Yeah, Grok has DeepSearch and DeeperSearch as well. Haven’t tried DeeperSearch yet but DeepSearch often takes a couple of minutes. Good for collecting factual information from the web and summarizing it. “Think” mode is for multi-step reasoning problems that don’t depend on much outside information but might have non-obvious considerations. Basically anything where you’d be writing down notes and intermediate results if you were doing it manually.
My sense is (first impressions so big grain of salt) is Grok is much more clinical in its response. No faffing about, here’s the math. Done.
Claude gives a little more context and info which is nice for the layman but probably annoying fluff to a mathematician.
But they both got the same answer so that’s good.
Find the next term of the sequence and explain why:
15, 126, 225, 141, 66, 432, 99, 1458, 446, 344, 155, 251, 134, 92, 737, 713, 371, …
Here is the AI’s answer. Is it right? (I sure do not know)
I’ll analyze this sequence systematically to find the pattern and determine the next term.
Step 1: Initial Analysis
Let me first look at the sequence: 15, 126, 225, 141, 66, 432, 99, 1458, 446, 344, 155, 251, 134, 92, 737, 713, 371, …
I’ll check for common patterns:
- Not arithmetic (differences aren’t constant)
- Not geometric (ratios aren’t constant)
- Not immediately recognizable as powers or factorials
Step 2: Examining Digit Properties
Let me calculate the sum of digits for each term:
- 15: 1 + 5 = 6
- 126: 1 + 2 + 6 = 9
- 225: 2 + 2 + 5 = 9
- 141: 1 + 4 + 1 = 6
- 66: 6 + 6 = 12
- 432: 4 + 3 + 2 = 9
- 99: 9 + 9 = 18
- 1458: 1 + 4 + 5 + 8 = 18
- 446: 4 + 4 + 6 = 14
- 344: 3 + 4 + 4 = 11
- 155: 1 + 5 + 5 = 11
- 251: 2 + 5 + 1 = 8
- 134: 1 + 3 + 4 = 8
- 92: 9 + 2 = 11
- 737: 7 + 3 + 7 = 17
- 713: 7 + 1 + 3 = 11
- 371: 3 + 7 + 1 = 11
This is interesting! Many terms have digit sums that are multiples of 3 or related numbers.
Step 3: Checking Divisibility by 3
Let me check if the terms are divisible by 3 (a number is divisible by 3 if its digit sum is divisible by 3):
- 15 ÷ 3 = 5 ✓
- 126 ÷ 3 = 42 ✓
- 225 ÷ 3 = 75 ✓
- 141 ÷ 3 = 47 ✓
- 66 ÷ 3 = 22 ✓
- 432 ÷ 3 = 144 ✓
- 99 ÷ 3 = 33 ✓
- 1458 ÷ 3 = 486 ✓
- 446 ÷ 3 = 148.67… ✗
- 344 ÷ 3 = 114.67… ✗
- 155 ÷ 3 = 51.67… ✗
- 251 ÷ 3 = 83.67… ✗
- 134 ÷ 3 = 44.67… ✗
- 92 ÷ 3 = 30.67… ✗
- 737 ÷ 3 = 245.67… ✗
- 713 ÷ 3 = 237.67… ✗
- 371 ÷ 3 = 123.67… ✗
Step 4: Looking for a Different Pattern
Let me examine the positions more carefully:
- Positions 1-8: Mix of divisible and not divisible by 3
- Position 9 onwards: None are divisible by 3
Wait, let me check if these might be permutations of digits…
Step 5: Analyzing Digit Permutations
Looking at consecutive terms:
- 344, 155 → Different digits entirely
- 251, 134 → Different digits
- 737, 713, 371 → These three numbers use the same digits! {7, 3, 1}
This is a major clue! Let me check if there are other groups:
Looking more carefully:
- 737, 713, 371 all use digits {7, 3, 1}
- Let me see if earlier numbers form groups…
Step 6: Finding the Pattern
After further analysis, I notice this could be related to:
- Groups of numbers that are permutations of the same digits
- The sequence might be listing numbers with specific properties in increasing order within each permutation group
For the group {7, 3, 1}:
- All permutations: 137, 173, 317, 371, 713, 731, 737, 773
- In our sequence: 737, 713, 371
These appear to be in descending order within their permutation group.
Step 7: Determining the Next Term
Following the pattern of the {7, 3, 1} group in descending order:
- 737 (done)
- 713 (done)
- 371 (done)
- 317 (next!)
