{"id":"gen-1785586785-4rM5T60qd58oBQA8VDq6","choices":[{"finish_reason":"stop","index":0,"logprobs":null,"message":{"content":"## Short answer\n\n**Binary splitting of the Taylor series** $e = \\sum_{k=0}^{\\infty} \\frac{1}{k!}$, evaluated in exact integer arithmetic on top of **FFT-based bignum multiplication** (e.g., GMP). This is quasi-linear in the number of digits and is exactly what every world-record computation of *e* (y-cruncher, currently at tens of trillions of digits) uses.\n\n## Why not the naive approaches\n\n| Method | Complexity for D digits | Practical limit |\n|---|---|---|\n| Spigot algorithm (Rabinowitz–Wagon) | O(D²) | ~10⁵ digits |\n| Incrementally summing 1/k! in big floats | ~O(D²) | ~10⁵ digits |\n| **Binary splitting + FFT multiply** | O(M(D)·log D) ≈ Õ(D) | trillions of digits |\n\nThe spigot algorithm is charming (only machine-word arithmetic, tiny memory), but quadratic — fine for 10⁴ digits, hopeless for 10⁸.\n\n## The algorithm\n\nThe series converges brutally fast: you need only N terms with N! > 10^D, i.e. N ≈ D·ln 10 / ln D (about **205,000 terms for 1,000,000 digits**). The trick is *how* you sum them.\n\nInstead of adding terms one by one, recursively split the sum in half and combine rationals:\n\n$$\\sum_{k=a}^{b-1} \\frac{1}{k!} = \\frac{P_l}{Q_l} + \\frac{P_r}{Q_r} = \\frac{P_l Q_r + P_r Q_l}{Q_l Q_r}$$\n\nThis makes all multiplications happen on balanced-size numbers, so the total cost is dominated by a few FFT multiplications at the top of the recursion tree instead of thousands of unbalanced ones.\n\n```python\nfrom math import log, lgamma\n\ndef bs(a, b):\n \"\"\"Return (P, Q) with P/Q = sum_{k=a}^{b-1} 1/k!, a >= 1.\"\"\"\n if b - a == 1:\n return 1, a\n m = (a + b) // 2\n Pl, Ql = bs(a, m)\n Pr, Qr = bs(m, b)\n return Pl * Qr + Pr * Ql, Ql * Qr\n\ndef e_digits(D):\n # smallest N with N! > 10^(D+10) (guard digits against tail error)\n N = max(10, int(D * log(10) / log(max(D, 3))))\n while lgamma(N + 1) < (D + 10) * log(10):\n N += 1\n P, Q = bs(1, N)\n digits = ((P + Q) * 10**D) // Q # e = 1 + P/Q, truncated\n s = str(digits)\n return s[0] + \".\" + s[1:]\n```\n\nNotes on making it actually fast:\n\n- **Use GMP integers** (`gmpy2.mpz` drop-in) — CPython only has Karatsuba; GMP adds Toom and Schönhage–Strassen FFT multiplication. That's the difference between minutes and seconds at a million digits.\n- The final `// Q` is a single big division — use Newton division (GMP does this internally).\n- Decimal string conversion is also sub-quadratic in GMP (`get_str`); naive digit-by-digit printing is O(D²) and becomes the bottleneck otherwise.\n- The recursion tree parallelizes trivially across cores (y-cruncher exploits this heavily).\n\n## Alternatives with the same asymptotics\n\nComputing exp(1) via **AGM-based ln + Newton inversion** (Brent's method) is also O(M(D)·log D), but for *e* specifically the Taylor series + binary splitting has smaller constants, so nothing beats it in practice.\n\n## If you want maximum digits with zero effort\n\n- **mpmath**: `mp.dps = 10**6; mp.e` (uses binary splitting internally)\n- **y-cruncher**: the record-holding program; multithreaded, disk-swapping, and it **verifies** results by computing 1/e via the alternating series $\\sum (-1)^k/k!$ and checking consistency — worth doing yourself for serious runs.\n\n**Bottom line:** binary splitting + FFT multiplication gives you ~O(D log² D) total work, and it's the only method that's ever been used at the trillion-digit scale.","refusal":null,"role":"assistant","annotations":null,"audio":null,"function_call":null,"tool_calls":null,"reasoning":"The user is asking about computing the digits of Euler's number (e ≈ 2.71828...) efficiently — they want to approximate the largest number of decimal digits as fast as possible.\n\nLet me think about what's known about computing e to high precision:\n\n**Classic series:**\ne = Σ 1/n! for n = 0 to ∞\n\nThis converges extremely fast — each term adds roughly log10(n!) ≈ n·log10(n/e)·log10(e) more digits. To get D decimal digits, you need roughly n terms where n! > 10^D, so n ≈ D·ln(10)/ln(n)... roughly n ≈ D/log10(D) terms. Actually solving n! > 10^D: using Stirling, n ln(n/e) > D ln 10, so n ≈ D ln(10) / ln(D) roughly. For D = 10^6 digits, n is about 10^6·2.3/13.8... let me not get bogged down.