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A091443
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Multiperfect numbers n which are divisible by sopfr(n) (multiperfect number: sigma(n) = k*n with k integer, sopfr: Sum of prime factors with repetition).
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4
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1379454720, 14182439040, 212517062615531520, 27099073228001299660800, 680489641226538823680000, 15229814702070563916152832000, 34111227434420791224041472000, 59023729003862626557345792000
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OFFSET
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1,1
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COMMENTS
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The sequence contains multiperfect numbers with multiplicity k from 3..8. They are extracted from a list with about 5000 multiperfect numbers with multiplicity from 2..11. Because of the size of these numbers, no numbers with multiplicity k > 8 were found, even though there were about 3000 of them in the list. 95% of the multiperfect numbers with multiplicity from 3..8 are known.
Conjecture: the sequence is finite.
There are 5255 multiperfect numbers known with multiplicity 3 to 11. No more findings for A091443 so we still have 33 multiperfect numbers divisible by their sopfr (without the trivial case 1). With multiplicity 3..8 quite surely all are found (only very few - if any - missing). It is estimated that there are about 2200 with multiplicity 9 and 2091 of them are already found. With multiplicity 10 of estimated 4500 1161 are known. So far no multiperfect number with multiplicity 9 or 10 is divisible by its sopfr (with repetition). Using sopfr without repetition (A114887), there is one number with multiplicity 9 (or more). - Sven Simon, Feb 12 2012
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LINKS
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EXAMPLE
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a(1): 1379454720 = 2^8*3*5*7*19*37*73, sopfr(n)= 2^5*5.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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