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Q: What is the n 1113122115?
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What is the next number in this sequence 5 15 1115 3115 132115?

1113122115 Each subsequent term of the sequence is the count of each digit in the previous term: 5 → one-"5" = 15 15 → one-"1"-one-"5" = 1115 1115 → three-"1"s-one-"5" = 3115 3115 → one-"3"-two-"1"s-one-"5" = 132115 the next term is: 132115 → one-"1"-one-"3"-one-"2"-two-"1"s-one-"5" = 1113122115


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Assuming you mean the first n counting numbers then: let S{n} be the sum; then: S{n} = 1 + 2 + ... + (n-1) + n As addition is commutative, the sum can be reversed to give: S{n} = n + (n-1) + ... + 2 + 1 Now add the two versions together (term by term), giving: S{n} + S{n} = (1 + n) + (2 + (n-1)) + ... + ((n-1) + 2) + (n + 1) → 2S{n} = (n+1) + (n+1) + ... + (n+1) + (n+1) As there were originally n terms, this is (n+1) added n times, giving: 2S{n} = n(n+1) → S{n} = ½n(n+1) The sum of the first n counting numbers is ½n(n+1).


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n+n-n-n-n+n-n-n squared to the 934892547857284579275348975297384579th power times 567896578239657824623786587346378 minus 36757544.545278789789375894789572356757583775389=n solve for n! the answer is 42