\n\n**Binary splitting:**\nThe key optimization is binary splitting of the series. Instead of computing each factorial term separately, you recursively split the sum into halves and combine using exact rational arithmetic with shared denominators. This reduces the complexity from O(n²) bit operations (naive) to roughly O(M(n) log n) where M(n) is the multiplication complexity. With FFT-based multiplication (Schönhage–Strassen, or modern n log n integer multiplication — Harvey & van der Hoeven 2019 proved integer multiplication is O(n log n)), you get close to O(D log² D) or so total time.\n\nBinary splitting for e:\ne = Σ 1/k!\nDefine P(a,b) and Q(a,b) recursively:\n- If b - a = 1: P(a, a+1) = 1, Q(a, a+1) = b (for a ≥ 1), or for a=0, Q = 1.\nActually the standard: T = P/Q where\n- P(a,a+1) = 1, Q(a,a+1) = a+1... let me recall.\n\nSum from k=0 to N of 1/k!. Write it as P(0,N)/Q(0,N) where:\n- base case: P(k, k+1) = 1, Q(k, k+1) = k! ... hmm, no.\n\nStandard binary splitting for hypergeometric series: term t_k with t_{k+1}/t_k = p(k)/q(k). For e, t_k = 1/k!, ratio = 1/(k+1).\n\nDefine for the sum S = Σ_{k=a}^{b-1} 1/k!:\nSplit at m: S = S_left + S_right = P_l/Q_l + P_r/Q_r = (P_l·Q_r + P_r·Q_l)/(Q_l·Q_r).\nBase case: S for single term k: P = 1, Q = k!.\n\nBut k! in the base case requires computing factorials; instead, better formulation: P(a,b), Q(a,b) with\n- base: P(k,k+1) = 1, Q(k,k+1) = k (with Q(0,1) = 1).\nThen the combined sum P/Q approximates Σ 1/k! appropriately... Let me verify: Σ_{k=1}^{N} 1/k! = P(1,N+1)/Q(1,N+1) with base P(k,k+1)=1, Q(k,k+1)=k. Check N=2: 1/1! + 1/2! = 1 + 1/2 = 3/2. Split: left = P(1,2)/Q(1,2) = 1/1, right = P(2,3)/Q(2,3) = 1/2. Combine: (1·2 + 1·1)/(1·2) = 3/2. ✓.\n\nSo the recursion: P(a,b) = P(a,m)·Q(m,b) + P(m,b)·Q(a,m); Q(a,b) = Q(a,m)·Q(m,b).\n\nThe final result: e ≈ 1 + P(1,N+1)/Q(1,N+1) (adding the k=0 term = 1).\n\nThe total bit complexity: the numbers at the top level have size about N log N bits (Q ≈ N!). With multiplication cost M(b) ~ b log b (modern) the total cost is O(M(D) log D) where D is digits — dominated by the top-level multiplications times log levels... Actually the standard analysis: total cost O(M(n log n) log n) where n is number of terms. In practice, this is dramatically faster than naive summation.\n\n**Even faster: use AGM-like or Newton iterations?**\nFor e specifically, the series method with binary splitting is the standard. Alternative: compute exp(1) via the exponential function using arithmetic-geometric mean (AGM) methods: ln can be computed via AGM in O(M(D) log D), then invert via Newton's method to get exp. But that's for general exp; for e = exp(1), the direct Taylor series with binary splitting is simpler and has the same asymptotic complexity, with smaller constants.\n\nActually, is there anything asymptotically better? Brent's method using AGM for elementary functions gives O(M(D) log D). Binary splitting of the Taylor series also gives O(M(D) log D)... hmm, actually let me think: binary splitting for a D-digit result needs N ≈ D/log D terms... the total work: at each level of the recursion tree, total sizes multiply... The overall complexity is O(M(N log N) log N)... with N log N ≈ D (bits needed ~ size of N! ≈ N log N bits ≈ D ln 10 · D... hmm, D digits = D·log2(10) bits ≈ 3.32 D bits; N! has ≈ N log2(N/e) bits; setting N log N ≈ 3.32 D gives N ≈ 3.32 D / log2 D).\n\nCost = O(M(N log N) · log N)? No — the standard binary splitting analysis: the recursion tree has log N levels; at each level the total bit size of all operands is O(N log N) total? Actually at level k, there are 2^k nodes each with operands of size ~ N log N / 2^k bits. Multiplication cost per node: M(size). Summing: 2^k · M(N log N / 2^k) ≤ M(N log N) (superlinearity of M). So each level costs ≤ M(N log N), total O(M(N log N) log N) = O(M(D) log D) since N log N ≈ D bits. With M(D) = D log D ( Harvey–van der Hoeven), total O(D log² D). \n\nSo: **binary splitting + FFT multiplication = O(D log² D)** roughly. That's the best known for e I believe. Record computations of e (e.g., to tens of trillions of digits) use y-cruncher, which uses binary splitting of the Taylor series.\n\nPractical advice:\n1. Use the Taylor series Σ 1/k! with **binary splitting**.