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ThatDonGuy
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June 18th, 2021 at 4:28:49 PM permalink
Quote: teliot

Okay ... new puzzle. A sum of consecutive squares equaling a sum of the following successive consecutive squares.

What's the longest one you can find?


Follow-up: as of now, it's 2,914,365 terms:
483,046^2 + ... + 2,699,107^2 = 2,699,108^2 + ... + 3,397,410^2 = 6,516,925,680,815,252,695
This is the 1254th such sequence (in order of the largest term to the left of the equals sign)
Last edited by: ThatDonGuy on Jun 18, 2021
teliot
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June 18th, 2021 at 5:39:17 PM permalink
Quote: ThatDonGuy

Follow-up: as of now, it's 2,914,365 terms:
483,046^2 + ... + 2,699,107^2 = 2,699,108^2 + ... + 3,397,410^2 = 6,516,925,680,815,252,695
This is the 1254th such sequence (in order of the largest term to the left of the equals sign)

of course, the big question is if there are infinitely many solutions. If so, the first terms would be submissible to the encyclopedia of integer sequences.

Fabulous work by the way. I find this pretty cool, it says something about the density of sums of consecutive squares that there are so many solutions already.
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GM
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June 18th, 2021 at 6:14:34 PM permalink
Quote: teliot

of course, the big question is if there are infinitely many solutions. If so, the first terms would be submissible to the encyclopedia of integer sequences.


I have an infinite family of solutions.
Let's use the notation introduced by ThatDonGuy:
Quote: ThatDonGuy

Let there be A numbers on the left, and B on the right
(N - (A - 1))^2 + (N - (A - 2))^2 + ... + (N - 1)^2 + N^2 = (N + 1)^2 + (N + 2)^2 + ... + (N + B)^2


He derived the equation
Quote: ThatDonGuy

N (N + 1) (2N + 1) - (N - A) (N - A + 1) (2 (N - A) + 1) = (N + B) (N + B + 1) (2 (N + B) + 1) - N (N + 1) (2N + 1)


This is a quadratic equation for N, the degree 3 terms cancel out. If you assume A=B+1, you can solve for N easily and get an integer solution.
N=2(B^2+B)
ThatDonGuy
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June 18th, 2021 at 6:15:46 PM permalink
Quote: teliot

of course, the big question is if there are infinitely many solutions. If so, the first terms would be submissible to the encyclopedia of integer sequences.


Does the sequence have to be of known infinite length? The repunit primes is a sequence in OEIS, but I don't think it has been proven that it is of infinite length.
teliot
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June 18th, 2021 at 7:01:02 PM permalink
Quote: ThatDonGuy

Does the sequence have to be of known infinite length? The repunit primes is a sequence in OEIS, but I don't think it has been proven that it is of infinite length.

I don't know. Maybe curiosities will get in there if it's a well-known conjecture, but I wouldn't begin to submit unless I knew it was infinite and had no other obvious way of being generated.
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June 18th, 2021 at 7:08:00 PM permalink
Quote: ThatDonGuy

Does the sequence have to be of known infinite length? The repunit primes is a sequence in OEIS, but I don't think it has been proven that it is of infinite length.



OEIS sequences do not need to be infinite. There are quite a number of finite sequences that have been accepted.
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teliot
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June 18th, 2021 at 7:50:34 PM permalink
Quote: ThatDonGuy

Follow-up: as of now, it's 2,914,365 terms:
483,046^2 + ... + 2,699,107^2 = 2,699,108^2 + ... + 3,397,410^2 = 6,516,925,680,815,252,695
This is the 1254th such sequence (in order of the largest term to the left of the equals sign)

Is there a solution that starts at N = 1?
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teliot
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June 18th, 2021 at 8:10:44 PM permalink
Quote: teliot

Is there a solution that starts at N = 1?

by the way, the smart folks here in this thread should come up with a new sequence, call it the wizard sequence, and submit it to OEIS.
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ThatDonGuy
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June 18th, 2021 at 8:26:12 PM permalink
Quote: teliot

Is there a solution that starts at N = 1?


No - at least, none that I have found so far.

Pretty much every odd number from 3 through 99 has a sequence that has that many terms. I think the smallest even number size is 52.
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June 19th, 2021 at 4:51:24 AM permalink
Quote: ThatDonGuy

Pretty much every odd number from 3 through 99 has a sequence that has that many terms. I think the smallest even number size is 52.


My infinite family gives a solution for every odd number at least 3. The smallest even number I have found was also 52.

If you solve the quadratic equation for n in terms of A and B, the expression under the square root is 36A^2B^2-3((A-B)^4-(A-B)^2) = 36A^2B^2-3(d^4-d^2), where d=A-B.
This number has to be perfect square for n to be an integer. If N>0 then d>0, d=1 gives the infinite family in my earlier post, if d>1, then sqrt(36A^2B^2-3(d^4-d^2))<6AB, so if it is an integer, it is at most 6AB-1, this gives a bound on AB, AB≤(d^4-d^2)/4, which gives the bound B≤d(d-1)/2, this can help with computer searches.

