cp's OEIS Frontend

This is to 4 as A058220 is to 2 and A140331 is to 3.

a(8) > 22174.

The total number of n-ary operators in a k-valued logic is T = k^(k^n), i.e. if S is a set of k elements, there are T ways of mapping an ordered subset of n elements taken from S to an element of S. Some operators are "degenerate": the operator has arity p, if only p of the n input values influence the output. = therefore the set of operators can be partitioned into n+1 disjoint subsets representing arities from 0 to n.

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.