n = a010d + a110d-1 + ... + ad100 with ai ≡ 1 (mod 2) ∀ i ∈ [1, 9], ∀ d > 0
} numbers with at least two digits, all of which are odd
[10, 21, 23, 45, 46, 59, 70] {
n is 0 or odd (mod 9)
} numbers that are odd when you repeatedly sum their digits
[6, 7, 8, 13, 18, 20, 24, 34, 40] n is the sum of the Scrabble point values of the letters in the US spelling of the numbers in the deck.
[12, 20, 35, 62, 85, 95, 100] numbers that are the sum of the proper divisors of some number < 1000 not in the deck for this game.
[11, 17, 19, 29, 43, 71, 83] primes
[5, 6, 13, 15, 16, 27, 100] numbers used in 0, including substrings of other numbers, but not including the deck spoiler
[10, 21, 56, 57, 64, 76, 729] integers n for which there exists some integer m such that (n-1)/3m and (n-2)/3m are each endpoints of intervals removed during (possibly different) steps of the usual construction of the Cantor set (i.e. the construction in which each step removes the middle third of intervals existing after the previous step)
[1, 10, 15, 28, 36, 78, 120] {
n*(n-1)/2)
}: triangular numbers
McMenno is inactive and has:
[14, 35, 343] {
7n
} divisible by 7
Implosion has 14 points and:
[30, 42, 65] Composite squarefree numbers where when you take the sum of prime factors and write it in english, at least 1/3 of the letters in the word are "e"
[9, 17, 69, 77] {
n ≡ 1 (mod 4) ∧ n ≥ 7 (mod 10)
} Numbers congruent to 1 mod 4 whose last digit, written in english, can have the letters "ty" appended to the end of it to multiply it by ten (e.g., "six" times ten is "sixty", but "fourty" is not a number, so the last digit cannot be four)
[5, 6, 50, 125] Numbers such that if you take the number of letters in the english spelling and add that to the number, and then repeat that process a second time, the result is in the range 11-15 mod 50 (inclusive).
[7, 9, 25, 37] {
pk | pk < 50, prime p, k > 0
}
\
{
19, 27
} numbers less than 50 with exactly 1 prime factor
DeathRowKitty has 23 points and:
[38, 82, 84] slots never touched by Ace, 5, or 9 in perfect out-shuffles of standard 52 card decks, mod 52
Felissan has 14 points and:
[2, 4, 32, 256] {
2n
} powers of two
[30, 40, 55] {
25 + (5n * (n + 1) / 2)
} 25 + 5n, where n is a triangular number
[79, 87, 92] {
maxdigit(n) > 7
} numbers that contain an 8 or 9
[5, 7, 64] the nth prime doesn't have any even digits
popsofctown has 7 points and:
[4, 20, 36, 68] {
16n + 4
} remainder is 4 when dividing by 16
[4, 16, 25, 81] {
n2
} squares
[55, 58, 60] {
n, k, c st n is composite; k - c is perfect; c|n; k = max(d) st d|n ∧d ≠ n
} composite number whose greatest non-trivial divisor minus any of its other divisors is a perfect number
[7, 10, 14] {
n*(n-1)/2) + 4
} triangular numbers + 4
StrangerCoug has 7 points and:
[13, 27, 72] {
n = a010d + a110d-1 + ... + ad100 with Σi∈[0,d]ai = k^2 | ai ≥ 0, d > 0, k ∈ ℤ
} numbers whose digit sum is a square
There are 72 cards left in the deck. It is Strangercoug's turn.
n = a010d + a110d-1 + ... + ad100 with ai ≡ 1 (mod 2) ∀ i ∈ [1, 9], ∀ d > 0
} numbers with at least two digits, all of which are odd
[10, 21, 23, 45, 46, 59, 70] {
n is 0 or odd (mod 9)
} numbers that are odd when you repeatedly sum their digits
[6, 7, 8, 13, 18, 20, 24, 34, 40] n is the sum of the Scrabble point values of the letters in the US spelling of the numbers in the deck.
