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A099147
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Iterated hexagonal numbers, starting at 1.
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4
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OFFSET
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1,2
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COMMENTS
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LINKS
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FORMULA
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a(n) = b(n) for n<=2, a(n) = b(a(n-1)) for n>2, where b(n) = A000384(n) = n*(2*n-1), the hexagonal numbers.
a(1) = 1, a(2) = 6, a(n) = 2*a(n-1)^2 - a(n-1) for n>2.
Let H(n) = n*(2*n-1) = the n-th hexagonal number. Define A(n, k) recursively by A(1, k) = H(k), A(n, k) = A(1, A(n-1, k)) for n>1. Then a(1) = A(1, 1), a(n) = A(n-1, 2) for n>1.
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EXAMPLE
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a(4) = b(a(3)) = b(b(a(2))) = b(b(b(2))) = b(b(6)) = b(66) = 8646, where b(n) = A000384(n).
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PROG
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(PARI) {hexagonal(n) = n*(2*n-1)} {a(n) = if(n<=2, hexagonal(n), hexagonal(a(n-1)))} \\ Klaus Brockhaus, Jan 10 2007
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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