

A130740


Numbers of the form 2^(4^n)+5.


0




OFFSET

1,1


COMMENTS

Numbers of the form 2^(4^n) + 5 are divisible by 7. We prove this by induction making use of the expansion (1) a^m  b^m = (ab)(a^(m1)+a^(m2)b +...+ b^(m1). For n = 1, we have 2^4 + 5 = 21 = 7*3. So the statement is true for n=1. Now assume the statement is true for some integer k and show that it is also true for k+1. Thus we have 2^(4^k) + 5 = 7h for some h. Let 2^(4^(k+1))+ 5 = h1. Now consider the difference h1  7h. If this is a multiple of 7 then so is h1. So we have 2^(4^(k+1))+ 5  (2^(4^k) + 5)) = 16^(4^k)2^(4^k). This is of the form (1) where n = 4^k, a = 16 and b = 2. So the difference, a^nb^n is divisible by (ab) = (162) = 14 = 7*2. This implies 2^(4^(k+1)+5 is divisible by 7. So we have assumed the statement was true for k and have shown it to be true for k+1. Therefore by the induction hypothesis, the statement is true for all n.


LINKS

Table of n, a(n) for n=1..5.


FORMULA

This function was derived from the even case of Fermat numbers of order 5 or F(m,5) = 2^(2^m) + 5. Let m = 2n to get 2^(2^(2n)) + 5 = 2^(4^n)+5.


PROG

(PARI) g(n) = for(x=0, n, y=2^(4^x)+ 5; print1(y", "))


CROSSREFS

Sequence in context: A220161 A213152 A082826 * A101119 A217014 A200886
Adjacent sequences: A130737 A130738 A130739 * A130741 A130742 A130743


KEYWORD

nonn


AUTHOR

Cino Hilliard, Jul 07 2007


STATUS

approved