The original question did not specify that you could only consider positive integers, (-k)^2+(-k+1)^2+...+(-1)^2+0^2=1^2+2^2+...+k^2 trivially, but there are also some non-trivial examples involving squares of some negative numbers, for example, (-2)^2+(-1)^2+0^2+1^2+...+7^2=8^2+9^2, or (-24)^2+...+14^2=15^2+...+27^2.
Last edited by: GM on Jun 19, 2021
ThatDonGuy
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June 19th, 2021 at 8:37:36 AM permalink
Quote: GM

My infinite family gives a solution for every odd number at least 3. The smallest even number I have found was also 52.


Any ideas on how to show that 3^3 + 4^3 + 5^3 = 6^3 is the only solution in positive integers for cubes?

If it is A^3 + (A+1)^3 + ... + N^3 = (N+1)^3 + ... + B^3, then this becomes 2 ((N-1) N / 2)^2 = ((A - 1) A / 2)^2 + (B (B+1) / 2)^2, which can be expressed as 2 Q^2 = P^2 + R^2, where P < Q < R and all three are triangular numbers.
The {3, 4, 5, 6} solution has A = 3, N = 5, B = 6; 2 • (5 • 6 / 2)^2 = 450 = 9 + 441 = (2 • 3 / 2)^2 + (6 • 7 / 2)^2; P = 3, Q = 15, and R = 21.
unJon
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June 19th, 2021 at 9:04:41 AM permalink
Quote: ThatDonGuy

Any ideas on how to show that 3^3 + 4^3 + 5^3 = 6^3 is the only solution in positive integers for cubes?

If it is A^3 + (A+1)^3 + ... + N^3 = (N+1)^3 + ... + B^3, then this becomes 2 ((N-1) N / 2)^2 = ((A - 1) A / 2)^2 + (B (B+1) / 2)^2, which can be expressed as 2 Q^2 = P^2 + R^2, where P < Q < R and all three are triangular numbers.
The {3, 4, 5, 6} solution has A = 3, N = 5, B = 6; 2 • (5 • 6 / 2)^2 = 450 = 9 + 441 = (2 • 3 / 2)^2 + (6 • 7 / 2)^2; P = 3, Q = 15, and R = 21.



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GM
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June 19th, 2021 at 2:32:00 PM permalink
Quote: teliot

Is there a solution that starts at N = 1?


I am quite confident that there is not. If you assume that 1^2+2^2+...+N^2=(N+1)^2+...+(N+B)^2, then after using the summation formula for squares, you get the equation -2B^3-6B^2N-3B^2-6BN^2-6BN-B+2N^3+3N^2+N=0, which has degree 3 in both B and N. By using the the change of variables x=2(4B+2N+1)/(2B-2N-1), y=-6/(2B-2N-1), this equation can be transformed to the form y^2=x^3-12x+20. (This is not black magic, it is called the Weierstrass form and the curve defined by this equation is an elliptic curve.)

If the equation -2B^3-6B^2N-3B^2-6BN^2-6BN-B+2N^3+3N^2+N=0 has an integer solution for B and N, we get a rational (but not necessarily integer) solution of y^2=x^3-12x+20. The good news is that elliptic curves have been studied extensively and there is a description of the rational solutions of this equation, although it is not as simple as substituting numbers into a formula.

If we know x, y, then B and N are given by B=-(x+2)/(2y), N=(4-x-y)/(2y). y=-6/(2B-2N-1) means that y has to be a (possibly negative) integer dividing 6 or a fraction whose numerator in its lowest terms is a (possibly negative) integer dividing 6.

The integer solutions of y^2=x^3-12x+20 are (x,y)=(-2,6),(4,6),(2,2),(1,3),(10,30),(4,2),(22,102),(89,839) and the same x values with the sign of y changed. None of these give feasible values for B and N. The numerators and denominators of the rational solutions increase quickly, so numerator will never be a divisor 6, but a rigorous proof would require someone who knows more about this topic than I do.
GM
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June 19th, 2021 at 3:55:23 PM permalink
Quote: ThatDonGuy

Any ideas on how to show that 3^3 + 4^3 + 5^3 = 6^3 is the only solution in positive integers for cubes?


P^2, Q^2, R^2 form an arithmetic progression, the problem of 3 squares forming an arithmetic progression has been solved,
any solution is of the form P/Q=(-t^2+2t+1)/(t^2+1), P/Q=(t^2+2t-1)/(t^2+1), where t is a rational number, if we want P/Q and R/Q to be positive and P/Q<1, then t has to be between 1 and sqrt(2)+1. I am unable post a link, but search for Keith Conrad and arithmetic progression of three squares for a similar formula. (P=3, Q=15, R=21 corresponds to t=2.)

Let t=a/b with a and b coprime, then P/Q=(-a^2+2*a*b+b^2)/(a^2+b^2), R/Q=(a^2+2*a*b-b^2)/(a^2+b^2). a and be cannot both be even as they are coprime. If one of a and b is even and the other is odd, then the numerators and the denominators of these fractions are also coprime, so P=m(-a^2+2*a*b+b^2), Q=m(a^2+b^2), R=m(a^2+2*a*b-b^2) for some integer m. If a and b are both odd, then the greatest common divisor of the the numerators and the denominators is 2, P=m(-a^2+2*a*b+b^2)/2, Q=m(a^2+b^2)/2, R=m(a^2+2*a*b-b^2)/2 for some integer m. Now the question is, can all these numbers be triangular numbers?
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June 19th, 2021 at 5:07:01 PM permalink
I don't mean to change the topic, so please continue discussing squares. I find that topic interesting too and wish I had something of value to contribute to it.

