6 added 551 characters in body
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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

Haskell, 29 bytes (Cracked)

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
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Haskell, 29 bytes (Cracked)

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.

Haskell, 29 bytes (Cracked)

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.

Haskell, 29 bytes (Cracked)

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
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Haskell, 29 bytes (Cracked)

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.

Haskell, 29 bytes (Cracked)

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.

Haskell, 29 bytes (Cracked)

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
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2 added 1 character in body
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1
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