I see here and there mention of the $\lambda_I$-Calculus (in which every variable must be used at least once) and the $\lambda_K$-Calculus (in which a variable can also be unused).  Are they equivalent?  Why has the latter kinda obscured the former?EDITBy equivalent, I mean they have the same expressive power, namely, being universal or Turing complete.
Are the $\lambda_I$-Calculus and the $\lambda_K$-Calculus equivalent?
lambda calculus
You've basically answered the question yourself.$\lambda K$ is just another name for the standard, untyped lambda calculus.$\lambda I$ is a strict subset of $\lambda K$.$\lambda I$ doesn't allow terms where one abstracts over a variable but doesn't use it. So$$K = \lambda xy.x \in \lambda K$$but$$ K \not\in \lambda I$$Thanks to this restriction, $\lambda I$ has some interesting properties, in particular if $M$ has a normal form then so do all its sub-terms.Barendregt, H. P. The Lambda Calculus: Its Syntax and Semantics contains some notes about $\lambda I$, namely:... the $\lambda I$ calculus is sufficient to define all recursive functions  (since $K_1 := \lambda xy . yIIx$ satisfies $K_1xc_n = x$ for each of Church's numerals $c_n := \lambda fz . f^n z$ - it is also the case that for each finite set $n$ of nf's, we can find a local $K$ for $n$ $K_n$ such that $K_nMN = M$ for each $N$ in $n$).... The $\lambda I$ calculus corresponds to the combinatory logic with primitive combinators $I$, $B$, $C$, and $S$. ...