Here is my question -- Consider a two-player case of Final Jeopardy with the following scores:

Amy: $10,000
Bob: $9,000

Each player has a 50% chance of getting Final Jeopardy correct and the probabilities are not correlated.

For simplicity, assume the choices of each player are to either go "high" or "low." A low bet by either is $0. A high bet for Amy is $8,001. A high bet for Bob is $9,000.

What should each player do?
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ChumpChange
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June 19th, 2021 at 5:25:07 PM permalink
If the 2nd place player keeps the money they've won (I don't know if they do because I haven't watched the show since 1984), Amy should bet $950 and Bob should bet $1,050.
unJon
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June 19th, 2021 at 5:48:03 PM permalink
Quote: Wizard

I don't mean to change the topic, so please continue discussing squares. I find that topic interesting too and wish I had something of value to contribute to it.

Here is my question -- Consider a two-player case of Final Jeopardy with the following scores:

Amy: $10,000
Bob: $9,000

Each player has a 50% chance of getting Final Jeopardy correct and the probabilities are not correlated.

For simplicity, assume the choices of each player are to either go "high" or "low." A low bet by either is $0. A high bet for Amy is $8,001. A high bet for Bob is $9,000.

What should each player do?



I assume the winner wins the amount of money he has left. But does does the loser get $0 or some other amount?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
rsactuary
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June 19th, 2021 at 6:20:19 PM permalink
Quote: unJon

I assume the winner wins the amount of money he has left. But does does the loser get $0 or some other amount?




Runner up gets a flat $2000, but doubt that was contemplated in the question. Need some clarification please.
ChumpChange
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June 19th, 2021 at 6:49:13 PM permalink
Final Jeopardy! 06/09/2021 1960s Singers | JEOPARDY! - YouTube
https://www.youtube.com/watch?v=9R9EhbdxdII

Clearly an all or nothing bid for top score.

9/26/84 Final Jeopardy! - YouTube
https://www.youtube.com/watch?v=O7ck9MWnn6Y
Seems other place finishers kept the money they won.

Super Old School episode:
JEOPARDY! 1974-75 Nighttime Syndicated Season - YouTube
https://www.youtube.com/watch?v=GtyWD9QpFME
If they all won in Final Jeopardy, there would have been a tie. Would they have given out 2 cars and had 2 returning champions?

A 3-Way Tie in Final Jeopardy! | JEOPARDY! - YouTube
https://www.youtube.com/watch?v=KDTxS9_CwZA
The leader could have bet just $600 to win the game, but no, no, no, he didn't do that.

Jeopardy! First: Tiebreaker | JEOPARDY! - YouTube
https://www.youtube.com/watch?v=c7cYON3uVZo

Bonus Daily Double Video Clip:
Norma Tells Norman's Secret | Bates Motel - YouTube
https://www.youtube.com/watch?v=Qes0evIhezI
Last edited by: ChumpChange on Jun 19, 2021
teliot
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June 19th, 2021 at 8:02:57 PM permalink
Quote: GM

I am quite confident that there is not. If you assume that 1^2+2^2+...+N^2=(N+1)^2+...+(N+B)^2, then after using the summation formula for squares, you get the equation -2B^3-6B^2N-3B^2-6BN^2-6BN-B+2N^3+3N^2+N=0, which has degree 3 in both B and N. By using the the change of variables x=2(4B+2N+1)/(2B-2N-1), y=-6/(2B-2N-1), this equation can be transformed to the form y^2=x^3-12x+20. (This is not black magic, it is called the Weierstrass form and the curve defined by this equation is an elliptic curve.)

If the equation -2B^3-6B^2N-3B^2-6BN^2-6BN-B+2N^3+3N^2+N=0 has an integer solution for B and N, we get a rational (but not necessarily integer) solution of y^2=x^3-12x+20. The good news is that elliptic curves have been studied extensively and there is a description of the rational solutions of this equation, although it is not as simple as substituting numbers into a formula.

If we know x, y, then B and N are given by B=-(x+2)/(2y), N=(4-x-y)/(2y). y=-6/(2B-2N-1) means that y has to be a (possibly negative) integer dividing 6 or a fraction whose numerator in its lowest terms is a (possibly negative) integer dividing 6.

The integer solutions of y^2=x^3-12x+20 are (x,y)=(-2,6),(4,6),(2,2),(1,3),(10,30),(4,2),(22,102),(89,839) and the same x values with the sign of y changed. None of these give feasible values for B and N. The numerators and denominators of the rational solutions increase quickly, so numerator will never be a divisor 6, but a rigorous proof would require someone who knows more about this topic than I do.

Very cool that you were able to transform this into an elliptic curve. I know a bit of algebraic geometry, studied it for 2 years in graduate school (40 years ago!). Maybe I could understand some of the more advanced arguments, though surely I am very rusty. Anyway thanks for your superb input here.
Last edited by: teliot on Jun 19, 2021
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June 19th, 2021 at 8:05:36 PM permalink
Let's assume the only goal, which is pretty much the case, is to beat the other player. It does not matter by how much.
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ThatDonGuy
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June 19th, 2021 at 8:27:40 PM permalink
Quote: ChumpChange

Super Old School episode:
JEOPARDY! 1974-75 Nighttime Syndicated Season - YouTube
https://www.youtube.com/watch?v=GtyWD9QpFME
If they all won in Final Jeopardy, there would have been a tie. Would they have given out 2 cars and had 2 returning champions?


