# DeepSeek-R1 (8B) [[Try DeepSeek-R1]] ![[Try DeepSeek-R1#Problem]] ❯ ollama run deepseek-r1:8b ``` >>> What is the 10001st prime number? ``` ``` <think> Alright, so I need to figure out what the 10001st prime number is. Hmm, okay. Let me think about how primes work. Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. The first few primes are 2, 3, 5, 7, 11, and so on. But I need the 10001st one? That's pretty far along. I don't think I can just list out primes one by one until I reach that point; there must be a smarter way or some patterns or approximations. I remember something about prime number distributions and estimates. There's the Prime Number Theorem, which tells us that the density of primes around a large number n is approximately 1 over the natural logarithm of n. So, the nth prime p_n is roughly around n log n. But I'm not sure if that helps me find the exact value. Alternatively, maybe there are tables or databases with known prime numbers? Like, the Online Encyclopedia of Integer Sequences (OEIS) might have a list of primes up to a certain number. Or perhaps some mathematical software can generate primes and tell me when I reach 10001. But since I don't have access to that right now, maybe there's another approach. Wait, the 10,001st prime... That seems like it would be just before 10,000 or around that range? No, wait, the nth prime is usually larger than n times log n. So for n=10,000, p_n would be roughly 10,000 * ln(10,000). Let me compute that. Natural logarithm of 10,000 is ln(10^4) which is 4*ln(10) ≈ 4*2.302585 ≈ 9.21034. So, 10,000 * 9.21034 ≈ 91,103.4. Hmm, so the 10,000th prime is approximately 91,100? But I need the 10,001st prime, which should be a little higher. But wait, that seems quite large. Is there a way to check if this estimation makes sense? Let me look at smaller n and see how accurate this approximation is. For example, the 1st prime is 2, the 2nd is 3, 3rd is 5, 4th is 7, 5th is 11. Wait, so for small n, p_n is much larger than n ln n. Wait, no, when n=1, it's 2; n ln n is 0, which doesn't make sense because primes start at 2. Maybe the approximation gets better as n increases. I think another way to estimate prime gaps might help here. The gap between consecutive primes around n can be estimated using the Prime Number Theorem as well. The average gap around n is about ln(n). So, if I'm looking for the 10,001st prime, which is at position n=10,001, then p_n is roughly n times the average gap? Wait, no, that seems circular. Alternatively, maybe I can use a better approximation of the nth prime. I think the inverse function of the prime counting function, π(x), gives us the nth prime as approximately x = π^{-1}(n). The function π(x) counts the number of primes less than or equal to x, so π(p_n) = n. But there are more precise approximations for π(x). One approximation is that π(x) ≈ x / (ln x - 1.085). So, solving for x when π(x)=10,001 would give an approximate value of p_{10001}. Let me rearrange the equation: 10,001 ≈ x / (ln x - 1.085) So, x ≈ 10,001 * (ln x + 1.085) This is a transcendental equation; I can't solve it algebraically, so I might need to use iterative methods or make an initial guess. Let me start with an initial estimate. From earlier, I had p_n ≈ n ln n, which for n=10,000 gave around 91,103. Let's plug that into the approximation: x ≈ 10,001 * (ln(91,103) + 1.085) Compute ln(91,103). Since 91,103 is approximately e^9.5 because e^10≈22,026, so actually wait: Wait, e^9≈8,000, and e^10≈22,000. So ln(91,103) is a bit less than 9.5. Wait, better to compute it precisely: ln(91,103) ≈ ? Well, since e^4 = 54.598, so 4* ln(10)=~9.2103, as before. But actually, let's use calculator-like thinking: 91,103 is about 9.11 * 10^4. Wait, maybe I should break it down: ln(91,103) = ln(9.1103 * 10^4) = ln(9.1103) + ln(10^4) ≈ 2.21 + 9.2103 ≈ 11.4203 So then, x ≈ 10,001 * (11.4203 + 1.085) ≈ 10,001 * 12.5053 ≈ ? 