\n2. Use a bignum library with fast multiplication (GMP, or MPIR; FLINT has direct binary splitting functions).\n3. y-cruncher holds world records and computes e.\n\nAlso there's a classic trick: e = 1/Σ (-1)^k/k!? No, that's 1/e = e^{-1} = Σ (-1)^k/k!. Alternating series allows error bounding easily but not faster.\n\nAnother historical trick: compute e via continued fraction e = [2; 1, 2, 1, 1, 4, 1, 1, 6, ...] — converges but only linearly per term; convergents give about... the CF of e gives roughly... each convergent adds a few digits; not competitive.\n\nMachin-like formulas? For e there's no Machin-like formula; but you can compute e = exp(1) where exp is computed via... no. Actually there are \"Machin-like\" approaches: e = exp(1) computed as exp evaluated via the series at points like 1/2^k with repeated squaring: exp(1) = (exp(1/2^k))^{2^k} — computing exp(1/2^k) via series is fast since terms shrink, then square k times (each squaring is a full-precision multiplication: k ≈ log2(D/...)). Choose k so series needs few terms. exp(2^{-k}): need ~D log 10 / (k ln 2 + ... ) terms. Balance: cost ≈ M(D)·k + (terms)·M(D)... This is basically equivalent; binary splitting is cleaner.\n\nAlso mention: the naive way (summing 1/k! in floating point with increasing precision, or even exact big rationals incrementally) is O(n²)-ish and works fine for thousands of digits but not millions.\n\nFor a practical implementation:\n- In Python: use `mpmath` — `mp.mp.dps = N; print(mp.e)`? mpmath computes e via... it has `mp.e` computed via exp(1) using binary splitting I think. Or use `gmpy2`.\n- Simple binary splitting code in Python (integers) is maybe 15 lines and computes a million digits in seconds (Python's multiplication is Karatsuba+FFT? CPython uses Karatsuba and, since 3.x, also has... CPython uses Karatsuba for large ints; for very large, CPython doesn't use FFT (it does not implement Schönhage–Strassen). GMP does. So use gmpy2 for millions of digits.)\n\nLet me sketch the algorithm:\n\n```\ndef bs(a, b):\n # returns P, Q for sum_{k=a}^{b-1} 1/k!\n if b - a == 1:\n return (1, a) if a > 0 else (1, 1)\n m = (a + b) // 2\n Pl, Ql = bs(a, m)\n Pr, Qr = bs(m, b)\n return Pl * Qr + Pr * Ql, Ql * Qr\n```\n\nWait check base case: for a=0: term 1/0! = 1; Q should be 1. With my formulation sum from k=1: handle a=0 separately: add 1 at the end. If a≥1: P=1, Q=a.\n\nThen e ≈ 1 + P(1, N)/Q(1, N)? Sum_{k=1}^{N-1} 1/k! = P(1,N)/Q(1,N). Add 2 (for k=0 term 1/0! = 1... wait 0! = 1 so term is 1, and k=1 term is 1). Hmm: e = Σ_{k≥0} 1/k! = 1 + 1 + 1/2 + 1/6 + ... So e = 1 + Σ_{k≥1} 1/k! where the k=1 term is 1. So e ≈ 1 + P/Q, or incorporate directly: e = (Q + P)/Q.\n\nNumber of terms: need N such that 1/N! < 10^{-D}, i.e., N ln(N/e) ≥ D ln 10. Solve: N ≈ D ln 10 / ln(D ln 10 / ...) — iterate once or twice. Or use inverse Stirling. For D = 10^6, N ≈ 205,000-ish? Let me estimate: D ln 10 ≈ 2.3026×10^6. N ln(N/e) = 2.3026e6. Try N = 205,000: ln(205000/e) = ln(75445) ≈ 11.23. 205000 × 11.23 ≈ 2.30e6. ✓. So ~205k terms for a million digits. Each leaf multiplications small; the top-level numbers are ~3.3 Mbits. Feasible.\n\nFinal division: one big division of P by Q to D digits — cost M(D) with Newton division.\n\nDecimal conversion: converting a binary bignum to decimal string costs O(M(D) log D) via divide-and-conquer (scaled remainder). GMP's `mpz_get_str` uses sub-quadratic conversion for huge numbers. For millions of digits this matters; naive digit-by-digit is O(D²).\n\nError control: tail Σ_{k≥N} 1/k! < (1/N!)·(1/(1 - 1/(N+1))) ≈ e·/... roughly < 2/N!. Use guard digits (compute with extra bits) or exact bounds: since we compute exact rational partial sum, error is bounded by tail; choose N with margin and also work with extra guard bits in the final division.\n\nVerification: compute via two different N, or check against known digits; record computations use a second formula (e.g., compute 1/e via alternating series) and compare — y-cruncher does that.\n\nFurther optimizations:\n- **Parallelism**: binary splitting tree can be evaluated in parallel; y-cruncher is multithreaded.