1,1

Numbers of the form 4^(4^n) + 3 are divisible by 7. We prove this by induction making use of the expansion (1) a^m - b^m = (a-b)(a^(m-1)+a^(m-2)b+...+b^(m-1). For n = 1, we have 4^4 + 3 = 259 = 7*37. 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 4^(4^k) + 3 = 7h for some h. Let 4^(4^(k+1))+ 3 = h1. Now consider the difference h1 - 7h. If this is a multiple of 7 then so is h1. So we have 4^(4^(k+1))+ 3 - (4^(4^k) + 3)) = 256^(4^k)-4^(4^k). This is of the form (1) where n = 4^k, a = 256 and b = 4. So the difference, a^n-b^n is divisible by (a-b) = (256-4) = 252 = 7*36. This implies 4^(4^(k+1)+3 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.

This function is derived from the odd case of Fermat numbers of order 3 or F(m,3) = 2^(2^m)+3. Let m = 2n+1, to get 2^(2^(2n+1)) + 3 = 2^(2*2^(2*n))+3 = 4^(2^(2*n))+3 = 4^(4^n)) + 3.

For n = 0, 4^(4^0) + 3 = 4 + 3 = 7, the first entry.

(PARI) g(n) = for(x=0, n, y=4^(4^x)+3; print1(y", "))

Sequence in context: A133589 A142805 A119942 this_sequence A003385 A129423 A048291

Adjacent sequences: A130738 A130739 A130740 this_sequence A130742 A130743 A130744

nonn

Cino Hilliard (hillcino368(AT)hotmail.com), Jul 07 2007