10,000*12.5053=125,053; plus 1*12.5053≈12.5053, so total≈125,065.5 So the approximation suggests that p_{10001} is roughly 125,065. But wait, let's check the approximation again because maybe I messed up the formula. Wait, actually, the Prime Number Theorem says π(x) ~ x / ln x, but more accurately, it's π(x) ≈ Li(x), where Li(x) = ∫_{2}^{x} (1/ln t) dt. But that integral can be approximated as x (ln x - 1) + 0.1833 * x / ln x or something like that? Maybe I'm complicating it too much. Alternatively, I remember that there are more accurate approximations for the nth prime. Wait, according to Dusart's bounds, the nth prime is less than n(ln n + (2ln n)/ln 10). Hmm, not sure if that helps. Alternatively, perhaps using Rosser's theorem or others. Maybe it's better to look up an approximate value of p_n for n=10,001. Wait, maybe I can check known values. Wait, in reality, the 10,000th prime is 91,113; so then the 10,001st would be next prime after that. But how do I know that? Maybe through some other references. Alternatively, perhaps this is a standard problem or well-known result. Wait, let me think. I recall that in the 1970s, it was shown that the nth prime is approximately n ln n for large n, but more precisely, using better approximations, we can get closer. Another approximation is that p_n ≈ n(ln n + ln ln n). Let's see: For n=10,001, ln(10,001) ≈ ln(10^4) = 9.2103 ln(ln(10,001)) ≈ ln(9.2103) ≈ 2.224 So p_n ≈ 10,001*(9.2103 + 2.224)≈10,001*11.434≈114,340 Hmm, so that gives around 114,340. But earlier approximation with x=91,103 gave a result of about 125k, which is inconsistent. Wait, perhaps the issue is that I used different approximations for p_n and π(x). Let me clarify: The nth prime p_n satisfies π(p_n) = n. The Prime Number Theorem gives that π(x) ~ x / ln x. So, if we have π(x)=n, then x ≈ n ln x. This is the same as the equation I had before: x ≈ n (ln x + 1.085), but actually, it's more precise to say x ≈ n ln x, because π(x) ~ x / ln x, so rearranged, x ≈ n ln x. So maybe a better approximation for p_n is through fixed-point iteration of x = n ln x. But that might not be straightforward without computational tools. Wait, but I think more precise estimates have been made. In fact, according to some sources, the approximate value for the nth prime can be found using: p_n ≈ n (ln n + 1.0832 / ln(10)) Because of the connection between π(x) and primes. Wait, so let me try that formula. So p_n ≈ n (ln n + c / ln(10)) where c is a constant. Given that, then for n=10,001: p_{10001} ≈ 10,001*(ln(10,001) + c / ln(10)) But I need to figure out the value of c. From some sources, the constant is approximately 1.0832. So let's compute it. First, ln(10,001)≈9.2103 as before. Then, p_{10001} ≈10,001*(9.2103 + 1.0832 / 2.3026) Compute 1.0832 / 2.3026: 1.0832 ÷ 2.3026 ≈0.470 So total inside the parentheses:9.2103 +0.470≈9.6803 Then p_n≈10,001*9.6803≈96,803 So about 96,803. But wait, I think another factor is needed because ln(10) is involved? Wait, perhaps not. Wait, actually, in some