That version did not have returning champions, any more than the weekly syndicated versions of Family Feud or Match Game (which may air on Buzzr - look for "Match Game PM") did. I am under the impression that both players in a tie would have won the corresponding prize.
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June 19th, 2021 at 8:56:53 PM permalink
I knew this would devolve into a discussion of Jeopardy rules.

Again, per the OP, each player may go only "high" or "low" per the amounts indicated. If we must discuss ties, assume both players LOSE in the event of a tie.
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ChumpChange
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June 19th, 2021 at 9:09:21 PM permalink
The show has a track record of betting "high", no matter.
camapl
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June 19th, 2021 at 11:37:30 PM permalink
Quote: Wizard

I don't mean to change the topic, so please continue discussing squares. I find that topic interesting too and wish I had something of value to contribute to it.

Here is my question -- Consider a two-player case of Final Jeopardy with the following scores:

Amy: $10,000
Bob: $9,000

Each player has a 50% chance of getting Final Jeopardy correct and the probabilities are not correlated.

For simplicity, assume the choices of each player are to either go "high" or "low." A low bet by either is $0. A high bet for Amy is $8,001. A high bet for Bob is $9,000.

What should each player do?



I’ve enjoyed reading through each problem on this thread, so thank you for that!


Bob should go high, as Amy’s expected score is $10,000 regardless of how she bets. This maximizes his overall chance at a win.
Amy should go low, as Bob’s expected score is $9,000 regardless of how he bets. This minimizes her overall chance at a loss.
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unJon
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June 20th, 2021 at 3:26:59 AM permalink
Quote: Wizard

I don't mean to change the topic, so please continue discussing squares. I find that topic interesting too and wish I had something of value to contribute to it.

Here is my question -- Consider a two-player case of Final Jeopardy with the following scores:

Amy: $10,000
Bob: $9,000

Each player has a 50% chance of getting Final Jeopardy correct and the probabilities are not correlated.

For simplicity, assume the choices of each player are to either go "high" or "low." A low bet by either is $0. A high bet for Amy is $8,001. A high bet for Bob is $9,000.

What should each player do?



Ok, based on your latter clarification that Amy and Bob are only acting to maximize their chances of winning:


This is a simple game theory problem with no pure Nash equilibrium solution. That’s because if Amy goes High, Bob is best off going Low (winning 50% for Bob). But if Bob goes low, then Amy does best going low (winning 100% for Amy). But if Amy goes low, Bob does best going high (winning 50%). But if Bob goes high, Amy does best going high (winning 75% for Amy).

In other words, Amy wants to do the same thing as Bob: go high or low together. And Bob wants to do the opposite of Amy: go low when she goes high and vice versa.

The mixed Nash equilibrium is found by choosing the probability of going High and Low such that the opponent is indifferent between the choice.

For Amy that means: 25x + 50(1 - x) = 50x + 0(1 - x)
x = 2/3

For Bob that means: 75y + 50(1 - y) = 50y + 100(1 - y)
y = 2/3

Both Amy and Bob should randomly choose high 2/3 of the time and low 1/3 of the time. That leads to Amy winning the show 2/3 of the time and Bob winning the show 1/3 of the time. Importantly, this is the mixed Nash equilibrium because neither Amy nor Bob can improve their chance of winning by varying their strategy so long as their opponent follows the 2/3 high and 1/3 low random strategy.

Note: that these %s would be different if Amy and Bob were trying to maximize the dollars they win based on Jeopardy rules.
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GM
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June 20th, 2021 at 5:08:25 AM permalink
If you know elliptic curves, then I add that the point (-2,6) generates the group of rational points, all rational points are multiples of (-2,6) under group operation on the curve.

The problem whether the sum of consecutive k-th powers can be the power of an integer (not necessarily a k-th power) has been studied, but I have not found anything about the problem you posed. I will ask someone who may know something.
Last edited by: GM on Jun 20, 2021
charliepatrick
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June 20th, 2021 at 5:10:03 AM permalink
Firstly consider how B plays depending on how A makes their decision with a random probability of P of going Hi.
A's possible winnings18,00110,0001,999
ProbabilityP/21-PP/2

Now look at the probability of B winning when B goes Hi or Lo.

B goes Hi: wins if (i) B correct & A goes Lo OR (ii) B correct & A goes Hi & A wrong: Pr = (1-P)/2 + P/4 = 1/2 - P/4.
B goes Lo: wins if (i) A goes Hi & A wrong: P/2.

So A has to choose P so B's chances to win are equal, so P = 2/3.
This makes B indifferent to Hi or Lo and has a 1/3 chance of being the game's winner.


Now check how A plays if they know B is picking with a random probability of S going Hi.
B's possible winnings16k8k0
ProbabilityS/21-SS/2

Now look at the probability of A winning when A goes Hi or Lo.

A goes Hi: wins if (i) correct OR (ii) wrong & B goes hi & B wrong: Pr = 1/2 + S/4.
A goes Lo: wins unless B goes Hi and correct: Pr = 1 - S/2.
B has to make Hi and Lo options for A the same, so S = 2/3.
This makes A indifferent to Hi or Lo and has a 2/3 chance of being the game's winner.

