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A152578
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Numbers of the form 5^(2^n) + 2.
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1
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OFFSET
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1,1
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COMMENTS
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Except for the first term, these numbers are divisible by 3. This follows from the identity I: a^n-b^n = (a+b)(a^(n-1) - a^(n-2)b + ... + b^(n-1)) for odd values of n. In this example, by expanding the binomial (3+2)^(2^n)+2, we get 3h + 2^(2^n)+2 for some h. Now 2^(2^n)+2 = 2*(2^(2^n)-1)+1). Since 2^n-1 is odd, by identity I, 3 divides 2^(2^n)+2 + 3h. Therefore 3 divides 5^(2^n)+2 for n > 0.
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LINKS
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PROG
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(PARI) g(a, n) = if(a%2, b=2, b=1); for(x=0, n, y=a^(2^x)+b; print1(y", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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