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A133454
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Chain of 6 highly composite numbers generated when subject to the recurrence relation tau(a(n+1)) = a(n), with a(0)=3, where tau(n) is the number-of-divisors function A000005(n).
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0
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OFFSET
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1,1
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COMMENTS
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We omit the seed a(0) from the sequence and keep the offset at 1, because 3 is not highly composite. - R. J. Mathar, Jun 20 2021
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LINKS
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EXAMPLE
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Since 4 is the HCN with 3 divisors, we have tau(4) = 3 and therefore a(1)=4; the HCN with 4 divisors is 6, so that tau(6) = 4 and hence a(2)=6; the HCN with 6 divisors is 12 so that tau(12) = 6, implying a(3)=12, ...
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CROSSREFS
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KEYWORD
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fini,full,nonn
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AUTHOR
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STATUS
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approved
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