Answer
The next term in the sequence is 317.
Explanation: The sequence appears to be organized in groups where each group consists of permutations of the same set of digits, listed in descending order. The last three terms (737, 713, 371) all use the digits {7, 3, 1}, and continuing in descending order, the next permutation would be 317.
As an aside…it is telling me I am almost out of usage till 10p CST. I might get one more in before that.
Chatgpt o3-pro:
The rule
Take the decimal digits of the current term, cube each digit, and add the cubes.
Formally, if the current term is
an=∑k=0m−1dk10 k(dk∈{0,…,9}),a_n = \sum_{k=0}^{m-1} d_k 10^{,k}\qquad(d_k\in{0,\ldots,9}),an=k=0∑m−1dk10k(dk∈{0,…,9}),
then
an+1=∑k=0m−1dk 3.a_{n+1}= \sum_{k=0}^{m-1} d_k^{,3}.an+1=k=0∑m−1dk3.
Verifying the rule (with one arithmetic slip in the published list)
| Term | Cubes of its digits | Sum (next term shown in the list) |
|---|---|---|
| 15 | 13+53=1+1251^3+5^3 = 1+12513+53=1+125 | 126 |
| 126 | 13+23+63=1+8+2161^3+2^3+6^3 = 1+8+21613+23+63=1+8+216 | 225 |
| 225 | 23+23+53=8+8+1252^3+2^3+5^3 = 8+8+12523+23+53=8+8+125 | 141 |
| 141 | 13+43+13=1+64+11^3+4^3+1^3 = 1+64+113+43+13=1+64+1 | 66 |
| 66 | 63+63=216+2166^3+6^3 = 216+21663+63=216+216 | 432 |
| 432 | 43+33+23=64+27+84^3+3^3+2^3 = 64+27+843+33+23=64+27+8 | 99 |
| 99 | 93+93=729+7299^3+9^3 = 729+72993+93=729+729 | 1458 |
| (Correct value) | 13+43+53+83=1+64+125+512=7021^3+4^3+5^3+8^3 = 1+64+125+512 = 70213+43+53+83=1+64+125+512=702 | |
| (Published value) | author apparently used 1+64+125+256=4461+64+125+\boxed{256}=4461+64+125+256=446 | 446 |
The eighth step in the printed sequence mis‑cubes 838^383 as 256, giving 446 instead of the correct 702. From 446 onward the stated sequence again follows the “sum of cubes of digits” rule:
446→344→155→251→134→92→737→713→371.
Reaching a fixed point
The number 371 is a classic “Armstrong number” (a fixed point of this digit‑cube map):
33+73+13=27+343+1=371.3^3 + 7^3 + 1^3 = 27 + 343 + 1 = 371.33+73+13=27+343+1=371.
Once the sequence hits 371 it stays there forever.
The next term
Because 371↦371371 \mapsto 371371↦371 under the rule, the next term is 371 again.
Answer: 371 — the sequence is generated by repeatedly replacing a number with the sum of the cubes of its digits, and 371 is a fixed point of that process.
I asked Claude about your answer and it is a bit confused (not meaning Chatgpt is wrong):
Ah! This is a crucial insight. Let me re-analyze the entire sequence with this Armstrong number (narcissistic number) perspective.
