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 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). 0
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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. LINKS EXAMPLE 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. PROG #(Sage) 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 CROSSREFS Cf. A051010, A072030, A278744. Sequence in context: A097884 A094818 A114233 * A063086 A077636 A215847 Adjacent sequences:  A279044 A279045 A279046 * A279048 A279049 A279050 KEYWORD nonn AUTHOR Adnan Baysal, Dec 04 2016 STATUS approved

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Last modified May 25 01:17 EDT 2019. Contains 323534 sequences. (Running on oeis4.)