\n- **FFT multiplication** (GMP uses Schönhage–Strassen / Toom variants). \n- **Working in fixed-point** instead of rationals: instead of exact rational arithmetic, multiply partial sums as fixed-point integers scaled by 10^D — the standard \"sum 1/n! by dividing a big integer successively\" trick: compute S = Σ 10^D/k!... Actually a neat classic trick: compute integer T = floor(10^D · Σ_{k=0}^{N} 1/k!) via Horner-like scheme: start from the tail: T_N = 1; T_{k-1} = 1 + T_k/k... Hmm, there's the classic \"spigot\" algorithm for e!\n\n**Spigot algorithm** for e: The famous one from Rabinowitz & Wagon — e's digits can be extracted one at a time using an array, based on the mixed-radix representation 1/k! in bases (1, 2, 3, ...). It computes D digits in O(D²) time with O(D) space, using only small integers (machine words!). That's the simplest to implement, memory-light, but quadratic — good for up to ~10^4–10^5 digits maybe.\n\nActually the spigot for e: e = 2 + 1/2!(1 + 1/3(1 + 1/4(1 + ...)))... The algorithm: array A[0..n] initialized to 1, repeatedly: carry = 0; for i from n down: x = A[i]*10 + carry; A[i] = x mod (i+1)... etc. Each iteration yields one digit. O(n) per digit → O(D²) total. Simple, but for \"biggest quantity fast\" not the winner.\n\nSo the answer structure:\n1. The math: Taylor series; number of terms needed.\n2. The key algorithmic idea: binary splitting (reduces complexity dramatically).\n3. Underlying multiplication: FFT-based (GMP/FLINT), complexity ~O(D log²D) overall.\n4. Practical options: y-cruncher (records), mpmath/gmpy2 snippet, ~15-line Python binary splitting code.\n5. Mention spigot as simple but O(n²), AGM/Newton for exp as alternative with same asymptotics.\n6. Verification trick.\n\nThe user asks \"most optimized and fast way to approximate the biggest quantity of decimals\" — so they want state of the art. Answer: binary splitting of the hypergeometric series with FFT multiplication, as done by y-cruncher. Give complexity, term count, code sketch, and practical tips.\n\nLet me write concise Python code:\n\n```python\nfrom gmpy2 import mpz # or plain int for < ~100k digits\n\ndef bs(a, b):\n \"\"\"Return (P, Q) such that P/Q = sum_{k=a}^{b-1} 1/k! (a >= 1).\"\"\"\n if b - a == 1:\n return mpz(1), mpz(a)\n m = (a + b) >> 1\n p1, q1 = bs(a, m)\n p2, q2 = bs(m, b)\n return p1 * q2 + p2 * q1, q1 * q2\n\ndef e_digits(d):\n # need N with N! > 10^d → sum tail < 10^-d\n import math\n N = max(4, int(d * math.log(10) / math.log(d))) # refine\n while math.lgamma(N + 1) < d * math.log(10): N += 1\n P, Q = bs(1, N)\n scale = 10**d\n return (Q * scale // Q) ... \n```\n\nHmm: e = 1 + P/Q. digits = floor((P+Q) * 10^d / Q). Let me write:\n\n```python\ndef e_decimal(d):\n P, Q = bs(1, N)\n num = (P + Q) * 10**d\n return num // Q # integer with d+1 digits: \"2\" followed by d decimals\n```\n\nThen format string: str(num) → \"2.\" + rest.\n\nChoosing N: solve N! ≥ 10^{d+guard}. Use Stirling via math.lgamma: while lgamma(N+1) < (d+10)*ln10: N += 1, starting from a Stirling estimate. Fine.\n\nPerformance: with plain Python ints, ~100k digits in a few seconds? CPython multiplication is Karatsuba (O(n^1.585)); a million-digit run may take minutes. With gmpy2, million digits in ~a second or a few seconds. y-cruncher: trillions of digits in hours/days on good hardware.\n\nRecord: e has been computed to 35 trillion digits (2023-2024?) — I recall y-cruncher records: e to 31.4 trillion digits (2022, by Jordan Ranous? no that was pi...). Let me recall: π reached 100 trillion (2022, Emma Haruka Iwao) and 105T (2024). For e: record around 35 trillion digits (2023?) — I think \"e to 35,000,000,000,000 digits\" computed... I'm not 100% sure; I should hedge. I recall e's record was 31.4 trillion around 2023? There was a 2024 record of 202.11 billion? no. I'll hedge: \"tens of trillions of digits (y-cruncher)\".\n\nAlternative asymptotically equivalent: compute ln via AGM then Newton invert, or exp via Brent's O(M(n) log n) algorithm — mention briefly that for exp(1) nothing beats the Taylor series + binary splitting in practice.\n\nAlso mention memory: binary splitting needs O(D log D)... memory ~ size of the top-level numbers times depth? Recursion holds O(log N) numbers of size O(D) → O(D log N) bits? Actually you hold P,Q at each recursion level: sizes decrease geometrically down the tree, total O(D) bits per level path... total memory O(D log N) bits worst case if you keep both branches? In the recursive implementation, at any moment you hold one (P,Q) per level of depth → sizes sum to O(D) bits total along a path? Top level D bits, next D/2... geometric → O(D). Fine: memory O(D) bits ~ few hundred MB for a trillion digits... no wait D digits = 3.3 D bits; a trillion digits → 415 GB. That's why records need big machines and disk swapping.