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A152583
Numbers of the form 11^(2^n) + 2.
1
13, 123, 14643, 214358883, 45949729863572163, 2111377674535255285545615254209923, 4457915684525902395869512133369841539490161434991526715513934826243
OFFSET
1,1
COMMENTS
Except for the first term, these numbers are divisible by 3. This follows from the binomial expansion of (9+2)^(2^n)+2 = 9h + 2^(2^n)+2. Now 2^(2^n)+2 can be written as 2*(2^(2^n-1)+1) and 2^(2^n-1)+1 is divisible by 3. This is evident from the identity, a^m+b^m = (a+b)(a^(m-1) - a(m-2)b + ... + b^(m-1)) for odd m and 2^n-1 is odd.
PROG
(PARI) g(a, n) = if(a%2, b=2, b=1); for(x=0, n, y=a^(2^x)+b; print1(y", "))
CROSSREFS
Sequence in context: A115204 A016277 A202131 * A304353 A305916 A134550
KEYWORD
nonn
AUTHOR
Cino Hilliard, Dec 08 2008
STATUS
approved