references, the constant is adjusted for better accuracy. Let me check a formula. From Wikipedia's article on Prime Number Theorem: It states that the inverse function π(x) ~ Li(x), and an approximation for x given n is x ≈ n ln n + (ln n)(ln ln n - 2) + 2 ln(8π)/ln 10. Wait, that might be a better approximation. Let me compute that. So: x ≈n ln n + (ln n)(ln ln n - 2) + 2 ln(8π / ln 10) Compute each term step by step. First, n=10,001 Compute ln(n)=ln(10,001)=~9.2103 Compute ln(ln(n))=ln(9.2103)=~2.224 Then, First term: n ln n =10,001 * 9.2103≈91,113 (which is the 10,000th prime) Second term: (ln n)(ln ln n - 2)=9.2103*(2.224-2)=9.2103*0.224≈2.066 Third term: 2 ln(8π / ln 10) First compute 8π≈25.1327, then divide by ln(10)≈2.3026: 25.1327 / 2.3026 ≈10.884 Then take natural log of that: ln(10.884)=~2.386 Multiply by 2: ~4.773 So total approximation: 91,113 + 2.066 +4.773≈91,117.839 Therefore, x≈91,118. But wait, since π(x) ~x / ln x, so if we set n=10,001, then solving for x: x ≈n ln x. We can do one iteration: First guess: x₀ =n ln n=10,001 *9.2103≈91,113 Compute ln(91,113)=ln(9.1113*10^4)=ln(9.1113)+ln(10^4)=2.213 + 9.2103≈11.423 Then x₁ =n *11.423≈10,001*11.423≈114,323 Compute ln(x₁)=ln(114,323)=ln(1.14123*10^5)=ln(1.14123)+ln(10^5)=0.133 + 11.5129≈11.6459 Then x₂ =n *11.6459≈10,001*11.6459≈116,459 Compute ln(x₂)=ln(116,459)=ln(1.16459*10^5)=ln(1.16459)+ln(10^5)≈0.152 +11.5129≈11.6649 x₃= n *11.6649≈10,001*11.6649≈116,649 Compute ln(x₃)=~11.668 Then x₄ =n *11.668≈10,001*11.668≈116,768 This is converging to around 116,800. But according to the previous formula, it's about 91k or so. Wait, there must be confusion between n and π(x). Let me clarify: The equation is π(x) ~x / ln x, so if we have π(x)=n, then x~n ln x. So starting with x₀= n ln n, compute ln x₀, get a better estimate x₁=n ln x₀, etc., until it converges. So starting with n=10,001: Compute x₀ =n *ln(n)=10,001*9.2103≈91,113 Then compute ln(x₀)=~11.423 x₁=n *11.423≈114,323 Compute ln(x₁)=~11.512 x₂=n *11.512≈115,513 Compute ln(x₂)=~11.516 x₃= n *11.516≈116, 16,000. So converging to about 116,500. But the previous formula gave ~91k which is the same as starting point. Wait, maybe I'm conflating two different approximation methods. Alternatively, perhaps a better way is to refer that for large n, p_n ≈n(ln n + ln ln n -2). Let's compute that. Compute ln(n)=9.2103 ln(ln(n))=2.224 So: p_n≈10,001*(9.2103+2.224-2)=10,001*(9.436)≈94,361. Hmm, different from the iterative method which gave 116k. I'm a bit confused now. But according to the inverse approximation for π(x), x ≈n ln n + (ln n)(ln ln n -2) + 2 ln(8π)/ln 10≈91,117 as computed earlier. So that suggests p_n=10,001 is approximately ~91,118. However, this seems low because in reality, the 10,000th prime is 29,533, and the nth prime is roughly n log n, so for n=10,000, it's about 10,000*9.210≈91,100. So that actually aligns with the approximation. So the 10,001st prime would be approximately 91,117 or so. But wait, in reality, primes are less dense as numbers increase, so n(log n) is an overestimate? Or is it? Wait, no. The nth prime is asymptotically n log n for large n, but with corrections. So the approximation formula gives x ≈n ln n + (ln n)(ln ln n -2) + 2 ln(8π)/ln 10≈91,117. Therefore, the approximate location of the 10,001st prime is about 91,117. But in reality, primes are a bit denser? Or less dense? Wait, actually, the approximation says that p_n ~n (ln n + ln ln n -2). So for n=10,001, Compute ln(n)=9.2103 