\n\nLet me also give the formula for choosing number of terms: N ≈ smallest n with n! > 10^{D} (Stirling: n ≈ D·ln10 / ln(D ln 10 / e)... just say use lgamma loop).\n\nComplexity summary:\n- Naive incremental summation (bignum floats, add 1/k! terms with running division): O(D²) word ops-ish → fine ≤ 10^5 digits.\n- Spigot: O(D²) but tiny memory and trivial code.\n- Binary splitting + FFT multiplication: Õ(D) (quasi-linear) — the winner; used by all record computations.\n\nOne more relevant point: \"biggest quantity of decimals\" fast — also worth noting that beyond algorithm, hardware (RAM, fast FFT, multithreading) dominates; y-cruncher is the practical answer for records.\n\nI should present:\n1. Short answer: binary splitting of Σ 1/k! with FFT-based bignum multiplication — quasi-linear in D; this is what record programs (y-cruncher) do.\n2. Why naive approaches are slower (complexity table).\n3. The algorithm in detail + minimal Python implementation.\n4. Term count & precision/error control.\n5. Practical tools (y-cruncher, mpmath, FLINT, gmpy2) and verification strategy.\n6. Optional: AGM/Newton alternative note.\n\nFormat: This is a technical question deserving a reasonably thorough answer with code. Use headers, a short complexity list, one code block, keep prose tight. Medium-long length. Avoid excessive bullet fragmentation; use a table maybe for complexity comparison. Code block essential. I'll keep it focused.","reasoning_details":[{"type":"reasoning.text","text":"The user is asking about computing the digits of Euler's number (e ≈ 2.71828...) efficiently — they want to approximate the largest number of decimal digits as fast as possible.\n\nLet me think about what's known about computing e to high precision:\n\n**Classic series:**\ne = Σ 1/n! for n = 0 to ∞\n\nThis converges extremely fast — each term adds roughly log10(n!) ≈ n·log10(n/e)·log10(e) more digits. To get D decimal digits, you need roughly n terms where n! > 10^D, so n ≈ D·ln(10)/ln(n)... roughly n ≈ D/log10(D) terms. Actually solving n! > 10^D: using Stirling, n ln(n/e) > D ln 10, so n ≈ D ln(10) / ln(D) roughly. For D = 10^6 digits, n is about 10^6·2.3/13.8... let me not get bogged down.\n\n**Binary splitting:**\nThe key optimization is binary splitting of the series. Instead of computing each factorial term separately, you recursively split the sum into halves and combine using exact rational arithmetic with shared denominators. This reduces the complexity from O(n²) bit operations (naive) to roughly O(M(n) log n) where M(n) is the multiplication complexity. With FFT-based multiplication (Schönhage–Strassen, or modern n log n integer multiplication — Harvey & van der Hoeven 2019 proved integer multiplication is O(n log n)), you get close to O(D log² D) or so total time.\n\nBinary splitting for e:\ne = Σ 1/k!\nDefine P(a,b) and Q(a,b) recursively:\n- If b - a = 1: P(a, a+1) = 1, Q(a, a+1) = b (for a ≥ 1), or for a=0, Q = 1.\nActually the standard: T = P/Q where\n- P(a,a+1) = 1, Q(a,a+1) = a+1... let me recall.\n\nSum from k=0 to N of 1/k!. Write it as P(0,N)/Q(0,N) where:\n- base case: P(k, k+1) = 1, Q(k, k+1) = k! ... hmm, no.\n\nStandard binary splitting for hypergeometric series: term t_k with t_{k+1}/t_k = p(k)/q(k). For e, t_k = 1/k!, ratio = 1/(k+1).\n\nDefine for the sum S = Σ_{k=a}^{b-1} 1/k!:\nSplit at m: S = S_left + S_right = P_l/Q_l + P_r/Q_r = (P_l·Q_r + P_r·Q_l)/(Q_l·Q_r).\nBase case: S for single term k: P = 1, Q = k!.\n\nBut k! in the base case requires computing factorials; instead, better formulation: P(a,b), Q(a,b) with\n- base: P(k,k+1) = 1, Q(k,k+1) = k (with Q(0,1) = 1).\nThen the combined sum P/Q approximates Σ 1/k! appropriately... Let me verify: Σ_{k=1}^{N} 1/k! = P(1,N+1)/Q(1,N+1) with base P(k,k+1)=1, Q(k,k+1)=k. Check N=2: 1/1! + 1/2! = 1 + 1/2 = 3/2. Split: left = P(1,2)/Q(1,2) = 1/1, right = P(2,3)/Q(2,3) = 1/2. Combine: (1·2 + 1·1)/(1·2) = 3/2. ✓.\n\nSo the recursion: P(a,b) = P(a,m)·Q(m,b) + P(m,b)·Q(a,m); Q(a,b) = Q(a,m)·Q(m,b).