What is h in this RSA variant? I am trying to implement a proposed improved algorithm of RSA . Here the author has increased the number of exponents. However I am unable to understand what $h$ is in the Key generation step. Can someone tell me what $h$ is exactly? 

The author's paper is : http://www.jacn.net/papers/47-N024.pdf
I had originally posted this paper : http://iosrjournals.org/iosr-jece/papers/Vol9-Issue1/Version-5/H09154347.pdf


  n = p x q
  Ø(n) = (p-1)(q-1)
  ϒ n, h = ph-p0 ph-p1…ph-ph-1 + qh-q0 qh- q1 … qh-qh-1
  r such as 1 &lt; r &lt; n
  and gcd r, Ø=1 and gcd r,ϒ=1 (r should be small integer)
  e such as r.e = 1 mod Ø(n) and 1&lt; e &lt; Ø(n)
  d such as d.e= 1mod ϒn and 1 &lt; d &lt; ϒn
  Encryption Key = {e,n}… public key
  Decryption Key = {r,d,n}… private key
The authors of the paper might be able to tell you; I suspect no one else will.

You are having problems understanding the paper; a large part of the reason is that the paper isn't very well written; it introduces terminology (such as "ph") without ever defining it, and includes things that don't make any sense, such as the first line of their "Encryption Process": "If M=3, such that M$&lt;$n ").

This might be an indication that they are not used to writing papers (along with their lack of fluency in English), or it might be an indication that they really don't know what they're talking about.

My best guess about their proposal (from making "corrections" in what they wrote, and in a way that's looks to be secure) would be:


Select primes $p$ and $q$ and a small value $r$
Compute $e = r^{-1} \bmod pq$ and $d = e^{-1} \bmod (p-1)(q-1)$ (there's no indication what to do if $e$ is not relatively prime to $(p-1)(q-1)$; perhaps pick a different $r$?)
To encrypt a message $M$, compute $C = eM^e \bmod pq$
To decrypt a message $C$, compute $M = (rC)^d \bmod pq$


This would work and also be secure (as it can be easily reduced to RSA); however it also begs the question "what advantages does it have over standard RSA?".  After all, they use a large public exponent $e$ (remember, it's $r$ that is small); and so it would appear they're making the public operation more expensive; what benefit are they gaining for that expense (perhaps some tolerance to side channel attacks?  If so, they would need to spell out the side-channel resistant implementation).