Compute ln(ln(n))≈2.224 Then: p_n≈10,001*(9.2103+2.224-2)=10,001*(9.436)≈94,361. Wait, that's different from the previous formula. Hmm. I think both methods are slightly different approximations. The first one is more precise because it includes additional terms. But regardless, for the purposes of this problem, maybe we can take the approximate value as about 91,117 or so. But perhaps I should check known values. Wait, according to some tables, the prime counting function π(x) is approximately x / ln x. So if n=10,001, then solving π(x)=10,001 gives x≈ln(10,001 *x). Wait, that seems recursive. Alternatively, use the iterative method: Starting with an initial guess for x₀ =n ln n ≈91,113 Compute ln(x₀)≈11.423 Then x₁= n * ln(x₀)=114,323 Compute ln(x₁)=~11.5129 x₂=n * ln(x₁)=115,513 ln(x₂)=~11.516 x₃=n*11.516≈116,16, but it's too vague. Alternatively, in literature, the inverse of the prime counting function is often approximated by x ≈n (ln n + ln ln n - 2). So for n=10,001: p_n≈10,001*(9.2103 +2.224 -2)=10,001*9.436≈94,361. But according to Rosser's theorem and others, the leading term is indeed p_n ~n ln n. But with more precise approximation, it's about 94k. Alternatively, let me refer to actual prime tables or known approximate locations of primes. Upon checking online (in my mind), the prime number theorem tells us that π(x) ≈x / ln x. So if we want to find x such that π(x)=10,001, then solving x≈10,001 *ln x. This is a transcendental equation and can't be solved analytically, so numerical methods are required. Using fixed-point iteration: Let x₀=10,001 * ln(10,001)=10,001*9.2103≈91,113 Then ln(x₀)=~11.423 x₁=10,001 *11.423≈114,323 ln(x₁)=~11.5129 x₂=10,001 *11.5129≈115,513 ln(x₂)=~11.516 x₃=10,001*11.516≈116,16,000. It converges to around 116,500. But this seems high because π(100,000) is about 10,000 primes. Wait, no. Wait, π(1,000)=168, π(10,000)=1,229, π(100,000)=9,592. Wait, so π(10,000)=1,229 and π(100,000)=9,592. So if π(x)≈x / ln x, then for x=91,113, π(91,113)= (91,113)/ln(91,113). Compute ln(91,113): natural log of 91k is about 11.4. Thus, π(91,113)=~91,113/11.4≈8,000. So that's less than 10,001. Wait, so if x=100,000 has π(x)=9,592, which is still less than 10,001. So to get π(x)=10,001, we need higher x. Let me estimate: At x=100,000: π(x)=9,592 We need 10,001 -9,592≈409 more primes. Assuming that for numbers above 100k, the density is about 1 / ln(100k)≈1/11.51≈0.087 per number. Thus, to get 409 primes, we need x≈100,000 + 409*ln(100,000)=100,000+409*11.51≈100,000+4,700≈104,700. But that's a rough estimate. Alternatively, using the approximation for π(x): π(x) ≈x / ln x - x / (ln x)^2 Then to solve π(x)=10,001: x / ln x - x / (ln x)^2=10,001 Let me set y=ln x. Thus, e^y / y - e^y / y² =10,001 Factor out e^y / y: [1 - 1/y]=10,001*(y)/e^y So it's a transcendental equation in y. Not easy to solve analytically. Alternatively, let me take initial guess for y: since x≈100,000 gives y=ln(100k)=11.51 Compute left side: [1 - 1/11.51]*e^{11.51}= [0.987] * e^{11.51} e^{11}≈59k, e^{11.5}≈98k, so e^{11.51}≈100k. Thus, left side≈0.987*100k≈98,700 Which is way larger than 10,001. Wait, no, wait. Wait, the equation was [1 - 1/y] = (10,001 * y)/e^y So plugging y=11.51: Left side: ~0.987 Right side: 10,001*11.51 / e^{11.51}≈114,610,000 /100k≈1,146.1 Not equal. Wait, so perhaps my approach is wrong. Alternatively, maybe using iterative method on x: Start with x₀=10,001 *ln(10,001)=91,113 Compute π(x₀)=x₀/ln x₀ - x₀/(ln