So, number of 8-digit numbers (with digits 2 and 3 only) containing at least one occurrence of 222 is:

["How Many 8-Digit Numbers Use Only the Digits 2 and 3 and Contain at Least One Occurrence of 222?", "If you’re fascinated by numbers and combinatorics, one intriguing question is: How many 8-digit numbers composed exclusively of the digits 2 and 3 contain at least one occurrence of the substring "222"? This seemingly simple query dives deep into counting constrained binary-like sequences and has implications in coding theory, combinatorics, and digital encryption.", "In this article, we explore the exact count of valid 8-digit numbers formed from digits 2 and 3, with a focus on those containing at least one "222" — a fascinating study of patterns in restricted digit strings.", "---", "### What Are Valid 8-Digit Numbers Using Only Digits 2 and 3?", "Since each digit in the number must be either 2 or 3, and the number has exactly 8 digits, we’re essentially counting all binary sequences of length 8 where each position is filled with either a 2 or a 3. For each of the 8 positions, there are 2 choices: so the total number of such 8-digit numbers is:", "[\n2^8 = 256\n]", "This means there are 256 total valid numbers. But not all contain "222." We want only those that contain at least one occurrence of three consecutive 2s.", "---", "### Why Counting “At Least One” Is Easier Than Counting “Exactly Some”", "Instead of calculating directly how many sequences contain "222", we can use the complement rule:", "[\n\ ext{# with at least one 222} = \ ext{Total} - \ ext{# with no 222}\n]", "So:", "[\n\ ext{Answer} = 2^8 - N(\ ext{no "222" in any 8-digit sequence})\n]", "Where ( N(\ ext{no "222"}) ) is the number of binary-like strings of length 8 using only 2 and 3 that do not contain three consecutive 2s anywhere.", "---", "### Model the Problem Using Recurrence Relations", "Define ( a_n ) as the number of n-digit numbers using only digits 2 and 3, with no three consecutive 2s.", "Each such valid string ends in one of these patterns:\n- Ends with 3 → preceding part is any valid string of length ( n-1 )\n- Ends with one or two 2s — but never three", "Let’s formalize:", "- Let ( a_n ) = number of valid n-digit sequences with no "222"", "We build valid sequences of length ( n ) from shorter ones:", "Let’s define a recurrence based on the last few digits:", "Let:\n- ( a_n ): total valid sequences of length ( n ) ending with:\n - a 3, or\n - ending in exactly one or two 2s", "We implement the recurrence via states:", "Let:\n- ( A_n ): sequences of length ( n ) ending in 3\n- ( B_n ): ending in exactly one 2\n- ( C_n ): ending in exactly two 2s", "Then:", "- ( A_n = a_{n-1} ) → any valid sequence of length ( n-1 ) followed by 3\n- ( B_n = A_{n-1} ) → only sequences ending in 3 can be extended by one 2\n- ( C_n = B_{n-1} ) → only sequences ending in a single 2 can be extended by another 2 (making two)", "And:\n[\na_n = A_n + B_n + C_n\n]", "But we can simplify by noting:", "[\na_n = a_{n-1} + a_{n-2} + a_{n-3}\n]", "Why? — Because:\n- If the sequence ends in 3 → the rest is any valid ( a_{n-1} )\n- If ends in "32" → the first ( n-2 ) must be valid: ( a_{n-2} )\n- If ends in "222" → invalid, so excluded — we split based on how many trailing 2s.", "Alternatively, recurrence with state tracking is clearer:", "Initialize base cases:", "- ( n = 1 ):\n - "2" → valid, ends in 2 one time → ( B_1 = 1 )\n - ( A_1 = 1 ) ("3")\n - ( C_1 = 0 ) (can’t have two 2s)\n So ( a_1 = 2 )", "- ( n = 2 ):\n Possible: 22, 23, 32, 33 — all valid (no 222 possible)\n "22" → ends in two 2s → ( C_2 = 1 )\n "32" → ( B_2 = 1 )\n "23", "33" → ( A_2 = 2 )\n So ( a_2 = 1+1+2 = 4 = 2^2 ) — all valid", "- ( n = 3 ):\n The only invalid sequence is "222"\n Total: 8, invalid: 1 → valid: 7\n Use recurrence:", "Compute step-by-step:", "[\n\begin{align}\na_1 &= 2 \quad \ ext{(2,3)} \\na_2 &= 4 \quad \ ext{(22,23,32,33)} \\na_3 &= \ ext{Total} - \ ext{"222"} = 8 - 1 = 7 \\n\end{align}\n]", "Now use recurrence:", "We define:", "- ( A_1 = 1 ), ( B_1 = 1 ), ( C_1 = 0 )\n- ( A_2 = a_1 = 2 ),\n ( B_2 = A_1 = 1 ),\n ( C_2 = B_1 = 1 ) → ( a_2 = 2 + 1 + 1 = 4 ) ✓\n- ( A_3 = a_2 = 4 )\n ( B_3 = A_2 = 2 )\n ( C_3 = B_2 = 1 )\n ( a_3 = 4 + 2 + 1 = 7 ) ✓", "Continue:", "- ( n = 4 ):\n ( A_4 = a_3 = 7 )\n ( B_4 = A_3 = 4 )\n ( C_4 = B_3 = 2 )\n ( a_4 = 7 + 4 + 2 = 13 )", "- ( n = 5 ):\n ( A_5 = a_4 = 13 )\n ( B_5 = A_4 = 7 )\n ( C_5 = B_4 = 4 )\n ( a_5 = 13 + 7 + 4 = 24 )", "- ( n = 6 ):\n ( A_6 = a_5 = 24 )\n ( B_6 = A_5 = 13 )\n ( C_6 = B_5 = 7 )\n ( a_6 = 24 + 13 + 7 = 44 )", "- ( n = 7 ):\n ( A_7 = a_6 = 44 )\n ( B_7 = A_6 = 24 )\n ( C_7 = B_6 = 13 )\n ( a_7 = 44 + 24 + 13 = 81 )", "- ( n = 8 ):\n ( A_8 = a_7 = 81 )\n ( B_8 = A_7 = 44 )\n ( C_8 = B_7 = 24 )\n ( a_8 = 81 + 44 + 24 = 149 )", "So, number of 8-digit numbers using only digits 2 and 3 with no "222" substring is ( 149 ).", "---", "### Compute Final Answer", "Total such 8-digit numbers: ( 2^8 = 256 )\nNumber with at least one "222":\n[\n256 - 149 = 107\n]", "---", "### Summary", "- Total 8-digit numbers with digits 2 and 3: ( 2^8 = 256 )\n- Number without "222": ( 149 )\n- Number with at least one "222":\n[\n\boxed{107}\n]", "This is a classic application of complementary counting in constrained digit strings. The recurrence ( a_n = a_{n-1} + a_{n-2} + a_{n-3} ), derived from state-based decomposition, efficiently counts forbidden patterns.", "---", "Keywords: 8-digit numbers, digits 2 and 3, count numbers with 222, combinatorics, restricted sequences, no three consecutive 2s, recurrence relations", "Meta Description: Discover how many 8-digit numbers made only from digits 2 and 3 contain at least one occurrence of the substring "222" using state-based recurrence in combinatorics. Total count: 107."]