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A059215
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Least number k such that k^n reversed is a prime.
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2
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2, 4, 5, 2, 2, 16, 2, 8, 32, 2, 17, 4, 25, 28, 8, 53, 2, 25, 79, 95, 47, 46, 28, 2, 19, 5, 85, 86, 541, 32, 104, 314, 25, 115, 4, 5, 2, 25, 67, 71, 142, 226, 5, 53, 2, 304, 14, 106, 85, 8, 238, 128, 185, 23, 2, 65, 565, 122, 136, 668, 23, 37, 28, 1117, 178, 5, 74
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OFFSET
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1,1
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COMMENTS
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a(2) = 2 since 1^2, 2^2, 3^2 and 4^2 reversed are 1, 4, 9 and 61 and 61 is the first prime.
a(3) = 5 since 1^3, 2^3, 3^3, 4^3, and 5^3 reversed are 1, 8, 72, 46 and 521 and 521 is the first prime.
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LINKS
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MATHEMATICA
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Do[ k = 2; While[ ! PrimeQ[ ToExpression[ StringReverse[ ToString[ k^n ] ] ] ], k++ ]; Print[ k ], {n, 1, 50} ]
lnk[n_]:=Module[{k=1}, While[!PrimeQ[IntegerReverse[k^n]], k++]; k]; Array[ lnk, 50] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Mar 20 2021 *)
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PROG
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(Python)
from itertools import count
from gmpy2 import digits, is_prime
def a(n): return next(k for k in count(2) if is_prime(int(digits(k**n)[::-1])))
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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