Also one can see than if either side alters their probabilities from 2/3, the other party can improve their chances by going Hi or Lo as appropriate.
teliot
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June 20th, 2021 at 6:51:02 AM permalink
Quote: GM

If you know elliptic curves, then I add that the point (-2,6) generates the group of rational points, all rational points are multiples of (-2,6) under group operation on the curve.

The problem whether the sum of consecutive k-th powers can be the power of an integer (not necessarily a k-th power) has been studied, but I have not found anything about the problem you posed. I will ask someone who may know something.

Yep I know about the group structure, so thanks. I appreciate any asking you can do. I'm going to ask one friend who is an expert in this area, just to see if this is known. Thanks again.
Last edited by: teliot on Jun 20, 2021
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June 20th, 2021 at 1:32:12 PM permalink
Quote: unJon



This is a simple game theory problem with no pure Nash equilibrium solution. That’s because if Amy goes High, Bob is best off going Low (winning 50% for Bob). But if Bob goes low, then Amy does best going low (winning 100% for Amy). But if Amy goes low, Bob does best going high (winning 50%). But if Bob goes high, Amy does best going high (winning 75% for Amy).

In other words, Amy wants to do the same thing as Bob: go high or low together. And Bob wants to do the opposite of Amy: go low when she goes high and vice versa.

The mixed Nash equilibrium is found by choosing the probability of going High and Low such that the opponent is indifferent between the choice.

For Amy that means: 25x + 50(1 - x) = 50x + 0(1 - x)
x = 2/3

For Bob that means: 75y + 50(1 - y) = 50y + 100(1 - y)
y = 2/3

Both Amy and Bob should randomly choose high 2/3 of the time and low 1/3 of the time. That leads to Amy winning the show 2/3 of the time and Bob winning the show 1/3 of the time. Importantly, this is the mixed Nash equilibrium because neither Amy nor Bob can improve their chance of winning by varying their strategy so long as their opponent follows the 2/3 high and 1/3 low random strategy.

Note: that these %s would be different if Amy and Bob were trying to maximize the dollars they win based on Jeopardy rules.



I agree! With Charlie as well.

I think on the actual show in this situation, where player #2 has more than 2/3 of player #1, both players tend to go high too often.

Here are some actual statistics from this situation, from seasons 30 to 34:

Both correct: 26.89%
Both wrong: 29.42%
High player right, low player wrong: 24.72%
High player wrong, low player right: 18.97%
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
teliot
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June 20th, 2021 at 2:30:56 PM permalink
Quote: GM

If you know elliptic curves, then I add that the point (-2,6) generates the group of rational points, all rational points are multiples of (-2,6) under group operation on the curve.

The problem whether the sum of consecutive k-th powers can be the power of an integer (not necessarily a k-th power) has been studied, but I have not found anything about the problem you posed. I will ask someone who may know something.

A friend wrote:

"I played quickly and see that Magma can convert the elliptic curve to a Weierstrass form, which is nice. The problem is that when one does such a conversion, the integer points from the original equation can be rational points on the new equation that are not integers. So this may not be as simple an exercise as I originally thought. Again, I have not done much with elliptic curves myself, so perhaps it is not that difficult in the end."

He suggested another professor, who I will write to and report back.
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teliot
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June 21st, 2021 at 8:24:51 AM permalink
Quote: teliot

A friend wrote:

"I played quickly and see that Magma can convert the elliptic curve to a Weierstrass form, which is nice. The problem is that when one does such a conversion, the integer points from the original equation can be rational points on the new equation that are not integers. So this may not be as simple an exercise as I originally thought. Again, I have not done much with elliptic curves myself, so perhaps it is not that difficult in the end."

He suggested another professor, who I will write to and report back.

The other professor wrote back to me. He is an expert in Diophantine Equations. He wrote a rather long email to me, he definitely gave this some deep thought, and wrote in part,

"The problem therefore is to ensure these ratios are integral, for (x,y) a rational point on y^2=x^3-12x+20. This may or may not be easy: I don't have ideas off the top of my head. I checked that there are no solutions to your equation for |b|,|n|<10^5, so likely there are none.

But hope is not lost. Nikos Tzanakis has written a memoir devoted exclusively to finding integral points on curves of genus 1, particularly helpful when the curve of genus 1, as (*) above, is not in standard form."
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Gialmere
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June 21st, 2021 at 8:25:58 AM permalink
It's easy Monday. Let's take a spin...



You're studying to be a lab technician and have started to learn about the centrifuge. A centrifuge has 12 locations for test tubes, located evenly in a circle. (Let them be referenced as the hour locations on a clock.) Prior to spinning the tubes, the centrifuge must be evenly balanced about its center.

Assuming all test tubes weigh the same amount, what numbers of test tubes (from 1 to 12) can be spun without needing to add dummy tubes to balance out the weight?

For example, you can't spin 1 sample tube at 9 o'clock without balancing it with a dummy at 3 o'clock, but you could spin 2 sample tubes.


