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COMMENT
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Suggestion from Rickey Bowers Jr, (math(AT)bitRAKE.com), May 18 2006:
a[0] = 5,
a[1] = 2 * Prime[5],
a[2] = [Prime[5]]^2,
a[3] = 2^4 * Prime[Prime[5]],
a[4] = [Prime[Prime[5]]]^(2^4),
...
a[2n-1] = (2^(n^2)) * Prime^n[a[0]],
a[2n] = Prime^n[a[0]]^(2^(n^2)).
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