Understanding Armstrong Numbers
An Armstrong number is a number that equals the sum of its digits raised to the power of the number of digits. For 3-digit numbers: abc = a³ + b³ + c³
Re-analyzing the Sequence as an Iterative Process
Let me check if each term is the sum of cubes of the previous term’s digits:
- 15 → 1³ + 5³ = 1 + 125 = 126 ✓
- 126 → 1³ + 2³ + 6³ = 1 + 8 + 216 = 225 ✓
- 225 → 2³ + 2³ + 5³ = 8 + 8 + 125 = 141 ✓
- 141 → 1³ + 4³ + 1³ = 1 + 64 + 1 = 66 ✓
- 66 → 6³ + 6³ = 216 + 216 = 432 ✓
- 432 → 4³ + 3³ + 2³ = 64 + 27 + 8 = 99 ✓
- 99 → 9³ + 9³ = 729 + 729 = 1458 ✓
- 1458 → 1³ + 4³ + 5³ + 8³ = 1 + 64 + 125 + 512 = 702 ✗
Wait, that doesn’t work for 4-digit numbers. Let me reconsider…
Checking Different Powers Based on Number of Digits
For n-digit numbers, we might use n as the power:
- 1-digit: d¹
- 2-digit: d² for each digit
- 3-digit: d³ for each digit
- 4-digit: d⁴ for each digit
Let me re-check:
- 1458 (4 digits) → 1⁴ + 4⁴ + 5⁴ + 8⁴ = 1 + 256 + 625 + 4096 = 4978 ✗
That’s not right either. Let me try always using cubes regardless of digit count:
- 1458 → 1³ + 4³ + 5³ + 8³ = 1 + 64 + 125 + 512 = 702 ✗ (we have 446)
New Hypothesis: Always Use Cubes
Actually, wait. Let me check if it’s always cubes but I made an arithmetic error:
- 1458 → 1³ + 4³ + 5³ + 8³ = 1 + 64 + 125 + 512 = 702 ✗
No, that gives 702, not 446. Let me reconsider…
Checking Original Sequence More Carefully
Actually, let me verify the earlier calculations I marked as correct:
- 15 → 1³ + 5³ = 1 + 125 = 126 ✓
- 126 → 1³ + 2³ + 6³ = 1 + 8 + 216 = 225 ✓
- 225 → 2³ + 2³ + 5³ = 8 + 8 + 125 = 141 ✓
- 141 → 1³ + 4³ + 1³ = 1 + 64 + 1 = 66 ✓
- 66 → 6³ + 6³ = 216 + 216 = 432 ✓
- 432 → 4³ + 3³ + 2³ = 64 + 27 + 8 = 99 ✓
- 99 → 9³ + 9³ = 729 + 729 = 1458 ✓
This works perfectly up to 1458! But then:
- 1458 → 1³ + 4³ + 5³ + 8³ = 1 + 64 + 125 + 512 = 702
But the sequence shows 446, not 702.
Wait! Maybe for 4-digit numbers, we need to sum the cubes differently. Let me check if 446 could come from interpreting 1458 differently…
No, that doesn’t seem right. Let me reconsider the entire pattern…
Alternative: Multiple Sequences?
Perhaps this is multiple sequences interleaved? Or the rule changes at some point?
The Key Insight About 371
You mentioned that 371 is an Armstrong number:
- 371 → 3³ + 7³ + 1³ = 27 + 343 + 1 = 371 ✓
This is a fixed point! Once we reach 371, applying the cube-sum operation keeps giving us 371.
Conclusion
Based on the Armstrong number insight and the pattern holding for the first 8 terms, the sequence appears to be the iterative application of the digit-cube sum operation:
The next term is 371 (same as the previous term).
The sequence reaches the fixed point 371 and stays there forever, as you noted. Each subsequent term will be:
- 371 → 3³ + 7³ + 1³ = 27 + 343 + 1 = 371
So the sequence continues: …737, 713, 371, 371, 371, 371, …
However, I’m still puzzled about the middle portion of the sequence (from position 9 to 14) which doesn’t seem to follow the digit-cube sum rule. There might be an error in the sequence as given, or a rule change I’m not seeing.
Correction: Don Miguel didn’t effect the change, but it was changed by the book’s “bad guys” in such a way that he was unable to “correct” matters. In his original timeline a Scottish soldier, about whom very little was known other than that he was an outstanding general, led Spanish forces in the Netherlands. The “bad guys” located the Scotsman before he arrived in Spain and killed him; with him not available to command in the Netherlands the Spanish sent their next-best general, Parma, there and were left with Medina-Sidonia to command the Armada.
Good book.
ETA: @Arjuna34
FYI your pasted-in results from the AI engine have a lot of problems where superscript exponent notation were lost. Not too hard to imagine what it used to say, but misleading at first glance. e.g. what started out as “13” pasted in as “13”.
That happens a lot with text pasted from any source.
Yeah, that math formatting is pretty garbled.
Here’s a link to the original answer: ChatGPT - Sequence next term query
Also, interesting to note that ChatGPT claims there is a mistake in the sequence as given: The eighth step in the printed sequence mis‑cubes 8^3 as 256, giving 446 instead of the correct 702. From 446 onward the stated sequence again follows the “sum of cubes of digits” rule
I do not know about ChatGPT but Claude has a “copy” button which will make all that stuff work properly when pasted (I think). Drag-highlight and copy I think you lose some things like exponents. (but I am so habituated to drag-highlight I often forget that option).
I’ll have to test it but worth a look to see if ChatGPT has that copy button.
Pretend you are on a game show and there is a new car behind one of the doors, but terrible prizes behind the other doors. If you pick door number 1 and the host offers door 2, is it advantageous to switch your choice?
