User blog comment:BlauesWasser/Phela Function (Infinitely powerful)/@comment-31663462-20180222114141/@comment-1605058-20180223113917

\(2^{\aleph_n}\neq\aleph_{n+1}\), at least not necessarily. The statement that this is so for all, even transfinite, n, is known as the generalized continuum hypothesis and it's known to be independent of the standard foundations. Phela(n) clocks somewhere between \(\beth_n\) and \(\beth_{2n}\) and no tighter bound is possible. Already Phela(1) can be far larger than anything we can sensibly describe using aleph hierarchy.