[12, 20, 35, 62, 85, 95, 100] numbers that are the sum of the proper divisors of some number < 1000 not in the deck for this game.
[11, 17, 19, 29, 43, 71, 83] primes
[5, 6, 13, 15, 16, 27, 100] numbers used in 0, including substrings of other numbers, but not including the deck spoiler
[10, 21, 56, 57, 64, 76, 729] integers n for which there exists some integer m such that (n-1)/3m and (n-2)/3m are each endpoints of intervals removed during (possibly different) steps of the usual construction of the Cantor set (i.e. the construction in which each step removes the middle third of intervals existing after the previous step)
[1, 10, 15, 28, 36, 78, 120] {
n*(n-1)/2)
}: triangular numbers
McMenno is inactive:
Implosion has 14 points and:
[30, 42, 65] Composite squarefree numbers where when you take the sum of prime factors and write it in english, at least 1/3 of the letters in the word are "e"
[9, 17, 69, 77] {
n ≡ 1 (mod 4) ∧ n ≥ 7 (mod 10)
} Numbers congruent to 1 mod 4 whose last digit, written in english, can have the letters "ty" appended to the end of it to multiply it by ten (e.g., "six" times ten is "sixty", but "fourty" is not a number, so the last digit cannot be four)
[5, 6, 50, 125] Numbers such that if you take the number of letters in the english spelling and add that to the number, and then repeat that process a second time, the result is in the range 11-15 mod 50 (inclusive).
[7, 9, 25, 37] {
pk | pk < 50, prime p, k > 0
}
\
{
19, 27
} numbers less than 50 with exactly 1 prime factor
DeathRowKitty has 23 points and:
[38, 82, 84] slots never touched by Ace, 5, or 9 in perfect out-shuffles of standard 52 card decks, mod 52
Felissan has 14 points and:
[2, 4, 32, 256] {
2n
} powers of two
[30, 40, 55] {
25 + (5n * (n + 1) / 2)
} 25 + 5n, where n is a triangular number
[79, 87, 92] {
maxdigit(n) > 7
} numbers that contain an 8 or 9
[5, 7, 64] the nth prime doesn't have any even digits
popsofctown has 7 points and:
[4, 20, 36, 68] {
16n + 4
} remainder is 4 when dividing by 16
[4, 16, 25, 81] {
n2
} squares
[55, 58, 60] {
n, k, c st n is composite; k - c is perfect; c|n; k = max(d) st d|n ∧d ≠ n
} composite number whose greatest non-trivial divisor minus any of its other divisors is a perfect number
[7, 10, 14] {
n*(n-1)/2) + 4
} triangular numbers + 4
StrangerCoug has 7 points and:
[13, 27, 72] {
n = a010d + a110d-1 + ... + ad100 with Σi∈[0,d]ai = k^2 | ai ≥ 0, d > 0, k ∈ ℤ
} numbers whose digit sum is a square
[14, 35, 42, 56, 343] {
7n
} divisible by 7
There are 70 cards left in the deck. It is implosion's turn.
Retroactive rule changes require all 5 players to agree to them. How do people feel about the rule "A sequence is a group of numbers that all have something in common with each other. Numbers may not be excluded from sequences they would naturally belong to."
Incentivising sequence theft
I think the "you can complete it in 6 if you stole it and it's currently in front of you" idea is interesting and worth a playtest. I definitely think we should brainstorm around the stealing mechanic to make it worthwhile to contribute to sequences you can't finish right away.
There would be more stealing if we let sequences grow indefinitely and only awarded points at the end, but in practice whoever had the evens or odds would win and that sucks.
Add [8, 47, 49] to the exactly one prime factor sequence, regardless of whether or not it gets retroactively changed :p
I wasn't under the impression that the 6-cards-for-stolen-sequences rule was being considered for being implemented now, but I'd mildly prefer it not be introduced mid-game with so many sequences active.
n = a010d + a110d-1 + ... + ad100 with ai ≡ 1 (mod 2) ∀ i ∈ [1, 9], ∀ d > 0
} numbers with at least two digits, all of which are odd
[10, 21, 23, 45, 46, 59, 70] {
n is 0 or odd (mod 9)
} numbers that are odd when you repeatedly sum their digits
[6, 7, 8, 13, 18, 20, 24, 34, 40] n is the sum of the Scrabble point values of the letters in the US spelling of the numbers in the deck.
