cp's OEIS Frontend

A279047 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).

%I A279047 #21 Jun 11 2024 15:45:57
%S A279047 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 A279047 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 A279047 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 A279047 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 A279047 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 A279047 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 A279047 (SageMath)
%o A279047 A = []
%o A279047 q = 1
%o A279047 for i in range(100):
%o A279047     q = next_prime(q)
%o A279047     p = next_prime(q)
%o A279047     r = p%q
%o A279047     ctr = 1
%o A279047     while r!=1:
%o A279047         r = p%r
%o A279047         ctr += 1
%o A279047     A.append(ctr)
%o A279047 print(A)
%Y A279047 Cf. A051010, A072030, A278744.
%K A279047 nonn
%O A279047 1,2
%A A279047 _Adnan Baysal_, Dec 04 2016