Thread Rating:

Poll

16 votes (50%)
12 votes (37.5%)
5 votes (15.62%)
2 votes (6.25%)
10 votes (31.25%)
3 votes (9.37%)
6 votes (18.75%)
5 votes (15.62%)
10 votes (31.25%)
7 votes (21.87%)

32 members have voted

teliot
teliot
Joined: Oct 19, 2009
  • Threads: 40
  • Posts: 2221
June 15th, 2021 at 3:43:24 PM permalink
Okay ... new puzzle. A sum of consecutive squares equaling a sum of the following successive consecutive squares.

3^2 + 4^2 = 5^2.

10^2 + 11^2 + 12^2 = 13^2 + 14^2

21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2

18^2 + 19^2 + 20^2 + ... + 34^2 = 35^2 + 36^2 + 37^2 + ... + 42^2

What's the longest one you can find?
Poetry website: www.totallydisconnected.com
Gialmere
Gialmere
Joined: Nov 26, 2018
  • Threads: 41
  • Posts: 2051
June 15th, 2021 at 4:03:46 PM permalink
Quote: aceside

I havenít got a correct confirmation, so I recalculated it. My new result is mu=0.18896447.


Correct!!

Very good.
---------------------------------------

Did you hear about the Star Wars fan who ate a truck?

They call him the The Mandalorryin.
Have you tried 22 tonight? I said 22.
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 24
  • Posts: 1046
June 15th, 2021 at 5:22:32 PM permalink
Let the record show that Miplet did already have the correct answer, which he had calculated on a large spreadsheet. Today we learned some new methods and also we got to the exact rational answer for the first time.

I owe all three of you a beer:

Miplet: First to solve
ThatDonGuy: First to post a rational answer
Wizard: In my opinion, he provided the most elegant/efficient solution...expressed as a single line formula

Incidentally, my method uses the same concept as the Wizard's but it's not exactly the same. I took 1 minus the integral over all time of:

(1-((1-1/e^(t/210))*(1-1/e^(t/60))*(1-1/e^(t/30))*(1-1/e^(4t/75))*(1-1/e^(5t/66)))^2)*1/e^(7303t/11550)*7303/11550 dt
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 105
  • Posts: 4925
Thanks for this post from:
teliot
June 15th, 2021 at 6:25:21 PM permalink
Quote: teliot

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

3^2 + 4^2 = 5^2.

10^2 + 11^2 + 12^2 = 13^2 + 14^2

21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2

18^2 + 19^2 + 20^2 + ... + 34^2 = 35^2 + 36^2 + 37^2 + ... + 42^2

What's the longest one you can find?



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
The left is the sum of 1^2 through N^2 - the sum of 1^2 through (N - A)^2
The right is the sum of 1^2 through (N + B)^2 - the sum of 1^2 through N^2
1^2 + 2^2 + ... + N^2 = N (N + 1) (2N + 1) / 6 for all positive integers N
1/6 N (N + 1) (2N + 1) - 1/6 (N - A) (N - A + 1) (2 (N - A) + 1) = 1/6 (N + B) (N + B + 1) (2 (N + B) + 1) - 1/6 N (N + 1) (2N + 1)
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)

The largest sum I have found so far has 6195 terms:
385^2 + 386^2 + ... + 5222^2 = 5223^2 + 5224^2 + ... + 6579^2 = 47,461,421,035

ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 105
  • Posts: 4925
June 15th, 2021 at 6:26:27 PM permalink
Quote: Ace2

Incidentally, my method uses the same concept as the Wizard's but it's not exactly the same. I took 1 minus the integral over all time of:

(1-((1-1/e^(t/210))*(1-1/e^(t/60))*(1-1/e^(t/30))*(1-1/e^(4t/75))*(1-1/e^(5t/66)))^2)*1/e^(7303t/11550)*7303/11550 dt


I understand the first five terms, but where does 7303 / 11550 come from?
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 24
  • Posts: 1046
June 15th, 2021 at 7:06:16 PM permalink
Quote: ThatDonGuy

I understand the first five terms, but where does 7303 / 11550 come from?

It's the probability of sevening out. Sum the first five terms, multiply by 2, then add 7303/11550 and it equals 1.

The formula says: while at least one point has not been won and no there have been no seven-outs, what's the chance that the next decision is a seven out?
Itís all about making that GTA
teliot
teliot
Joined: Oct 19, 2009
  • Threads: 40
  • Posts: 2221
June 15th, 2021 at 7:16:15 PM permalink
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
The left is the sum of 1^2 through N^2 - the sum of 1^2 through (N - A)^2
The right is the sum of 1^2 through (N + B)^2 - the sum of 1^2 through N^2
1^2 + 2^2 + ... + N^2 = N (N + 1) (2N + 1) / 6 for all positive integers N
1/6 N (N + 1) (2N + 1) - 1/6 (N - A) (N - A + 1) (2 (N - A) + 1) = 1/6 (N + B) (N + B + 1) (2 (N + B) + 1) - 1/6 N (N + 1) (2N + 1)
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)

The largest sum I have found so far has 6195 terms:
385^2 + 386^2 + ... + 5222^2 = 5223^2 + 5224^2 + ... + 6579^2 = 47,461,421,035

Clever! And beats my largest by (ehem) a few terms.
Poetry website: www.totallydisconnected.com
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 105
  • Posts: 4925
Thanks for this post from:
teliot
June 15th, 2021 at 7:27:34 PM permalink
Quote: teliot

Quote: ThatDonGuy



The largest sum I have found so far has 6195 terms:
385^2 + 386^2 + ... + 5222^2 = 5223^2 + 5224^2 + ... + 6579^2 = 47,461,421,035

Clever! And beats my largest by (ehem) a few terms.



8964 terms: 2931^2 + ... + 9487^2 = 9488^2 + ... + 11,894^2 = 276,276,145,995

Last edited by: ThatDonGuy on Jun 15, 2021
Wizard
Administrator
Wizard
Joined: Oct 14, 2009
  • Threads: 1387
  • Posts: 23294
June 15th, 2021 at 8:32:21 PM permalink
Quote: Ace2

I owe all three of you a beer:



Very gracious of you! I proudly accept.
It's not whether you win or lose; it's whether or not you had a good bet.
aceside
aceside
Joined: May 14, 2021
  • Threads: 0
  • Posts: 48
June 16th, 2021 at 8:01:24 AM permalink
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
The left is the sum of 1^2 through N^2 - the sum of 1^2 through (N - A)^2
The right is the sum of 1^2 through (N + B)^2 - the sum of 1^2 through N^2
1^2 + 2^2 + ... + N^2 = N (N + 1) (2N + 1) / 6 for all positive integers N
1/6 N (N + 1) (2N + 1) - 1/6 (N - A) (N - A + 1) (2 (N - A) + 1) = 1/6 (N + B) (N + B + 1) (2 (N + B) + 1) - 1/6 N (N + 1) (2N + 1)
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)

The largest sum I have found so far has 6195 terms:
385^2 + 386^2 + ... + 5222^2 = 5223^2 + 5224^2 + ... + 6579^2 = 47,461,421,035


Can you also find some successive consecutive cubes for fun? Just replace all of the second power to the third power.

  • Jump to: