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Search: id:A085298
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| A085298 |
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x=a(n) is the smallest exponent such that p[n]^x when reversed is a prime. |
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+0
5
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| 1, 1, 1, 1, 1, 1, 1, 2, 8, 7, 1, 1, 2, 5, 15, 10, 12, 4, 39, 1, 1, 1, 11, 2, 1, 1, 10, 1, 23, 1, 5, 1, 243, 2, 1, 1, 1, 23, 1, 34, 1, 1, 1, 2, 58, 1, 3, 9, 166, 17, 68, 8, 8, 3, 7, 5, 5, 2, 2, 2, 61, 11, 97, 1, 1, 10, 2, 1, 1, 41, 1, 1, 66, 1, 5, 1, 1, 2, 2, 8, 40, 2, 8, 19, 2, 2, 723
(list; graph; listen)
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OFFSET
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1,8
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COMMENT
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It is conjectured that for every n such exponent exists.
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FORMULA
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a(n)=Min{x; reversed[p(n)^x] is a prime}
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EXAMPLE
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a(n)=1 means that rev[p(n)^1]=rev[p(n)] is prime i.e. p(n) is in A007500;
a(n)=2 means that rev[p(n)^2] is prime but rev[p(n)] is not,like n=8:p=19 and 91 is not a prime but rev[19^2]=rev[361]=163 is a prime;
For n, the first k exponent providing rev[p(n)^k] prime can be quite large, like at n=87:
rev[p(87)^723]=rev[449^723] is the first [probably] prime has 1918 decimal digits: 948......573.
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MATHEMATICA
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nd[x_, y_] := 10*x+y tn[x_] := Fold[nd, 0, x] bac[x_] := tn[Reverse[IntegerDigits[x]]] Table[f=1; Do[s=bac[Prime[n]^k]; If[PrimeQ[s]&&Equal[f, 1], Print[{k, n, Prime[n], s}]; f=0], {k, 1, 1000}], {n, 1, 256}]
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CROSSREFS
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Cf. A057708, A058993, A056994, A059695, A003459, A007500, A055387, A061461, A068652.
Sequence in context: A021781 A130820 A134877 this_sequence A011060 A093624 A021352
Adjacent sequences: A085295 A085296 A085297 this_sequence A085299 A085300 A085301
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KEYWORD
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base,nonn
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu), Jun 24 2003
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