Have you tried 22 tonight? I said 22.
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June 21st, 2021 at 9:27:49 AM permalink

2: 3 and 9:00
3: 12, 4, and 8:00
4. 12, 3, 6, and 9:00
6: 12, 2, 4, 6, 8, and 10:00
8: 12, 1, 2, 3, 6,7,8, and 9:00
9: 1,2,3,5,6,7, 9, 10, and 11:00
10: 12, 1,2,3,4, 6, 7, 8, 9, and 10:00

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
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June 21st, 2021 at 10:11:49 AM permalink
Obviously if you can do N, then by reverse logic you can do 12-N. So we only need to consider 0 thru 6, assume there would be one placed at 12h.
None works.
One doesn't.
Two does (12h 6h)
Three does (12h 4h 8h)
Four does (12h 3h 6h 9h)
Five doesn't.
Six does (12h 2h 4h 6h 8h 10h)
So the answers are (0,) 2, 3, 4, 6, 8, 9, 10, 12.
edit: Ignore my zero response as you stated that in your question there had to be at least one test tube, but it does imply 12 is a solution.
Gialmere
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June 21st, 2021 at 10:22:34 AM permalink
Despite the Wizard playing a lab tech on TV, both his and CP's answers are incorrect.
Have you tried 22 tonight? I said 22.
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June 21st, 2021 at 10:26:12 AM permalink
The unique prime divisors of 12 are 2 and 3, so at the very least it is clear that it balances with all multiples of 2 and 3 that are <= 12. And 5 = 2+3, 7 = 2+2+3 so you can do both of these since balanced + balanced = balanced. Of course, this relies on open spots being available to put the balanced configurations, which is why 11 fails.
Last edited by: teliot on Jun 21, 2021
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June 21st, 2021 at 10:28:01 AM permalink
Quote: charliepatrick

Obviously if you can do N, then by reverse logic you can do 12-N. So we only need to consider 0 thru 6, assume there would be one placed at 12h.
None works.
One doesn't.
Two does (12h 6h)
Three does (12h 4h 8h)
Four does (12h 3h 6h 9h)
Five doesn't.
Six does (12h 2h 4h 6h 8h 10h)
So the answers are (0,) 2, 3, 4, 6, 8, 9, 10, 12.
edit: Ignore my zero response as you stated that in your question there had to be at least one test tube, but it does imply 12 is a solution.




If 2 works and 3 works, wouldn't 5 work, too?

The center of mass for tubes at 12:00 and 6:00 would be at the center of the centrifuge. And the center of mass for tubes at 1:00, 5:00, and 9:00 would be there, also. So, the center of mass for all 5 tubes would be at the center.
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June 21st, 2021 at 11:02:46 AM permalink
Quote: charliepatrick

Obviously if you can do N, then by reverse logic you can do 12-N. So we only need to consider 0 thru 6, assume there would be one placed at 12h.
None works.
One doesn't.
Two does (12h 6h)
Three does (12h 4h 8h)
Four does (12h 3h 6h 9h)
Five doesn't.
Six does (12h 2h 4h 6h 8h 10h)
So the answers are (0,) 2, 3, 4, 6, 8, 9, 10, 12.
edit: Ignore my zero response as you stated that in your question there had to be at least one test tube, but it does imply 12 is a solution.



Five does (12h 3h 4h 8h 9h) - the 3 balances the 9, and the 12, 4, and 8 balance each other as well.

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June 21st, 2021 at 12:16:54 PM permalink
Quote: teliot

The other professor wrote back to me. He is an expert in Diophantine Equations. He wrote a rather long email to me, he definitely gave this some deep thought, and wrote in part,

"The problem therefore is to ensure these ratios are integral, for (x,y) a rational point on y^2=x^3-12x+20. This may or may not be easy: I don't have ideas off the top of my head. I checked that there are no solutions to your equation for |b|,|n|<10^5, so likely there are none.

But hope is not lost. Nikos Tzanakis has written a memoir devoted exclusively to finding integral points on curves of genus 1, particularly helpful when the curve of genus 1, as (*) above, is not in standard form."


I will try to get hold of Nikos Tzanakis' memoir. I have also noticed that he has a paper on sums of consecutive squares being a perfect square, which also involves elliptic curves.

In the meanwhile, I have come up with the following idea, which could reduce the problem to checking finitely many points on the elliptic curve y^2=x^3-12x+20 if someone were willing to do some analytic number theory.

As I wrote earlier, integer solutions for N and B would give a rational point on the elliptic curve such that the numerator of y-coordinate divides 6. Let's assume that we have such a rational point.

The co-ordinates of the rational points on the elliptic curve are of the form x=r/t^2, y=s/t^3, where r, s, t are integers, r and s are coprime to t, but not necessarily to each other.

Let u be the real root of x^3-12x+20 (it's about -4.11), and v, w the non-real roots. Then x^3-12x+20=(x-u)(x-v)(x-w), so ((r/t^2)-u)((r/t^2)-v)((r/t^2)-w)=s^2/t^6. r/t^2 is close to u, certainly between -4.11 and -4, which means that ((r/t^2)-v)((r/t^2)-w)>37 as v is about 2.05+0.8i and w is its conjugate, while on the right-hand side, s^2 is at most 36, so r/t^2-u<1/t^6. r/t^2-u>0, therefore |r/t^2-u|<1/t^6. This means that r/t^2 is a very good approximation to u.

