A278744 -id:A278744 - 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).

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, 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, 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

Offset: 1

Adnan Baysal, Dec 04 2016

nonn

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.

			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.

Cf. A051010, A072030, A278744.

A = []
q = 1
for i in range(100):
    q = next_prime(q)
    p = next_prime(q)
    r = p%q
    ctr = 1
    while r!=1:
        r = p%r
        ctr += 1
    A.append(ctr)
print(A)

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).

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