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A288641 Define the sequence {b_n(k)} as the solutions of the recursion (k+1) * b_n(k+1) = b_n(k) * (b_n(k)^(n-1) + k) with b_n(0) = 1. a(n) is the least prime p where p * b_n(p) is not 0 mod p.

%I A288641 #36 Feb 16 2025 08:33:47
%S A288641 43,89,97,251,19,239,37,79,83,239,31,431,19,79,23,827,43,173,31,103,
%T A288641 179,73,19,431,193,101,53,811,47,1427,19,251,29,311,137,71,23,499,43,
%U A288641 47,19,419,31,191,83,337,59,1559,19,127,109,163,67,353,83,191,83,107
%N A288641 Define the sequence {b_n(k)} as the solutions of the recursion (k+1) * b_n(k+1) = b_n(k) * (b_n(k)^(n-1) + k) with b_n(0) = 1. a(n) is the least prime p where p * b_n(p) is not 0 mod p.
%C A288641 If A108394(n) is a prime, a(n) = A108394(n).
%H A288641 Seiichi Manyama, <a href="/A288641/b288641.txt">Table of n, a(n) for n = 2..1000</a>
%H A288641 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/GoebelsSequence.html">Goebel's Sequence</a>
%e A288641 (k+1) * b_2(k+1) = b_2(k) * (b_2(k) + k) with b_2(0) = 1.
%e A288641 b_2(1) == 2, b_2(2) == 3, b_2(3) == 5, ... , b_2(42) == 33 mod 43.
%e A288641 So 43 * b_2(43) == b_2(42) * (b_2(42) + 42) == 24 (> 0) mod 43.
%Y A288641 Cf. A003504 ({b_2(n+1)}), A005166 ({b_3(n)}), A005167 ({b_4(n)}), A108394, A288676.
%K A288641 nonn
%O A288641 2,1
%A A288641 _Seiichi Manyama_, Jun 13 2017