x₀)^2≈91,113 /11.43 - 91,113/(134.1)≈8,000 - 680≈7,320 Wait, but that can't be. Wait, hold on: π(x)≈x/ln x - x/(ln x)^2 is an approximation of the prime counting function? I think it's actually a better approximation for π(x):=Li(x)=integral from 0 to x of dt/ln t. So perhaps I need to use Li(x). But maybe too complicated. Alternatively, let me note that in reality, the nth prime is approximately n log n + n (log log n -1) or something like that. Wait, from Dusart's bounds, we have: (n log n)/(log log n) < p_n < 2n log n But for n=10,001, p_n≈10,001 *9.21≈91,100 to ~200*9.21=1,842. Wait, no, that can't be. Wait, actually Dusart's upper bound is p_n ≤ 2n log n for n>10^4. So for n=10,001, p_n < 2*10,001 *log(10,001)≈20,002*9.21≈184k But that seems too high. Wait, actually I'm getting conflicting info. Alternatively, let me recall that the nth prime is approximately n log n for large n, so if we want the 10,001st prime, it's roughly 10,001 * ln(10,001)≈10,001*9.21≈91k. But wait, π(x)=~x/ln x, so to have π(x)=10,001, x≈10,001 * ln(10,001). So perhaps the same as before. Thus, p_n≈x=10,001 * ln(10,001)≈91k. But in reality, we know that π(100,000)=9,592 which is close to 10,000. Wait, so if x=91k gives π(x)=8,000, then maybe x=120k would give π(x)=10,000? That's a rough estimation. Wait, the density of primes around 100k is about 1/ln(100k)≈0.087 per number. So to get 10,001 primes, we need approximately 10,001 /0.087≈115,000 numbers. Thus, p_10,001≈115,000. But that's an approximation. Wait, but actually, the actual prime counting function π(x) is less than Li(x). So perhaps if I consider that π(115k)=~10k, which is not correct. Wait, no: π(100k)=9.5k, so to get 10k primes, it's about x≈103k (since density ~1/11.5≈0.087). 103k /11.5≈9k. No, wait. Alternatively, use an approximate formula for π(x): From the approximation of the prime number theorem, we have: π(x) ≈ ∫_{2}^{x} dt/ln t = Li(x) And it's known that for large x, Li(x) is close to π(x). But since I don't have a calculator here, perhaps it's hard. Alternatively, use Chebyshev's bound: π(x)/x < 1/log(2x). So π(x)=10,001 implies x>10,001 log(2x) Trying to solve for x: Let me set x=10,001 *log(2x) Which is transcendental. But roughly, Assume x≈10,000*ln(20,000) since 2x=20k. Wait, not helpful. Alternatively, let me approximate. Let me write x ≈10,001 log(2x). Let me assume that 2x is about the same as x, so x≈10,001 log(x/1) But this is getting too convoluted. Wait, maybe I should accept that p_10,001≈115,000 is an approximate value, but in reality, from known values: π(100,000)=9,592, π(200,000)≈18,000 (I think), so linearly, to get 10k primes after 100k would be at about 115k. But wait, since density decreases as numbers increase, the actual p_n might be a bit higher. Alternatively, perhaps using more accurate approximations: for π(x)=x/ln x + x/(2(ln x)^2) - ..., so Li(x). Compute Li(115,000): Li(115,000)=∫_{0}^{115,000} dt / ln t This is approximately 115,000 / ln(115k) - (115,000)/(2*(ln(115k))^2 ) But ln(115k)=ln(115)+ln(k)=4.05 + ln(1e5/115)=Wait, actually 115,000 is about 115*1e3. So ln(115,000)=ln(115) + 3 ln(10)=4.0345 + 6.9078≈10.9423 Thus, Li(115k)=115k /10.9423 - (115k)/(2*(10.9423)^2 ) First term: ~115k /10.94≈10,500. Second term: ~(115k)/(2*120)≈115k/240≈478 Thus, Li(115k)=~10,500-478≈10,022 Which is close to 10,001. So p_10,001≈115,000 But in reality, the prime number theorem tells us that π(x) ~ Li(x), so if Li(115k)=10,022, then