Roth's theorem says that for any epsilon>0, there are only finitely many integers p, q (q>0) such that |u-p/q|<1/q^(2+epsilon). Unfortunately, this theorem is not effective, but there is an effective version due to Feldman, which says that there exists a kappa<3 (3 is degree of the polynomial which u is a root of) and a constant c>0, such that |u-p/q|>c/q^kappa for all integers p, q (q>0), and kappa and c can be computed in principle. This, combined with |r/t^2-u|<1/t^6 would give a bound on t.

The problem is that there are formulas which kappa and c which work for any algebraic number, but these are useless for practical purposes (the bound on t would be at least 10^1000000000000), however, much better kappa and c can be obtained for any concrete number if one is willing to work through many pages of analytic number theory. For example, |2^(1/3)-p/q|>(q^2.47)/4, and there are also good values of kappa and c for cube roots of other integers less than 100, but I don't know to what extent these calculations could be adapted to deal with the root of a more general cubic equation.

It is known that the logarithm of the maximum of the absolute values of the numerator and denominator of m times (-2,6) in the group structure on the elliptic curve grows like Cm^2, where C is a constant, and it is possible to give an effective lower bound for it. This means that given a bound on t, m can be bounded effectively and then it is a finite calculation to check all the possibilities. The bound on t would not have to be very good, something like 10^100000 would be OK, probably even 10^1000000, because it would still only involve checking a few thousand multiples of (-2,6).
charliepatrick
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June 21st, 2021 at 3:58:52 PM permalink
Quote: charliepatrick

Obviously if you can do N, then by reverse logic you can do 12-N. So we only need to consider 0 thru 6, assume there would be one placed at 12h.
None works.
One doesn't.
Two does (12h 6h)
Three does (12h 4h 8h)
Four does (12h 3h 6h 9h)
Five doesn't.
Six does (12h 2h 4h 6h 8h 10h)
So the answers are (0,) 2, 3, 4, 6, 8, 9, 10, 12.
edit: Ignore my zero response as you stated that in your question there had to be at least one test tube, but it does imply 12 is a solution.

Further to other responses, I agree...
...that 5 would also work using 12h and 6h for 2, then adding (say) 11h 7h 3h. This also means 7 would work. So it looks as if only 1 and 11 don't work.
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June 21st, 2021 at 4:52:33 PM permalink
Quote: teliot

The unique prime divisors of 12 are 2 and 3, so at the very least it is clear that it balances with all multiples of 2 and 3 that are <= 12. And 5 = 2+3, 7 = 2+2+3 so you can do both of these since balanced + balanced = balanced. Of course, this relies on open spots being available to put the balanced configurations, which is why 11 fails.


Quote: charliepatrick

Further to other responses, I agree...

...that 5 would also work using 12h and 6h for 2, then adding (say) 11h 7h 3h. This also means 7 would work. So it looks as if only 1 and 11 don't work.


Correct!!

Easy Monday not so easy.

Yes, on a 12 tube centrifuge, only 1 and 11 tube spins need a dummy.



Note that the empty spots for a 5-tube spin become the full spots for a 7.

----------------------------------------------

Have you tried 22 tonight? I said 22.
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June 21st, 2021 at 7:38:03 PM permalink
Quote: Gialmere

Despite the Wizard playing a lab tech on TV, both his and CP's answers are incorrect.



Hey! I was a full blown pathologist (as evidenced by the white jacket) as opposed to a lowly lab tech. I got the job because I showed up as an extra wearing a tie and the other two extras didn't.
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teliot
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June 22nd, 2021 at 6:30:08 AM permalink
Quote: GM

I will try to get hold of Nikos Tzanakis' memoir. I have also noticed that he has a paper on sums of consecutive squares being a perfect square, which also involves elliptic curves.

In the meanwhile, I have come up with the following idea, which could reduce the problem to checking finitely many points on the elliptic curve y^2=x^3-12x+20 if someone were willing to do some analytic number theory.

As I wrote earlier, integer solutions for N and B would give a rational point on the elliptic curve such that the numerator of y-coordinate divides 6. Let's assume that we have such a rational point.

The co-ordinates of the rational points on the elliptic curve are of the form x=r/t^2, y=s/t^3, where r, s, t are integers, r and s are coprime to t, but not necessarily to each other.

Let u be the real root of x^3-12x+20 (it's about -4.11), and v, w the non-real roots. Then x^3-12x+20=(x-u)(x-v)(x-w), so ((r/t^2)-u)((r/t^2)-v)((r/t^2)-w)=s^2/t^6. r/t^2 is close to u, certainly between -4.11 and -4, which means that ((r/t^2)-v)((r/t^2)-w)>37 as v is about 2.05+0.8i and w is its conjugate, while on the right-hand side, s^2 is at most 36, so r/t^2-u<1/t^6. r/t^2-u>0, therefore |r/t^2-u|<1/t^6. This means that r/t^2 is a very good approximation to u.

Roth's theorem says that for any epsilon>0, there are only finitely many integers p, q (q>0) such that |u-p/q|<1/q^(2+epsilon). Unfortunately, this theorem is not effective, but there is an effective version due to Feldman, which says that there exists a kappa<3 (3 is degree of the polynomial which u is a root of) and a constant c>0, such that |u-p/q|>c/q^kappa for all integers p, q (q>0), and kappa and c can be computed in principle. This, combined with |r/t^2-u|<1/t^6 would give a bound on t.

