Lösung: Wir suchen die Anzahl von 8-stelligen Zahlen, bei denen jede Ziffer entweder 2 oder 3 ist, und es mindestens drei aufeinanderfolgende 2er gibt (also „222 als Teilstring).

["Solution: Counting 8-Digit Numbers with Only Digits 2 or 3 and at Least Three Consecutive 2s", "When solving combinatorial problems involving digit constraints, clarity and systematic counting are essential. One intriguing question is: How many 8-digit numbers consist only of the digits 2 and 3, and contain at least three consecutive 2s?", "This article presents a detailed, step-by-step solution using combinatorial reasoning and recursive counting to efficiently determine the valid numbers.", "---", "### Problem Recap", "We seek 8-digit numbers where:\n- Each digit is either 2 or 3\n- The number contains the substring "222" (three consecutive 2s)", "The total number of unrestricted 8-digit strings using digits 2 and 3 is (2^8 = 256), but we restrict to only those including at least one run of three or more consecutive 2s.", "A brute-force check of all 256 strings is possible but inefficient. Instead, we use recursive counting with constraints to determine valid strings compactly.", "---", "### Strategy Overview", "We model the counting using states representing the current suffix of consecutive 2s. Define:", "- ( f(n, k) ): number of valid (n)-digit strings ending with exactly (k) consecutive 2s (with (k = 0, 1, 2)), containing only digits 2 and 3, and containing no "222" so far.", "We want:\n[\n\ ext{Total} = \sum_{k=0}^{2} f(8, k) + N_{\ ext{ends-with-222}}\n]\nwhere (N_{\ ext{ends-with-222}}) counts strings where a run of at least three 2s has already occurred — these must be added to avoid double-counting.", "But a more direct and elegant approach tracks all strings up to 8 digits excluding those containing "222", then subtracts from total to get the desired count.", "Let:\n- ( T = 2^8 = 256 ): total 8-digit strings with digits 2 and 3\n- ( S ): number of 8-digit strings with digits 2,3 and containing at least one "222"", "Then:\n[\nS = T - C\n]\nwhere (C) is the number of 8-digit strings using only 2s and 3s that do not contain three consecutive 2s.", "---", "### Counting Strings Without "222" Using Recurrence", "Let ( a_n ) be the number of (n)-digit strings (digits 2 or 3) with no "222" substring.", "We build a recurrence based on how the string ends:", "Let:\n- ( a_n^{(0)} ): number of valid strings of length (n) ending in 3\n- ( a_n^{(1)} ): ending in exactly one 2\n- ( a_n^{(2)} ): ending in exactly two consecutive 2s", "Then:\n- ( a_n = a_n^{(0)} + a_n^{(1)} + a_n^{(2)} )", "Transition rules:\n- Append 3 to any string:\n - ( a_{n}^{(0)} = a_{n-1}^{(0)} + a_{n-1}^{(1)} + a_{n-1}^{(2)} = a_{n-1} )\n- Append 2 (only allowed if not forming "222"):\n - Can append 2 only if previous end was not "22"\n - So:\n - ( a_n^{(1)} = a_{n-1}^{(0)} ) (append 2 to strings ending in 3)\n - ( a_n^{(2)} = a_{n-1}^{(1)} ) (append 2 to single 2 → now two)\n - Cannot append 2 to two 2s (would make "222")", "Initial conditions ((n = 1)):\n- ( a_1^{(0)} = 1 ) → "3"\n- ( a_1^{(1)} = 1 ) → "2"\n- ( a_1^{(2)} = 0 ) → cannot have two 2s", "We compute iteratively:", "---", "n = 2:\n- ( a_2^{(0)} = a_1 = 1 + 1 + 0 = 2 )\n- ( a_2^{(1)} = a_1^{(0)} = 1 )\n- ( a_2^{(2)} = a_1^{(1)} = 1 )\n- ( a_2 = 2 + 1 + 1 = 4 ) (Total 4-letter strings without "222": all possible — valid)", "n = 3:\n- ( a_3^{(0)} = a_2 = 4 )\n- ( a_3^{(1)} = a_2^{(0)} = 2 )\n- ( a_3^{(2)} = a_2^{(1)} = 1 )\n- ( a_3 = 4 + 2 + 1 = 7 )\n→ Only "222" is missing; total possible is 8 → confirms correct.", "n = 4:\n- ( a_4^{(0)} = a_3 = 7 )\n- ( a_4^{(1)} = a_3^{(0)} = 4 )\n- ( a_4^{(2)} = a_3^{(1)} = 2 )\n- ( a_4 = 7 + 4 + 2 = 13 )", "n = 5:\n- ( a_5^{(0)} = a_4 = 13 )\n- ( a_5^{(1)} = a_4^{(0)} = 7 )\n- ( a_5^{(2)} = a_4^{(1)} = 4 )\n- ( a_5 = 13 + 7 + 4 = 24 )", "n = 6:\n- ( a_6^{(0)} = a_5 = 24 )\n- ( a_6^{(1)} = a_5^{(0)} = 13 )\n- ( a_6^{(2)} = a_5^{(1)} = 7 )\n- ( a_6 = 24 + 13 + 7 = 44 )", "n = 7:\n- ( a_7^{(0)} = a_6 = 44 )\n- ( a_7^{(1)} = a_6^{(0)} = 24 )\n- ( a_7^{(2)} = a_6^{(1)} = 13 )\n- ( a_7 = 44 + 24 + 13 = 81 )", "n = 8:\n- ( a_8^{(0)} = a_7 = 81 )\n- ( a_8^{(1)} = a_7^{(0)} = 44 )\n- ( a_8^{(2)} = a_7^{(1)} = 24 )\n- ( a_8 = 81 + 44 + 24 = 149 )", "So, number of 8-digit strings using only 2 and 3 with no "222" is (a_8 = 149).", "---", "### Final Calculation", "Total 8-digit strings with digits 2 or 3:\n[\nT = 2^8 = 256\n]", "Strings containing at least one "222":\n[\nS = T - a_8 = 256 - 149 = 107\n]", "---", "### Important Clarification: Counting All Valid "222"-containing Strings", "Our recurrence (a_n) counts all 8-digit strings over {2,3} avoiding "222". Subtracting from total gives strings with at least one "222" — exactly the count we want.", "We do not need to separately count strings that end with "222", because this inclusion-exclusion is fully captured by subtracting the avoid strings.", "---", "### Why This Method Works", "By decomposing based on suffix runs, we avoid double-counting and efficiently count complex substring constraints using dynamic programming. The recurrence ensures correctness by building valid strings step-by-step and disallowing forbidden patterns.", "---", "### Conclusion", "There are 107 eight-digit numbers composed only of digits 2 and 3 that contain at least three consecutive 2s.", "This solution demonstrates how combinatorial reasoning and recursive state modeling enable efficient counting in permutation-constrained problems — a valuable technique for algorithms, dynamic programming, and mathematical enumeration.", "---", "Keywords: 8-digit numbers, 2 or 3 digits, at least three consecutive 2s, string counting, combinatorics, dynamic programming, substring "222", recursive counting, total combinations minus invalid.", "---", "Stay tuned for more insights into enumeration techniques and combinatorial puzzles!"]