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%I #17 Dec 07 2016 11:15:26
%S 1,2,2,4,2,2,2,4,4,2,2,2,2,4,5,6,2,2,4,2,2,4,4,2,2,2,4,2,2,2,4,4,2,5,
%T 2,2,2,4,5,6,2,2,2,2,2,3,4,4,2,2,6,2,2,4,5,4,2,2,2,2,4,5,4,2,2,5,2,6,
%U 2,2,6,4,2,2,4,4,4,2,2,7,2,2,2,2,4,4,2,2,2,4,8,5,4,3,4,4,3,2,2,2,4,5,4,2,2,6,5,6
%N Number k of modular reductions at which the recurrence relation x(i+1) = x(0) mod x(i) terminates with x(k) = 1, where x(0) = prime(n+1), x(1) = prime(n).
%C x(i) is a strictly decreasing sequence of nonnegative integers by definition of modular reduction. So at some point x(t) = 0. Let j be the previous positive value, i.e., x(t-1) = j. Then as x(0) mod j = prime(n+1) mod j = x(t) = 0, j|prime(n+1). Since j < prime(n+1), j = 1.
%e For n=4, x(0) = p(5) = 11, x(1) = p(4) = 7. 11 mod 7 = 4 ==> 11 mod 4 = 3 ==> 11 mod 3 = 2 ==> 11 mod 2 = 1. Since there are four modular reductions, a(4) = 4.
%o #(Sage)
%o A = []
%o q = 1
%o for i in range(100):
%o q = next_prime(q)
%o p = next_prime(q)
%o r = p%q
%o ctr = 1
%o while r!=1:
%o r = p%r
%o ctr += 1
%o A.append(ctr)
%o print A
%Y Cf. A051010, A072030, A278744.
%K nonn
%O 1,2
%A _Adnan Baysal_, Dec 04 2016
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