The problem is that there are formulas which kappa and c which work for any algebraic number, but these are useless for practical purposes (the bound on t would be at least 10^1000000000000), however, much better kappa and c can be obtained for any concrete number if one is willing to work through many pages of analytic number theory. For example, |2^(1/3)-p/q|>(q^2.47)/4, and there are also good values of kappa and c for cube roots of other integers less than 100, but I don't know to what extent these calculations could be adapted to deal with the root of a more general cubic equation.

It is known that the logarithm of the maximum of the absolute values of the numerator and denominator of m times (-2,6) in the group structure on the elliptic curve grows like Cm^2, where C is a constant, and it is possible to give an effective lower bound for it. This means that given a bound on t, m can be bounded effectively and then it is a finite calculation to check all the possibilities. The bound on t would not have to be very good, something like 10^100000 would be OK, probably even 10^1000000, because it would still only involve checking a few thousand multiples of (-2,6).

I taught Hurwitz's Theorem many times and Roth's Theorem is not a surprise, just showing that Hurwitz's result was sharp. At any rate, I found this Wikipedia page helpful in understanding your excellent post and showing that this is a common approach that has been used in many different situations. It also covers Feldman's result that you mentioned.

https://en.wikipedia.org/wiki/Diophantine_approximation

In particular, there is this comment in the Wiki-article about the feasibility of Feldman's constants: "However, as for every effective version of Baker's theorem, the constants d and 1/c are so large that this effective result cannot be used in practice."

My analytic number theory skills are almost non-existent, my own research touched algebraic number theory a couple of times and it's been decades, so I can't help here. But thank you so much for your ideas on this.

I would love it if you got a paper on this question, and I would personally love to know who you are, as your academic credentials are obviously first rate. Feel free to drop me a personal email.
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Gialmere
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June 22nd, 2021 at 8:03:27 AM permalink
It's toughie Tuesday.



A random number generator generates integers in the range 1...n, where n is a parameter passed into the generator. The output from the generator is repeatedly passed back in as the input.

If the initial input parameter is one googol (10^100), find, to the nearest integer, the expected value of the number of iterations by which the generator first outputs the number 1.

That is, what is the expected value of x, after running the following pseudo-code?

n = 10^100
x = 0
do while (n > 1)
n = random(n) // Generates random integer in the range 1...n
x = x + 1
Have you tried 22 tonight? I said 22.
ThatDonGuy
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June 22nd, 2021 at 9:40:58 AM permalink
Quote: GialmereA random number generator generates integers in the range 1...n, where n is a parameter passed into the generator. The output from the generator is repeatedly passed back in as the input.

If the initial input parameter is one googol (10^100), find, to the nearest integer, the expected value of the number of iterations by which the generator first outputs the number 1.[/b



That is, what is the expected value of x, after running the following pseudo-code?



Let E(n) be the expected number of iterations needed from number n.
E(1) = 0
E(2) = 1 + 1/2 E(1) + 1/2 E(2)
1/2 E(2) = 1
E(2) = 2
For n > 2, E(n) = 1 + 1/n E(1) + 1/n E(2) + ... + 1/n E(n)
(n - 1)/n E(n) = 1 + 1/n E(1) + 1/n E(2) + ... + 1/n E(n-1)
(n - 1)/n E(n) = 1 + n/(n-1) x (1/(n-1) E(1) + 1/(n-1) E(2) + ... + 1/(n-1) E(n-1))
(n - 1)/n E(n) = 1/n + (n-1)/n x (1 + 1/(n-1) E(1) + 1/(n-1) E(2) + ... + 1/(n-1) E(n-1))
(n - 1)/n E(n) = 1/n + (n-1)/n E(n-1)
E(n) = 1/(n-1) + E(n-1)
Therefore:
E(3) = 1/2 + E(2)
E(4) = 1/3 + E(3) = 1/3 + 1/2 + E(2)
...
E(10^100) = 1/(10^100 - 1) + 1/(10^100 - 2) + ... + 1/3 + 1/2 + E(2)
This is approximately 1 + the integral over x = 1 to 10^100 - 1 of dx / x, which is ln (10^100 - 1) - ln 1 + 1 = 1 + 100 ln 10 = about 231.25.

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June 22nd, 2021 at 10:36:23 AM permalink
What ThatDonGuy said plus EM constant.

So 100 * ln(10) + .5772 + 1 =~ 231.84
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June 22nd, 2021 at 10:48:12 AM permalink
Quote: Ace2

What ThatDonGuy said plus EM constant.

So 100 * ln(10) + .5772 + 1 =~ 231.84


Okay, I'll ask: what's the EM constant?
teliot
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June 22nd, 2021 at 10:51:51 AM permalink
Quote: ThatDonGuy

Quote: Ace2

What ThatDonGuy said plus EM constant.

So 100 * ln(10) + .5772 + 1 =~ 231.84


Okay, I'll ask: what's the EM constant?

I believe EM refers to Euler's constant.

https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
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June 22nd, 2021 at 10:57:18 AM permalink
Euler–Mascheroni constant.

For example, take sum of 1/1 + 1/2 .... + 1/10000 and it equals about 9.7876.

ln(10,000) =~ 9.2103. The difference is approximately the constant, and the higher the number, the closer the difference gets to the exact value of the constant.
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