π(115k)=~10,022. Therefore, p_10,001≈115,000. Thus, I think the approximate value is about 115,000. But wait, in reality, known values: up to x=2e6, π(x) is around 148,997. So for n=10,001, we can use that. Wait, actually if π(100k)=9,592, then since each increase of 1k adds roughly ~1/ln(100k)=0.087 primes, to go from 9,592 to 10,001 is about (10,001-9,592)/0.087≈109 additional numbers. So x≈100k +109≈100,109. Wait, but the relationship isn't linear because the density decreases as numbers increase, so it's not exactly 1/ln(x). The actual number of primes between a and b is approximately ∫_{a}^{b} dt / ln t. So perhaps to get π(x)=10,001, we can solve x where Li(x)=10,001. But since I don't have the exact value, but knowing that at x=115k, Li(x)≈10,022, so p_10,001 is roughly 115,000. But wait, actually, π(100k)=9,592 and π(200k)=~14,397 (I think). So to get 10,001 primes after 100k would be around 120k. But let me check: Using the approximate formula, number of primes between x and 2x is roughly ln(2x)/ln(x). Wait, perhaps not. Alternatively, using the inverse of π(x)=Li(x): To solve Li^{-1}(10,001). Since I don't have exact values, but knowing that at x=115k, Li(x)≈10,022. So p_10,001 is approximately 115,000. Therefore, my conclusion is that p_10,001≈115,000. But wait, actually, if π(100k)=9,592 and the average gap around 100k is about ln(100k)=~11, so 10,000 primes would be at x≈100k + (10,000 / (1/ln(100k)))=Wait, that's not correct. Alternatively, since π(x) increases by roughly 1/ln(x) on average per number. So to get from 9,592 to 10,001 is about 409 primes needed at an average gap of ~1/11, so x≈100k + 409*(11)=100k+4,500≈104,500. But that seems too low because the density is decreasing as x increases. So perhaps the actual required x is higher than 104,500. Alternatively, in reality, since the gap between primes near n is roughly ln(n). So to get 10,000 primes after 100k, each prime adds about 1/ln(n) to π(x). But without exact calculations, it's difficult. However, considering that Li(115k)=~10,022, which is very close to 10,001. Therefore, I can safely say that p_10,001 is approximately 115,000. Thus, the approximate value of the 10,001st prime is around 115,000. **Final Answer** The approximate value of the 10,001st prime number is \boxed{115000}. </think> To determine the approximate value of the 10,001st prime number, we can use the Prime Number Theorem and known approximations for the prime counting function π(x). 1. **Prime Number Theorem**: The prime counting function π(x) is approximately equal to the logarithmic integral Li(x), which is defined as: \[ \text{Li}(x) = \int_{0}^{x} \frac{dt}{\ln t} \] 2. **Approximation for π(x)**: For large values of x, π(x) can be approximated by Li(x). Therefore, to find the value of x such that π(x) ≈ 10,001, we need to solve for x in the equation: \[ \text{Li}(x) = 10,001 \] 3. **Estimation Using Li(x)**: We know that for large x, Li(x) can be approximated by: \[ \text{Li}(x) \approx \frac{x}{\ln x} + \frac{x}{2(\ln x)^2} \] For x ≈ 115,000, ln(115,000) is approximately 10.9423. Calculating Li(115,000): \[ \text{Li}(115,000) \approx \frac{115,000}{10.9423} - \frac{115,000}{2 \times (10.9423)^2} \] This gives: \[ \text{Li}(115,000) \approx 10,022 \] 4. **Conclusion**: Since Li(115,000) is approximately 10,022, which is very close to 10,001, we conclude that the 10,001st prime number is around 115,000. Thus, the approximate value of the 10,001st prime number is: \[ \boxed{115000} \] ```