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A212328
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Smallest k such that k^3 + 17 is divisible by 3^n.
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1
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1, 1, 4, 4, 58, 139, 139, 1597, 1597, 8158, 8158, 67207, 67207, 598648, 2192971, 6975940, 21324847, 21324847, 21324847, 408745336, 1571006803, 8544575605, 29465282011, 29465282011, 217751639665, 500181176146, 1347469785589, 6431201442247, 6431201442247
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OFFSET
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1,3
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COMMENTS
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This sequence is generalizable : the smallest k such that k^3 + p is divisible by 3^n exists if the prime p is congruent to + - 1 mod 18. For example, the sequence with p = 19 is given by {2, 2, 2, 20, 20, 20, 263, 992, 3179, 16301, 55667, 173765, 528059, …}. (See A129805). This sequence is given with the smallest p = 17.
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LINKS
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EXAMPLE
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a(4) = 4 because 4^3 + 17 = 81 is divisible by 3^4.
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MAPLE
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with(numtheory):for n from 1 to 20 do:i:=0:for x from 1 to 10^8 while(i=0) do: z:= x^3 + 17:if irem(z, 3^n)=0 then i:=1: printf ( "%d %d \n", n, x):else fi:od:od:
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PROG
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(PARI) print1(k=1); for(n=2, 100, if(Mod(k, 3^n)^3!=-17, k+=3^(n-2)* if(Mod(k+3^(n-2), 3^n)^3==-17, 1, 2)); print1(", "k)) \\ Charles R Greathouse IV, May 14 2012
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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