[12, 20, 35, 62, 85, 95, 100] numbers that are the sum of the proper divisors of some number < 1000 not in the deck for this game.
[11, 17, 19, 29, 43, 71, 83] primes
[5, 6, 13, 15, 16, 27, 100] numbers used in 0, including substrings of other numbers, but not including the deck spoiler
[10, 21, 56, 57, 64, 76, 729] integers n for which there exists some integer m such that (n-1)/3m and (n-2)/3m are each endpoints of intervals removed during (possibly different) steps of the usual construction of the Cantor set (i.e. the construction in which each step removes the middle third of intervals existing after the previous step)
[1, 10, 15, 28, 36, 78, 120] {
n*(n-1)/2)
}: triangular numbers
[1, 4, 9, 16, 25, 64, 81] {
n2
} squares
[7, 8, 9, 25, 37, 47, 49] {
pk | pk < 50, prime p, k > 0
}
\
{
19, 27
} numbers less than 50 with exactly 1 prime factor
McMenno is inactive:
Implosion has 21 points and:
[30, 42, 65] Composite squarefree numbers where when you take the sum of prime factors and write it in english, at least 1/3 of the letters in the word are "e"
[9, 17, 69, 77] {
n ≡ 1 (mod 4) ∧ n ≥ 7 (mod 10)
} Numbers congruent to 1 mod 4 whose last digit, written in english, can have the letters "ty" appended to the end of it to multiply it by ten (e.g., "six" times ten is "sixty", but "fourty" is not a number, so the last digit cannot be four)
[5, 6, 50, 125] Numbers such that if you take the number of letters in the english spelling and add that to the number, and then repeat that process a second time, the result is in the range 11-15 mod 50 (inclusive).
DeathRowKitty has 30 points and:
[38, 82, 84] slots never touched by Ace, 5, or 9 in perfect out-shuffles of standard 52 card decks, mod 52
Felissan has 14 points and:
[2, 4, 32, 256] {
2n
} powers of two
[30, 40, 55] {
25 + (5n * (n + 1) / 2)
} 25 + 5n, where n is a triangular number
[79, 87, 92] {
maxdigit(n) > 7
} numbers that contain an 8 or 9
[5, 7, 64] the nth prime doesn't have any even digits
popsofctown has 7 points and:
[4, 20, 36, 68] {
16n + 4
} remainder is 4 when dividing by 16
[55, 58, 60] {
n, k, c st n is composite; k - c is perfect; c|n; k = max(d) st d|n ∧d ≠ n
} composite number whose greatest non-trivial divisor minus any of its other divisors is a perfect number
[7, 10, 14] {
n*(n-1)/2) + 4
} triangular numbers + 4
StrangerCoug has 7 points and:
[13, 27, 72] {
n = a010d + a110d-1 + ... + ad100 with Σi∈[0,d]ai = k^2 | ai ≥ 0, d > 0, k ∈ ℤ
} numbers whose digit sum is a square
[14, 35, 42, 56, 343] {
7n
} divisible by 7
There are 64 cards left in the deck. It is Felissan's turn.
Completing stolen sequences at 6 cards doesn't have unanimous support for this round and will only be considered for future games.
New rule under consideration:
"A sequence is a set of numbers that all have something in common with each other. Numbers may not be excluded from sequences they would naturally belong to."
No active sequences are in violation of this rule so now I'm even more inclined towards implementing it.
People who support this rule change: DeathRowKitty, implosion, StrangerCoug, popsofctown (putting her here based on 124 even though it was before the discussion), Plotinus
People who haven't weighed in yet:
Felissan
Last edited by Plotinus on Sun Oct 27, 2019 7:31 pm, edited 1 time in total.
n = a010d + a110d-1 + ... + ad100 with ai ≡ 1 (mod 2) ∀ i ∈ [1, 9], ∀ d > 0
} numbers with at least two digits, all of which are odd
[10, 21, 23, 45, 46, 59, 70] {
n is 0 or odd (mod 9)
} numbers that are odd when you repeatedly sum their digits
[6, 7, 8, 13, 18, 20, 24, 34, 40] n is the sum of the Scrabble point values of the letters in the US spelling of the numbers in the deck.
