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 A130740 Numbers of the form 2^(4^n)+5. 0
 7, 21, 65541, 18446744073709551621, 115792089237316195423570985008687907853269984665640564039457584007913129639941 (list; graph; refs; listen; history; text; internal format)
 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 = (a-b)(a^(m-1)+a^(m-2)b +...+ b^(m-1). 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^n-b^n is divisible by (a-b) = (16-2) = 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 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

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Last modified March 29 21:39 EDT 2023. Contains 361599 sequences. (Running on oeis4.)