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A336257 a(n) = Catalan(n) mod (2*n+1).

%I A336257 #19 Jul 16 2020 18:59:11
%S A336257 0,1,2,5,5,9,2,9,2,17,17,21,12,22,2,29,18,30,2,30,2,41,30,45,9,21,2,
%T A336257 54,53,57,2,28,38,65,42,69,2,64,70,77,5,81,80,33,2,14,27,45,2,36,2,
%U A336257 101,87,105,2,78,2,34,75,6,101,45,62,125,39,129,74,60,2,137,90
%N A336257 a(n) = Catalan(n) mod (2*n+1).
%H A336257 Seiichi Manyama, <a href="/A336257/b336257.txt">Table of n, a(n) for n = 0..10000</a>
%H A336257 Arturo Merino, Ondrej Micka, and Torsten Mütze, <a href="https://arxiv.org/abs/2007.07164">On a combinatorial generation problem of Knuth</a>, arXiv:2007.07164 [math.CO], 2020. See p. 43.
%F A336257 a(n) = 2 for n in A104636.
%F A336257 a(n) = 2*n-1 for n in A104635.
%p A336257 a:= n-> binomial(2*n, n)/(n+1) mod (2*n+1):
%p A336257 seq(a(n), n=0..80);  # _Alois P. Heinz_, Jul 16 2020
%o A336257 (PARI) C(n)=binomial(2*n, n)/(n+1);
%o A336257 a(n) = C(n) % (2*n+1);
%o A336257 (Python)
%o A336257 A336257_list, c = [0,1], 1
%o A336257 for n in range(2,10001):
%o A336257     c = c*(4*n-2)//(n+1)
%o A336257     A336257_list.append(c % (2*n+1)) # _Chai Wah Wu_, Jul 16 2020
%Y A336257 Cf. A000108, A059288, A067770, A246714.
%Y A336257 Cf. A104635, A104636.
%K A336257 nonn,look
%O A336257 0,3
%A A336257 _Michel Marcus_, Jul 15 2020