[12, 20, 35, 62, 85, 95, 100] numbers that are the sum of the proper divisors of some number < 1000 not in the deck for this game.
[11, 17, 19, 29, 43, 71, 83] primes
[5, 6, 13, 15, 16, 27, 100] numbers used in 0, including substrings of other numbers, but not including the deck spoiler
[10, 21, 56, 57, 64, 76, 729] integers n for which there exists some integer m such that (n-1)/3m and (n-2)/3m are each endpoints of intervals removed during (possibly different) steps of the usual construction of the Cantor set (i.e. the construction in which each step removes the middle third of intervals existing after the previous step)
[1, 10, 15, 28, 36, 78, 120] {
n*(n-1)/2)
}: triangular numbers
[1, 4, 9, 16, 25, 64, 81] {
n2
} squares
[7, 8, 9, 25, 37, 47, 49] {
pk | pk < 50, prime p, k > 0
}
\
{
19, 27
} numbers less than 50 with exactly 1 prime factor
[30, 42, 65] Composite squarefree numbers where when you take the sum of prime factors and write it in english, at least 1/3 of the letters in the word are "e"
[9, 17, 69, 77] {
n ≡ 1 (mod 4) ∧ n ≥ 7 (mod 10)
} Numbers congruent to 1 mod 4 whose last digit, written in english, can have the letters "ty" appended to the end of it to multiply it by ten (e.g., "six" times ten is "sixty", but "fourty" is not a number, so the last digit cannot be four)
[5, 6, 50, 125] Numbers such that if you take the number of letters in the english spelling and add that to the number, and then repeat that process a second time, the result is in the range 11-15 mod 50 (inclusive).
DeathRowKitty has 30 points and:
[38, 82, 84] slots never touched by Ace, 5, or 9 in perfect out-shuffles of standard 52 card decks, mod 52
Felissan has 21 points and:
[2, 4, 32, 256] {
2n
} powers of two
[30, 40, 55] {
25 + (5n * (n + 1) / 2)
} 25 + 5n, where n is a triangular number
[79, 87, 92] {
maxdigit(n) > 7
} numbers that contain an 8 or 9
[5, 7, 64] the nth prime doesn't have any even digits
popsofctown has 7 points and:
[4, 20, 36, 68] {
16n + 4
} remainder is 4 when dividing by 16
[55, 58, 60] {
n, k, c st n is composite; k - c is perfect; c|n; k = max(d) st d|n ∧d ≠ n
} composite number whose greatest non-trivial divisor minus any of its other divisors is a perfect number
[7, 10, 14] {
n*(n-1)/2) + 4
} triangular numbers + 4
StrangerCoug has 7 points and:
[13, 27, 72] {
n = a010d + a110d-1 + ... + ad100 with Σi∈[0,d]ai = k^2 | ai ≥ 0, d > 0, k ∈ ℤ
} numbers whose digit sum is a square
[14, 35, 42, 56, 343] {
7n
} divisible by 7
There are 57 cards left in the deck. It is popsofctown's turn.
New rule under consideration:
"A sequence is a set of numbers that all have something in common with each other. Numbers may not be excluded from sequences they would naturally belong to."
People who support this rule change: DeathRowKitty, implosion, StrangerCoug, popsofctown (putting her here based on 124 even though it was before the discussion), Plotinus
People who haven't weighed in yet: Felissan
My presumptive rule support is accurate and I support trying out the steal incentivizer in a subsequent game not this one
"Let us say that you are right and there are two worlds. How much, then, is this 'other world' worth to you? What do you have there that you do not have here? Money? Power? Something worth causing the prince so much pain for?'"
"Well, I..."
"What? Nothing? You would make the prince suffer over... nothing?"