\n\nThe final result: e ≈ 1 + P(1,N+1)/Q(1,N+1) (adding the k=0 term = 1).\n\nThe total bit complexity: the numbers at the top level have size about N log N bits (Q ≈ N!). With multiplication cost M(b) ~ b log b (modern) the total cost is O(M(D) log D) where D is digits — dominated by the top-level multiplications times log levels... Actually the standard analysis: total cost O(M(n log n) log n) where n is number of terms. In practice, this is dramatically faster than naive summation.\n\n**Even faster: use AGM-like or Newton iterations?**\nFor e specifically, the series method with binary splitting is the standard. Alternative: compute exp(1) via the exponential function using arithmetic-geometric mean (AGM) methods: ln can be computed via AGM in O(M(D) log D), then invert via Newton's method to get exp. But that's for general exp; for e = exp(1), the direct Taylor series with binary splitting is simpler and has the same asymptotic complexity, with smaller constants.\n\nActually, is there anything asymptotically better? Brent's method using AGM for elementary functions gives O(M(D) log D). Binary splitting of the Taylor series also gives O(M(D) log D)... hmm, actually let me think: binary splitting for a D-digit result needs N ≈ D/log D terms... the total work: at each level of the recursion tree, total sizes multiply... The overall complexity is O(M(N log N) log N)... with N log N ≈ D (bits needed ~ size of N! ≈ N log N bits ≈ D ln 10 · D... hmm, D digits = D·log2(10) bits ≈ 3.32 D bits; N! has ≈ N log2(N/e) bits; setting N log N ≈ 3.32 D gives N ≈ 3.32 D / log2 D).\n\nCost = O(M(N log N) · log N)? No — the standard binary splitting analysis: the recursion tree has log N levels; at each level the total bit size of all operands is O(N log N) total? Actually at level k, there are 2^k nodes each with operands of size ~ N log N / 2^k bits. Multiplication cost per node: M(size). Summing: 2^k · M(N log N / 2^k) ≤ M(N log N) (superlinearity of M). So each level costs ≤ M(N log N), total O(M(N log N) log N) = O(M(D) log D) since N log N ≈ D bits. With M(D) = D log D ( Harvey–van der Hoeven), total O(D log² D). \n\nSo: **binary splitting + FFT multiplication = O(D log² D)** roughly. That's the best known for e I believe. Record computations of e (e.g., to tens of trillions of digits) use y-cruncher, which uses binary splitting of the Taylor series.\n\nPractical advice:\n1. Use the Taylor series Σ 1/k! with **binary splitting**.\n2. Use a bignum library with fast multiplication (GMP, or MPIR; FLINT has direct binary splitting functions).\n3. y-cruncher holds world records and computes e.\n\nAlso there's a classic trick: e = 1/Σ (-1)^k/k!? No, that's 1/e = e^{-1} = Σ (-1)^k/k!. Alternating series allows error bounding easily but not faster.\n\nAnother historical trick: compute e via continued fraction e = [2; 1, 2, 1, 1, 4, 1, 1, 6, ...] — converges but only linearly per term; convergents give about... the CF of e gives roughly... each convergent adds a few digits; not competitive.\n\nMachin-like formulas? For e there's no Machin-like formula; but you can compute e = exp(1) where exp is computed via... no. Actually there are \"Machin-like\" approaches: e = exp(1) computed as exp evaluated via the series at points like 1/2^k with repeated squaring: exp(1) = (exp(1/2^k))^{2^k} — computing exp(1/2^k) via series is fast since terms shrink, then square k times (each squaring is a full-precision multiplication: k ≈ log2(D/...)). Choose k so series needs few terms. exp(2^{-k}): need ~D log 10 / (k ln 2 + ... ) terms. Balance: cost ≈ M(D)·k + (terms)·M(D)... This is basically equivalent; binary splitting is cleaner.\n\nAlso mention: the naive way (summing 1/k! in floating point with increasing precision, or even exact big rationals incrementally) is O(n²)-ish and works fine for thousands of digits but not millions.\n\nFor a practical implementation:\n- In Python: use `mpmath` — `mp.mp.dps = N; print(mp.e)`? mpmath computes e via... it has `mp.e` computed via exp(1) using binary splitting I think. Or use `gmpy2`.\n- Simple binary splitting code in Python (integers) is maybe 15 lines and computes a million digits in seconds (Python's multiplication is Karatsuba+FFT? CPython uses Karatsuba and, since 3.x, also has... CPython uses Karatsuba for large ints; for very large, CPython doesn't use FFT (it does not implement Schönhage–Strassen). GMP does. So use gmpy2 for millions of digits.)