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Search: id:A086758
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| A086758 |
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a(n) is the smallest m such that the integer part of the first n powers of m^(1/n) are primes. |
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1
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OFFSET
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1,1
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COMMENT
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All terms of this sequence must be primes because Floor[(a(n)^(1/n))^n]=a(n). Floor[(a(8)^(1/8))^k]=Floor[(1287/545)^k] for k=1,...,10 (see puzzle 227). If a(9) exist it must be greater than 22000000.
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REFERENCES
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R. Crandall and C. Pomerance, Prime Numbers: A Computational Perspective, Springer, NY, 2001; see Exercise 1.75, p. 69.
Puzzle 227 of www.primepuzzles.net.
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LINKS
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C.Rivera Puzzle 227
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FORMULA
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For[m=1, Union[Table[PrimeQ[Floor[Prime[m]^(k/n)]], {k, n}]]!={True}, m++ ]; Prime[m]
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EXAMPLE
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a(5)=631 because Floor[631^(1/5)]=3,Floor[631^(2/5)]=13,Floor[631^(3/5)]=47,
Floor[631^(4/5)]=173&Floor[631^(5/5)]=631 are primes and 631 is the smallest
m with this property.
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MATHEMATICA
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Do[Print[For[m=1, Union[Table[PrimeQ[Floor[Prime[m]^(k/n)]], {k, n}]]!={True}, m++ ]; Prime[m]], {n, 8}]
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CROSSREFS
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Sequence in context: A056367 A082733 A095134 this_sequence A116702 A098156 A098586
Adjacent sequences: A086755 A086756 A086757 this_sequence A086759 A086760 A086761
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KEYWORD
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more,nonn
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AUTHOR
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Farideh Firoozbakht (f.firoozbakht(AT)sci.ui.ac.ir), Aug 01 2003
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