"Let us say that you are right and there are two worlds. How much, then, is this 'other world' worth to you? What do you have there that you do not have here? Money? Power? Something worth causing the prince so much pain for?'"
"Well, I..."
"What? Nothing? You would make the prince suffer over... nothing?"
I add 7 and 63 to the 7n sequence to steal and score it?!?!
"Let us say that you are right and there are two worlds. How much, then, is this 'other world' worth to you? What do you have there that you do not have here? Money? Power? Something worth causing the prince so much pain for?'"
"Well, I..."
"What? Nothing? You would make the prince suffer over... nothing?"
n = a010d + a110d-1 + ... + ad100 with ai ≡ 1 (mod 2) ∀ i ∈ [1, 9], ∀ d > 0
} numbers with at least two digits, all of which are odd
[10, 21, 23, 45, 46, 59, 70] {
n is 0 or odd (mod 9)
} numbers that are odd when you repeatedly sum their digits
[6, 7, 8, 13, 18, 20, 24, 34, 40] n is the sum of the Scrabble point values of the letters in the US spelling of the numbers in the deck.
[12, 20, 35, 62, 85, 95, 100] numbers that are the sum of the proper divisors of some number < 1000 not in the deck for this game.
[11, 17, 19, 29, 43, 71, 83] primes
[5, 6, 13, 15, 16, 27, 100] numbers used in 0, including substrings of other numbers, but not including the deck spoiler
[10, 21, 56, 57, 64, 76, 729] integers n for which there exists some integer m such that (n-1)/3m and (n-2)/3m are each endpoints of intervals removed during (possibly different) steps of the usual construction of the Cantor set (i.e. the construction in which each step removes the middle third of intervals existing after the previous step)
[1, 10, 15, 28, 36, 78, 120] {
n*(n-1)/2)
}: triangular numbers
[1, 4, 9, 16, 25, 64, 81] {
n2
} squares
[7, 8, 9, 25, 37, 47, 49] {
pk | pk < 50, prime p, k > 0
}
\
{
19, 27
} numbers less than 50 with exactly 1 prime factor
[30, 42, 65] Composite squarefree numbers where when you take the sum of prime factors and write it in english, at least 1/3 of the letters in the word are "e"
[9, 17, 69, 77] {
n ≡ 1 (mod 4) ∧ n ≥ 7 (mod 10)
} Numbers congruent to 1 mod 4 whose last digit, written in english, can have the letters "ty" appended to the end of it to multiply it by ten (e.g., "six" times ten is "sixty", but "fourty" is not a number, so the last digit cannot be four)
[5, 6, 50, 125] Numbers such that if you take the number of letters in the english spelling and add that to the number, and then repeat that process a second time, the result is in the range 11-15 mod 50 (inclusive).
DeathRowKitty has 30 points and:
[38, 82, 84] slots never touched by Ace, 5, or 9 in perfect out-shuffles of standard 52 card decks, mod 52
Felissan has 21 points and:
[2, 4, 32, 256] {
2n
} powers of two
[30, 40, 55] {
25 + (5n * (n + 1) / 2)
} 25 + 5n, where n is a triangular number
[5, 7, 64] the nth prime doesn't have any even digits
popsofctown has 14 points and:
[4, 20, 36, 68] {
16n + 4
} remainder is 4 when dividing by 16
[55, 58, 60] {
n, k, c st n is composite; k - c is perfect; c|n; k = max(d) st d|n ∧d ≠ n
} composite number whose greatest non-trivial divisor minus any of its other divisors is a perfect number
[7, 10, 14] {
n*(n-1)/2) + 4
} triangular numbers + 4
StrangerCoug has 14 points and:
[13, 27, 72] {
n = a010d + a110d-1 + ... + ad100 with Σi∈[0,d]ai = k^2 | ai ≥ 0, d > 0, k ∈ ℤ
} numbers whose digit sum is a square
There are 51 cards left in the deck. It is imposion's turn.
The new rule has been added to the OP.