\n\nLet me sketch the algorithm:\n\n```\ndef bs(a, b):\n # returns P, Q for sum_{k=a}^{b-1} 1/k!\n if b - a == 1:\n return (1, a) if a > 0 else (1, 1)\n m = (a + b) // 2\n Pl, Ql = bs(a, m)\n Pr, Qr = bs(m, b)\n return Pl * Qr + Pr * Ql, Ql * Qr\n```\n\nWait check base case: for a=0: term 1/0! = 1; Q should be 1. With my formulation sum from k=1: handle a=0 separately: add 1 at the end. If a≥1: P=1, Q=a.\n\nThen e ≈ 1 + P(1, N)/Q(1, N)? Sum_{k=1}^{N-1} 1/k! = P(1,N)/Q(1,N). Add 2 (for k=0 term 1/0! = 1... wait 0! = 1 so term is 1, and k=1 term is 1). Hmm: e = Σ_{k≥0} 1/k! = 1 + 1 + 1/2 + 1/6 + ... So e = 1 + Σ_{k≥1} 1/k! where the k=1 term is 1. So e ≈ 1 + P/Q, or incorporate directly: e = (Q + P)/Q.\n\nNumber of terms: need N such that 1/N! < 10^{-D}, i.e., N ln(N/e) ≥ D ln 10. Solve: N ≈ D ln 10 / ln(D ln 10 / ...) — iterate once or twice. Or use inverse Stirling. For D = 10^6, N ≈ 205,000-ish? Let me estimate: D ln 10 ≈ 2.3026×10^6. N ln(N/e) = 2.3026e6. Try N = 205,000: ln(205000/e) = ln(75445) ≈ 11.23. 205000 × 11.23 ≈ 2.30e6. ✓. So ~205k terms for a million digits. Each leaf multiplications small; the top-level numbers are ~3.3 Mbits. Feasible.\n\nFinal division: one big division of P by Q to D digits — cost M(D) with Newton division.\n\nDecimal conversion: converting a binary bignum to decimal string costs O(M(D) log D) via divide-and-conquer (scaled remainder). GMP's `mpz_get_str` uses sub-quadratic conversion for huge numbers. For millions of digits this matters; naive digit-by-digit is O(D²).\n\nError control: tail Σ_{k≥N} 1/k! < (1/N!)·(1/(1 - 1/(N+1))) ≈ e·/... roughly < 2/N!. Use guard digits (compute with extra bits) or exact bounds: since we compute exact rational partial sum, error is bounded by tail; choose N with margin and also work with extra guard bits in the final division.\n\nVerification: compute via two different N, or check against known digits; record computations use a second formula (e.g., compute 1/e via alternating series) and compare — y-cruncher does that.\n\nFurther optimizations:\n- **Parallelism**: binary splitting tree can be evaluated in parallel; y-cruncher is multithreaded.\n- **FFT multiplication** (GMP uses Schönhage–Strassen / Toom variants). \n- **Working in fixed-point** instead of rationals: instead of exact rational arithmetic, multiply partial sums as fixed-point integers scaled by 10^D — the standard \"sum 1/n! by dividing a big integer successively\" trick: compute S = Σ 10^D/k!... Actually a neat classic trick: compute integer T = floor(10^D · Σ_{k=0}^{N} 1/k!) via Horner-like scheme: start from the tail: T_N = 1; T_{k-1} = 1 + T_k/k... Hmm, there's the classic \"spigot\" algorithm for e!\n\n**Spigot algorithm** for e: The famous one from Rabinowitz & Wagon — e's digits can be extracted one at a time using an array, based on the mixed-radix representation 1/k! in bases (1, 2, 3, ...). It computes D digits in O(D²) time with O(D) space, using only small integers (machine words!). That's the simplest to implement, memory-light, but quadratic — good for up to ~10^4–10^5 digits maybe.\n\nActually the spigot for e: e = 2 + 1/2!(1 + 1/3(1 + 1/4(1 + ...)))... The algorithm: array A[0..n] initialized to 1, repeatedly: carry = 0; for i from n down: x = A[i]*10 + carry; A[i] = x mod (i+1)... etc. Each iteration yields one digit. O(n) per digit → O(D²) total. Simple, but for \"biggest quantity fast\" not the winner.\n\nSo the answer structure:\n1. The math: Taylor series; number of terms needed.\n2. The key algorithmic idea: binary splitting (reduces complexity dramatically).\n3. Underlying multiplication: FFT-based (GMP/FLINT), complexity ~O(D log²D) overall.\n4. Practical options: y-cruncher (records), mpmath/gmpy2 snippet, ~15-line Python binary splitting code.\n5. Mention spigot as simple but O(n²), AGM/Newton for exp as alternative with same asymptotics.\n6. Verification trick.\n\nThe user asks \"most optimized and fast way to approximate the biggest quantity of decimals\" — so they want state of the art. Answer: binary splitting of the hypergeometric series with FFT multiplication, as done by y-cruncher. Give complexity, term count, code sketch, and practical tips.