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A365771 a(n) = binomial(2*n+1, n)/(2*n+1) * binomial(3*n-1, n) for n >= 0.

%I A365771 #19 Aug 25 2025 11:42:14
%S A365771 1,2,20,280,4620,84084,1633632,33256080,701149020,15191562100,
%T A365771 336424047960,7584833081280,173575987821600,4022766574898400,
%U A365771 94247674040476800,2228957491057276320,53150802525726081660,1276661433215969608500,30863850087221160009000
%N A365771 a(n) = binomial(2*n+1, n)/(2*n+1) * binomial(3*n-1, n) for n >= 0.
%C A365771 Equals the central terms of triangle A365770.
%C A365771 Conjectures: given A033042 is the sums of distinct powers of 5, then
%C A365771 (1) a(5*A033042(n)) == 4 (mod 5) for n > 0,
%C A365771 (2) a(5*A033042(n) + 1) == 2 (mod 5) for n > 0,
%C A365771 (3) a(n) == 0 (mod 5) for n > 0 except when n or n-1 equals 5*A033042(k) for some k >= 0.
%H A365771 Paolo Xausa, <a href="/A365771/b365771.txt">Table of n, a(n) for n = 0..500</a>
%F A365771 a(n) = A365770(2*n,n) for n >= 0.
%F A365771 a(n) = A000108(n) * A165817(n) for n >= 0.
%F A365771 a(n) = 2*A319578(n) = (2/3) * A007004(n) for n >= 1. - _Peter Bala_, Aug 25 2025
%t A365771 A365771[n_] := Binomial[2*n + 1, n]/(2*n + 1)*Binomial[3*n - 1, n];
%t A365771 Array[A365771, 20, 0] (* _Paolo Xausa_, Oct 12 2024 *)
%o A365771 (PARI) {a(n) = binomial(2*n+1, n)/(2*n+1) * binomial(3*n-1, n)}
%o A365771 for(n=0,30,print1(a(n),", "))
%o A365771 (Python)
%o A365771 from math import comb
%o A365771 def A365771(n): return comb(m:=(n<<1)+1,n)*comb(m+n-2,n)//m if n else 1 # _Chai Wah Wu_, Oct 11 2023
%Y A365771 Cf. A007004, A000108, A165817, A033042, A319578, A356770.
%K A365771 nonn,easy,changed
%O A365771 0,2
%A A365771 _Paul D. Hanna_, Oct 10 2023