I think I didn't spell this out in the rules but I was thinking that after we all run out of cards to draw, we can continue playing the cards in our hands without drawing new ones, like at the end of Scrabble. At that stage you probably won't be making new sequences because if you can't make a bingo then you know for sure you'll never finish that sequence but you may still be able to finish some pre-existing ones. and then if nobody can take a turn then it's over. Ties can be broken by how many points worth of unfinished sequences you have in front of you.
And then we can have an intermission to discuss and finalise rule changes / deck changes. I'm pretty sure I want the next game to have 3 teams of 2 players each, and instead of having "inactive players" I'll just replace people when they disappear, like in mafia.
(If anyone wants McMenno replaced in this game, let me know. I know ErrantParabola was interested in playing.)
n = a010d + a110d-1 + ... + ad100 with ai ≡ 1 (mod 2) ∀ i ∈ [1, 9], ∀ d > 0
} numbers with at least two digits, all of which are odd
[10, 21, 23, 45, 46, 59, 70] {
n is 0 or odd (mod 9)
} numbers that are odd when you repeatedly sum their digits
[6, 7, 8, 13, 18, 20, 24, 34, 40] n is the sum of the Scrabble point values of the letters in the US spelling of the numbers in the deck.
[12, 20, 35, 62, 85, 95, 100] numbers that are the sum of the proper divisors of some number < 1000 not in the deck for this game.
[11, 17, 19, 29, 43, 71, 83] primes
[5, 6, 13, 15, 16, 27, 100] numbers used in 0, including substrings of other numbers, but not including the deck spoiler
[10, 21, 56, 57, 64, 76, 729] integers n for which there exists some integer m such that (n-1)/3m and (n-2)/3m are each endpoints of intervals removed during (possibly different) steps of the usual construction of the Cantor set (i.e. the construction in which each step removes the middle third of intervals existing after the previous step)
[1, 10, 15, 28, 36, 78, 120] {
n*(n-1)/2)
}: triangular numbers
[1, 4, 9, 16, 25, 64, 81] {
n2
} squares
[7, 8, 9, 25, 37, 47, 49] {
pk | pk < 50, prime p, k > 0
}
\
{
19, 27
} numbers less than 50 with exactly 1 prime factor
[3, 4, 5, 7, 21, 22, 64] the nth prime doesn't have any even digits
McMenno is inactive:
Implosion has 28 points and:
[30, 42, 65] Composite squarefree numbers where when you take the sum of prime factors and write it in english, at least 1/3 of the letters in the word are "e"
[9, 17, 69, 77] {
n ≡ 1 (mod 4) ∧ n ≥ 7 (mod 10)
} Numbers congruent to 1 mod 4 whose last digit, written in english, can have the letters "ty" appended to the end of it to multiply it by ten (e.g., "six" times ten is "sixty", but "fourty" is not a number, so the last digit cannot be four)
[5, 6, 50, 125] Numbers such that if you take the number of letters in the english spelling and add that to the number, and then repeat that process a second time, the result is in the range 11-15 mod 50 (inclusive).
DeathRowKitty has 30 points and:
[38, 82, 84] slots never touched by Ace, 5, or 9 in perfect out-shuffles of standard 52 card decks, mod 52
Felissan has 21 points and:
[2, 4, 32, 256] {
2n
} powers of two
[30, 40, 55] {
25 + (5n * (n + 1) / 2)
} 25 + 5n, where n is a triangular number
popsofctown has 14 points and:
[4, 20, 36, 68] {
16n + 4
} remainder is 4 when dividing by 16
[55, 58, 60] {
n, k, c st n is composite; k - c is perfect; c|n; k = max(d) st d|n ∧d ≠ n
} composite number whose greatest non-trivial divisor minus any of its other divisors is a perfect number
[7, 10, 14] {
n*(n-1)/2) + 4
} triangular numbers + 4
StrangerCoug has 14 points and:
[13, 27, 72] {
n = a010d + a110d-1 + ... + ad100 with Σi∈[0,d]ai = k^2 | ai ≥ 0, d > 0, k ∈ ℤ
} numbers whose digit sum is a square
There are 47 cards left in the deck. It is DeathRowKitty's turn.