n\ len(3^n) = n \lceil log_{10}(1+3^n)\rceil\$$. Thanks to the ceiling, both functions overlap for integer arguments in the range from $$\0\$$ to $$\22\$$:

source

5 added 8 characters in body

a n=n*ceiling(realToFrac n/2)


Try it online!

This is A093005: $$\ a(n)=n\lceil \frac{n}{2} \rceil\$$.

Test cases for $$\0 \leq n \leq 20 \$$, that is map a [0..20]:

[0,1,2,6,8,15,18,28,32,45,50,66,72,91,98,120,128,153,162,190,200]


Edit:

BMO found an 22 byte crack by noticing that ceiling(n/2) can be computed as div(n+1)2. However, my intended crack is only 20 bytes and uses a different approachaccording to the rules for the robbers, so it still mightthis means BMO's crack can be fun to try to figure it outcracked again.

a n=n*ceiling(realToFrac n/2)


Try it online!

This is A093005: $$\ a(n)=n\lceil \frac{n}{2} \rceil\$$.

Test cases for $$\0 \leq n \leq 20 \$$, that is map a [0..20]:

[0,1,2,6,8,15,18,28,32,45,50,66,72,91,98,120,128,153,162,190,200]


Edit:

BMO found an 22 byte crack by noticing that ceiling(n/2) can be computed as div(n+1)2. However, my intended crack is only 20 bytes and uses a different approach, so it still might be fun to try to figure it out.

a n=n*ceiling(realToFrac n/2)


Try it online!

This is A093005: $$\ a(n)=n\lceil \frac{n}{2} \rceil\$$.

Test cases for $$\0 \leq n \leq 20 \$$, that is map a [0..20]:

[0,1,2,6,8,15,18,28,32,45,50,66,72,91,98,120,128,153,162,190,200]


Edit:

BMO found an 22 byte crack by noticing that ceiling(n/2) can be computed as div(n+1)2. However, my intended crack is only 20 bytes and according to the rules for the robbers, this means BMO's crack can be cracked again.

4 edited body

a n=n*ceiling(realToFrac n/2)


Try it online!

This is A093005: $$\ a(n)=n\lceil \frac{n}{2} \rceil\$$.

Test cases for $$\0 \leq n \leq 20 \$$, that is map a [0..20]:

[0,1,2,6,8,15,18,28,32,45,50,66,72,91,98,120,128,153,162,190,200]


Edit:

BMO found an 2822 byte crack by noticing that ceiling(n/2) can be computed as div(n+1)2. However, my intended crack is only 20 bytes and uses a different approach, so it still might be fun to try to figure it out.

a n=n*ceiling(realToFrac n/2)


Try it online!

This is A093005: $$\ a(n)=n\lceil \frac{n}{2} \rceil\$$.

Test cases for $$\0 \leq n \leq 20 \$$, that is map a [0..20]:

[0,1,2,6,8,15,18,28,32,45,50,66,72,91,98,120,128,153,162,190,200]


Edit:

BMO found an 28 byte crack by noticing that ceiling(n/2) can be computed as div(n+1)2. However, my intended crack is only 20 bytes and uses a different approach, so it still might be fun to try to figure it out.

a n=n*ceiling(realToFrac n/2)


Try it online!

This is A093005: $$\ a(n)=n\lceil \frac{n}{2} \rceil\$$.

Test cases for $$\0 \leq n \leq 20 \$$, that is map a [0..20]:

[0,1,2,6,8,15,18,28,32,45,50,66,72,91,98,120,128,153,162,190,200]


Edit:

BMO found an 22 byte crack by noticing that ceiling(n/2) can be computed as div(n+1)2. However, my intended crack is only 20 bytes and uses a different approach, so it still might be fun to try to figure it out.

3 cracked
2 added 1 character in body
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