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A365771
a(n) = binomial(2*n+1, n)/(2*n+1) * binomial(3*n-1, n) for n >= 0.
2
1, 2, 20, 280, 4620, 84084, 1633632, 33256080, 701149020, 15191562100, 336424047960, 7584833081280, 173575987821600, 4022766574898400, 94247674040476800, 2228957491057276320, 53150802525726081660, 1276661433215969608500, 30863850087221160009000
OFFSET
0,2
COMMENTS
Equals the central terms of triangle A365770.
Conjectures: given A033042 is the sums of distinct powers of 5, then
(1) a(5*A033042(n)) == 4 (mod 5) for n > 0,
(2) a(5*A033042(n) + 1) == 2 (mod 5) for n > 0,
(3) a(n) == 0 (mod 5) for n > 0 except when n or n-1 equals 5*A033042(k) for some k >= 0.
LINKS
FORMULA
a(n) = A365770(2*n,n) for n >= 0.
a(n) = A000108(n) * A165817(n) for n >= 0.
MATHEMATICA
A365771[n_] := Binomial[2*n + 1, n]/(2*n + 1)*Binomial[3*n - 1, n];
Array[A365771, 20, 0] (* Paolo Xausa, Oct 12 2024 *)
PROG
(PARI) {a(n) = binomial(2*n+1, n)/(2*n+1) * binomial(3*n-1, n)}
for(n=0, 30, print1(a(n), ", "))
(Python)
from math import comb
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
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 10 2023
STATUS
approved