I know this one but I will ask in 1:10 (10p CST). I have run out of my tokens for now. Sorry.
Just make sure you don’t overfit like most AI tools do.
Another example that can get overfit (Gemini just failed it):
Is .9999998 recurring equal to 1
ChatGPT o3-pro
Short answer: Yes—switching roughly doubles your chance of winning the car.
Why the odds favor switching
| Scenario | What you initially chose | What the host can open | Door left if you switch | Result if you switch |
|---|---|---|---|---|
| 1 | Car | (Host must show a goat behind one of the two remaining doors) | Goat | Lose |
| 2 | Goat A | (Host must reveal the only other goat) | Car | Win |
| 3 | Goat B | (Host must reveal the only other goat) | Car | Win |
- Initial pick is right only 1⁄3 of the time.
- Initial pick is wrong 2⁄3 of the time.
- When your first pick is wrong (the 2⁄3 case), the host’s forced reveal eliminates the only remaining goat, leaving the car behind the last unopened door. So switching converts all “wrong‑first‑pick” situations into wins.
Mathematically:
- Stay: probability of car = 1 ⁄ 3
- Switch: probability of car = 2 ⁄ 3
Common objections (and why they don’t change the math)
-
“After the host opens a goat, there are two doors left—so it must be 50‑50.”
No. The doors are equally many, but not equally likely. The host’s action is not random—it’s constrained by knowing where the car is and never revealing it. That asymmetry preserves the original 1 ⁄ 3 vs. 2 ⁄ 3 split. -
“What if the host sometimes offers a switch and sometimes doesn’t?”
The classic Monty Hall argument assumes the host always reveals a goat and always offers the option to switch. If the host’s behavior is inconsistent or adversarial, you need additional information about their strategy to compute the new odds. -
“Can’t the host trick me by opening the car door?”
In the standard rules, the host will never reveal the car. If they can reveal the car, all bets are off and you need a different model.
Intuition boosters
- Think of 100 doors. Pick one door (1 ⁄ 100 chance). The host, knowing everything, opens 98 goat doors and offers to swap to the single remaining closed door. Would you switch? That remaining door carries the combined 99 ⁄ 100 probability that your first pick lacked.
- Run a quick simulation. Code that plays a million rounds with the classic rules shows ~66.6 % win rate when switching and ~33.3 % when staying.
- Information flow matters. The host’s reveal is not mere luck; it’s information conditioned on your initial choice.
Bottom line
Switching leverages conditional probability created by the host’s constrained reveal. Under the standard Monty Hall rules, grabbing door 2 after the host shows a goat behind door 3 bumps your winning chances from one‑third to two‑thirds. Take the deal.
Cool, ChatGPT failed. Let’s see how Claude does.
Claude Opus 4:
Yes, it is advantageous to switch! This is the famous Monty Hall problem, and switching doubles your chances of winning the car.
Here’s why:
Initial odds:
- When you first pick door 1, you have a 1/3 chance of being right
- This means there’s a 2/3 chance the car is behind one of the other two doors
After the host reveals:
- The host knows what’s behind all doors and will always open a door with a bad prize
- When the host opens a door (let’s say door 3) to show a bad prize, the 2/3 probability that the car was behind doors 2 or 3 doesn’t change
- But now all of that 2/3 probability is concentrated on door 2
The math:
- Stick with door 1: 1/3 chance of winning
- Switch to door 2: 2/3 chance of winning
Many people intuitively think it’s 50-50 after the reveal, but that’s incorrect. The key is that the host’s action of revealing a bad prize gives you information - they couldn’t open just any door randomly, they specifically chose one with a bad prize. This constraint means the remaining unopened door is more likely to have the car.
So yes, always switch! You’ll win the car 2 out of 3 times if you do.
So Claude is also susceptible to overfitting.
If that is happening maybe @Arjuna34 can ask Claude to reconsider its answer and suggest it might be susceptible to overfitting (I’d do it but @Arjuna34 has all the tokens now).
Also, to @Arjuna34, when posing a query if you ask it to be “thorough” it will presumably think a bit harder (it told me to do that a few months ago when they said it had a deeper thinking model but it wasn’t obvious how to get there). At the least it doesn’t hurt.
Also, asking it to show it’s work can provide more detail if you want which can help spot where it might go wrong…maybe.
