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    "# [1000-digit Fibonacci Number](https://projecteuler.net/problem=25)\n",
    "\n",
    "This problem can be solved with a scientific calculator!\n",
    "\n",
    "We're looking at [Fibonacci numbers](https://en.wikipedia.org/wiki/Fibonacci_sequence) again, after last seeing them in [problem 2](https://projecteuler.net/problem=2). Our goal is to find the index of the first 1000 digit number in the sequence. Mathematically, we can state this is as the lowest $n$ such that\n",
    "$$1+\\log{F_n} \\geq 1000$$\n",
    "(Here, $\\log$ is the base 10 logarithm).\n",
    "\n",
    "To assist in finding $n$, we can employ Binet's formula:\n",
    "$$F_n = \\frac{\\phi^n - (-\\phi)^{-n}}{\\sqrt{5}}$$\n",
    "Here, $\\phi$ is the [golden ratio](https://en.wikipedia.org/wiki/Golden_ratio).\n",
    "\n",
    "Binet's formula is exact, but we can make our work a little easier by approximating.\n",
    "$$F_n \\approx \\frac{\\phi^n}{\\sqrt{5}}$$\n",
    "This approximation works since $(-\\phi)^{-n}$ approaches 0 as $n$ increases. This also means this approximation gets more accurate as $n$ increases, which is especially convenient since we're looking for a large $n$. Substituting this approximation into the above inequality, we have\n",
    "$$1 + \\log{\\frac{\\phi^n}{\\sqrt{5}}} \\geq 1000$$\n",
    "\n",
    "Now with a little bit of algebra, we can solve for $n$.\n",
    "$$n \\geq 999\\log_{\\phi}10 + \\log_{\\phi}\\sqrt{5}$$\n",
    "If you're trying to solve this with a scientific calculator, you probably don't have a $\\log_{\\phi}$ button (*please let me know if you do*), but we can just use the [logarithmic change of base formula](https://en.wikipedia.org/wiki/List_of_logarithmic_identities). This ends up getting us\n",
    "$$n \\geq 4781.85927\\ldots$$\n",
    "Therefore, we want $n=4782$.\n",
    "\n",
    "## Relevant sequences\n",
    "* Fibonacci numbers: [A000045](https://oeis.org/A000045)\n",
    "\n",
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
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