\n\nLet me write concise Python code:\n\n```python\nfrom gmpy2 import mpz # or plain int for < ~100k digits\n\ndef bs(a, b):\n \"\"\"Return (P, Q) such that P/Q = sum_{k=a}^{b-1} 1/k! (a >= 1).\"\"\"\n if b - a == 1:\n return mpz(1), mpz(a)\n m = (a + b) >> 1\n p1, q1 = bs(a, m)\n p2, q2 = bs(m, b)\n return p1 * q2 + p2 * q1, q1 * q2\n\ndef e_digits(d):\n # need N with N! > 10^d → sum tail < 10^-d\n import math\n N = max(4, int(d * math.log(10) / math.log(d))) # refine\n while math.lgamma(N + 1) < d * math.log(10): N += 1\n P, Q = bs(1, N)\n scale = 10**d\n return (Q * scale // Q) ... \n```\n\nHmm: e = 1 + P/Q. digits = floor((P+Q) * 10^d / Q). Let me write:\n\n```python\ndef e_decimal(d):\n P, Q = bs(1, N)\n num = (P + Q) * 10**d\n return num // Q # integer with d+1 digits: \"2\" followed by d decimals\n```\n\nThen format string: str(num) → \"2.\" + rest.\n\nChoosing N: solve N! ≥ 10^{d+guard}. Use Stirling via math.lgamma: while lgamma(N+1) < (d+10)*ln10: N += 1, starting from a Stirling estimate. Fine.\n\nPerformance: with plain Python ints, ~100k digits in a few seconds? CPython multiplication is Karatsuba (O(n^1.585)); a million-digit run may take minutes. With gmpy2, million digits in ~a second or a few seconds. y-cruncher: trillions of digits in hours/days on good hardware.\n\nRecord: e has been computed to 35 trillion digits (2023-2024?) — I recall y-cruncher records: e to 31.4 trillion digits (2022, by Jordan Ranous? no that was pi...). Let me recall: π reached 100 trillion (2022, Emma Haruka Iwao) and 105T (2024). For e: record around 35 trillion digits (2023?) — I think \"e to 35,000,000,000,000 digits\" computed... I'm not 100% sure; I should hedge. I recall e's record was 31.4 trillion around 2023? There was a 2024 record of 202.11 billion? no. I'll hedge: \"tens of trillions of digits (y-cruncher)\".\n\nAlternative asymptotically equivalent: compute ln via AGM then Newton invert, or exp via Brent's O(M(n) log n) algorithm — mention briefly that for exp(1) nothing beats the Taylor series + binary splitting in practice.\n\nAlso mention memory: binary splitting needs O(D log D)... memory ~ size of the top-level numbers times depth? Recursion holds O(log N) numbers of size O(D) → O(D log N) bits? Actually you hold P,Q at each recursion level: sizes decrease geometrically down the tree, total O(D) bits per level path... total memory O(D log N) bits worst case if you keep both branches? In the recursive implementation, at any moment you hold one (P,Q) per level of depth → sizes sum to O(D) bits total along a path? Top level D bits, next D/2... geometric → O(D). Fine: memory O(D) bits ~ few hundred MB for a trillion digits... no wait D digits = 3.3 D bits; a trillion digits → 415 GB. That's why records need big machines and disk swapping.\n\nLet me also give the formula for choosing number of terms: N ≈ smallest n with n! > 10^{D} (Stirling: n ≈ D·ln10 / ln(D ln 10 / e)... just say use lgamma loop).\n\nComplexity summary:\n- Naive incremental summation (bignum floats, add 1/k! terms with running division): O(D²) word ops-ish → fine ≤ 10^5 digits.\n- Spigot: O(D²) but tiny memory and trivial code.\n- Binary splitting + FFT multiplication: Õ(D) (quasi-linear) — the winner; used by all record computations.\n\nOne more relevant point: \"biggest quantity of decimals\" fast — also worth noting that beyond algorithm, hardware (RAM, fast FFT, multithreading) dominates; y-cruncher is the practical answer for records.\n\nI should present:\n1. Short answer: binary splitting of Σ 1/k! with FFT-based bignum multiplication — quasi-linear in D; this is what record programs (y-cruncher) do.\n2. Why naive approaches are slower (complexity table).\n3. The algorithm in detail + minimal Python implementation.\n4. Term count & precision/error control.\n5. Practical tools (y-cruncher, mpmath, FLINT, gmpy2) and verification strategy.\n6. Optional: AGM/Newton alternative note.\n\nFormat: This is a technical question deserving a reasonably thorough answer with code. Use headers, a short complexity list, one code block, keep prose tight. Medium-long length. Avoid excessive bullet fragmentation; use a table maybe for complexity comparison. Code block essential. 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Which is the most optimized and fast way to aproximate the b...