Edwin Clark (eclark@math.usf.edu) asked:
Let E=[e_1,e_2,...,e_k] be a list of integers satisfying 1 <= e_1 <= e_2 <= . . . <= e_k.
Let S(E) be the set of all positive integers n such that n = Prod{p_i^e_i} where the p_i's are distinct primes.
Say that the integers in S(E) have factorization pattern E. I believe this is a known concept. ... Is it the case that if 1 is not in E then S(E) contains no consecutive pair of integers?
No. There are at least two counterexamples among the solutions of 23 x^2 + 1 = 8 y^2. The positive solutions of this are given by x(0) = 23; y(0) = 39; x(n) = 24335 x(n-1) + 14352 y(n-1); y(n) = 41262 x(n-1) + 24335 y(n-1). (Note that x(n) is always divisible by 23, so 23 x(n)^2 and 8 y(n)^2 are consecutive powerful numbers.) For n=8, 23 x(n)^2 and 8 y(n)^2 are in S([2,2,2,2,3]); their factorizations are 23^3 1361^2 39157289^2 13475265289^2 43842128059343041^2 and 2^3 3^2 13^2 509^2 61853469184829429986091801293807909^2 For n=9, they're in S([2,2,2,2,2,3]), with factorizations 23^3 151^2 2281^2 23447^2 6461369^2 29366147657538184938924047^2 and 2^3 3^2 13^2 35872892507^2 5281732523824487^2 8087241110442161^2. Dean Hickerson dean@math.ucdavis.edu