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A099129
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Let T(n) be the n-th triangular number n*(n+1)/2; then a(n) = n-th iteration T(T(T(...(n)))).
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2
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OFFSET
| 0,2
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COMMENTS
| The growth of this sequence is faster than exponential. This can be derived from the exponential generating function for triangular numbers: g(x) = [ 1 + 2x + x^2/2 ] (e^x) = 1 + 3x/1! + 6x^2/2! + 10x^3/3! + 15x^4/4! + 21x^5/5! + ...
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REFERENCES
| J. V. Post, "Iterated Triangular Numbers", preprint.
J. V. Post, "Iterated Polygonal Numbers", preprint.
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FORMULA
| a(n) = A007501(n, n)
The sequence grows like O(n^2^n*1/2^n). This can be derived from the growth O(n^2*1/2) of the triangle sum by iteration. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jan 21 2006
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EXAMPLE
| a(3) = 231 because we can write the 3-time iterated expression on T(3), the triangular number sequence n*(n+1)/2, namely: T(T(T(3))) = 231.
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CROSSREFS
| Cf. A007501, A000217.
Sequence in context: A112001 A099124 A172862 * A194482 A145180 A077231
Adjacent sequences: A099126 A099127 A099128 * A099130 A099131 A099132
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KEYWORD
| easy,nonn
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AUTHOR
| Jonathan Vos Post (jvospost3(AT)gmail.com), Nov 14 2004
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