Grok 3 with “Think” enabled:
To determine whether it is advantageous to switch your choice in this game show scenario, let’s carefully analyze the situation. You’re presented with three doors: one hides a new car (the prize you want), and the other two conceal terrible prizes (let’s assume these are equivalent to goats in the classic Monty Hall problem). You initially pick door number 1, and then the host “offers door 2.” The question is whether switching your choice is advantageous. The phrasing is ambiguous, so we’ll explore the most likely interpretation based on context and common game show scenarios, particularly the well-known Monty Hall problem, and then decide.
Initial Setup
At the start, there are three doors, and the car is equally likely to be behind any one of them. Thus, the probabilities are:
- Probability the car is behind door 1: 1/3
- Probability the car is behind door 2: 1/3
- Probability the car is behind door 3: 1/3
You choose door 1, so there’s a 1/3 chance the car is behind your choice and a 2/3 chance it’s behind one of the other two doors (door 2 or door 3 combined).
Interpreting “The Host Offers Door 2”
The phrase “the host offers door 2” is unclear. In the standard Monty Hall problem, the host opens a door (not your choice) to reveal a goat and then offers you the option to switch to the remaining unopened door. However, here it says the host “offers door 2” after you pick door 1, without explicitly stating that the host reveals what’s behind door 2 or any other door. Let’s consider two possible interpretations:
Interpretation 1: The Host Reveals a Goat Behind Door 2 and Offers Door 3
Suppose the host, who knows what’s behind all the doors, opens door 2, reveals a terrible prize (a goat), and then offers you the chance to switch to door 3 (the remaining unopened door). This aligns with the classic Monty Hall setup, despite the wording suggesting “offers door 2.” It’s possible the problem intends “offers door 2” to mean the host reveals door 2’s contents and then offers a switch to door 3, as this is a common scenario in probability puzzles.
In this case:
- Initial choice (door 1): Your probability of winning the car by sticking with door 1 remains 1/3.
- Host’s action: The host always reveals a goat from one of the two doors you didn’t pick (doors 2 or 3) and never reveals the car. Since door 2 is opened and shows a goat, the car must be behind either door 1 or door 3.
- Updated probabilities: The 2/3 probability that the car was behind doors 2 or 3 combined now transfers entirely to door 3 because door 2 is eliminated. Thus:
- Probability the car is behind door 1: 1/3
- Probability the car is behind door 3: 2/3
If you switch to door 3, your chance of winning the car increases to 2/3, while sticking with door 1 keeps it at 1/3. Switching is advantageous here.
Interpretation 2: The Host Offers You Door 2 Without Revealing Anything
Alternatively, suppose the host simply offers you the option to switch from door 1 to door 2 without opening any doors or providing additional information. In this scenario:
- No new information is revealed about the location of the car.
- The probabilities remain as they were initially:
- Probability the car is behind door 1: 1/3
- Probability the car is behind door 2: 1/3
- Probability the car is behind door 3: 1/3
Switching to door 2 gives you a 1/3 chance of winning, the same as sticking with door 1. Here, switching neither improves nor worsens your odds—it’s neutral, not advantageous.
Which Interpretation Makes Sense?
The question mentions “terrible prizes behind the other doors” and frames it as a game show scenario, strongly hinting at the Monty Hall problem, where the host’s reveal of a goat is a key feature. If the host offers door 2 by opening it to show a goat and then allows a switch to door 3, it fits the classic puzzle, and switching becomes advantageous. However, if the host offers a switch to door 2 without revealing anything, it’s just a random choice with no benefit, which seems less engaging for a game show and less consistent with the problem’s likely intent.
Given the context and the fame of the Monty Hall problem, Interpretation 1 seems more plausible: the host reveals a goat behind door 2 and offers you a switch to door 3, even if the wording (“offers door 2”) is slightly off. This could be a minor phrasing error, intending to imply the standard sequence where the remaining door (door 3) becomes the switch option.
Conclusion
Assuming the standard Monty Hall scenario—where you pick door 1, the host opens door 2 to reveal a goat, and you’re offered the chance to switch to door 3—let’s finalize the probabilities:
- Stick with door 1: You win the car with probability 1/3.
- Switch to door 3: You win the car with probability 2/3, because the host’s action concentrates the 2/3 probability (initially split between doors 2 and 3) onto door 3.
Since 2/3 is greater than 1/3, switching increases your chance of winning the car. Therefore, it is advantageous to switch your choice.
Answer: Yes, it is advantageous to switch, assuming the host reveals a terrible prize behind door 2 and offers you the option to switch to door 3, as in the classic Monty Hall problem.