This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A130741 #14 Sep 22 2024 02:02:40 %S A130741 7,259,4294967299,340282366920938463463374607431768211459 %N A130741 a(n) = 4^(4^n) + 3. %C A130741 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. %H A130741 Andrew Howroyd, <a href="/A130741/b130741.txt">Table of n, a(n) for n = 0..5</a> %F A130741 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. %F A130741 a(n) = 3+A137840(n). - _R. J. Mathar_, Sep 10 2016 %e A130741 For n = 0, 4^(4^0) + 3 = 4 + 3 = 7, the first entry. %o A130741 (PARI) a(n) = 4^(4^n) + 3 %K A130741 nonn,easy %O A130741 0,1 %A A130741 _Cino Hilliard_, Jul 07 2007