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A212490
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Least number m > 1 such that A000203(x)*x = m has exactly n solutions.
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6
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OFFSET
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1,1
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COMMENTS
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6 = 6.
336 = 6*28*2.
333312 = 6*28*496*2*2.
5418319872 = 6*28*496*8128*2*2*2.
sigma(n)/n increases to a record in A004394. This can be used to limit the checked divisors of some candidate m.
For n >= 6, If gcd(a(4), a(5)) | a(n) then a(n) > 1.1*10^17. If (gcd(a(4), a(5)) * 2047) | a(n) then a(n) > 1.8 * 10^20. (End)
a(6) <= 6*28*496*8128*33550336*137438691328*2*2*2*2*2. - Michel Marcus, Nov 01 2020
Using the same as above, a(7) <= 1716908124551996896669734276042690920448.
a(8) <= 7917841189233800244470292555938612387093638081493952626688. (End)
a(6) <= 7089671638182002688000,
a(7) <= 106345074572730040320,
a(9) <= 1826980530660612389572800675840. (End)
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REFERENCES
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R. K. Guy, Unsolved Problems in Theory of Numbers, Springer-Verlag, Third Edition, 2004, B11.
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LINKS
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EXAMPLE
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For n=3, 333312 has exactly 3 solutions: sigma(434)*434 = 333312, sigma(372)*372 = 333312, and sigma(336)*336 = 333312; therefore a(3) = 333312.
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PROG
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(PARI) isok(k, n) = sumdiv(k, d, d*sigma(d) == k) == n;
a(n) = my(k=2); while (! isok(k, n), k++); k; \\ Michel Marcus, Oct 28 2020
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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