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A352654 a(0) = 0, a(n) = [x^n] P(n-1,(1 + x)/(1 - x)) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial.

%I A352654 #31 Dec 26 2024 04:22:37
%S A352654 0,0,2,18,162,1500,14240,137886,1356194,13507416,135916002,1379301990,
%T A352654 14097919968,144977296932,1498750896708,15564971674518,
%U A352654 162298598439330,1698353774374704,17828728267167326,187693442844729174,1981038544180652162,20957881615473229620
%N A352654 a(0) = 0, a(n) = [x^n] P(n-1,(1 + x)/(1 - x)) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial.
%C A352654 Compare with A103882(n) = [x^n] P(n,(1 + x)/(1 - x)).
%C A352654 Using binomial(-n,k) = (-1)^k*binomial(n+k-1,k), valid for nonnegative k, we can extend the binomial sum a(n) = Sum_{k} binomial(n-1,k)*binomial(n-1,k-1)*binomial(n+k-1,k) to negative values of n to find a(-n) = Sum_{k} binomial(-n-1,k)*binomial(-n-1,k-1)* binomial(-n+k-1,k) = Sum_{k} (-1)^(k-1)*binomial(n+k,k)*binomial(n+k-1,k-1)*binomial(n,k) = (-1)^(n+1)*A103882(n) (n != 0).
%F A352654 a(n) = Sum_{k = 1..n-1} binomial(n-1,k)*binomial(n-1,k-1)*binomial(n+k-1,k).
%F A352654 a(n) = Sum_{k = 1..n-1} (-1)^(n+k-1)*binomial(n-1,k)*binomial(n+k-1,k)* binomial(n+k-1,k-1).
%F A352654 a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n,k)*binomial(n+k-1,k)^2 for n >= 1.
%F A352654 a(n) = (1/5)*(A005258(n) - 3*A005258(n-1)) for n >= 1. It follows that the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) hold for primes p >= 5 and positive integers n and k.
%F A352654 a(n) = A108628(n-1) - A208675(n) for n >= 1.
%F A352654 a(n) ~ (1/10)*5^(3/4)*(sqrt(5) + 1)/(Pi*sqrt(22 + 10*sqrt(5))*n)*((1/2)*(11 + 5*sqrt(5)))^n.
%F A352654 n^2*(5*n - 7)*(n - 2)*a(n) = (n - 1)*(55*n^3 - 187*n^2 + 190*n - 48)*a(n-1) + (n - 2)^2*(5*n -2 )*(n - 1)*a(n-2) with a(1) = 0 and a(2) = 2.
%F A352654 Conjecture: a(n) = [(x*y)^(n-1)*z^n] 1/(1 - x - x*y - y*z - x*z - x*y*z) = [(x*z)^(n-1)*y^n] 1/(1 - x - x*y - y*z - x*z - x*y*z) for n >= 1.
%F A352654 a(n) = n*(n-1)*hypergeom([2-n, 1-n, 1+n], [2, 2], 1). - _Peter Luschny_, Jan 03 2024
%p A352654 seq(add(binomial(n-1,k)*binomial(n-1,k-1)*binomial(n+k-1,k), k = 1..n-1), n = 0..20);
%o A352654 (PARI) a(n) = polcoef(pollegendre(n-1, (1 + x)/(1 - x)) + O(x^(n+1)), n); \\ _Michel Marcus_, Apr 17 2022
%o A352654 (Python)
%o A352654 def A352654(n):
%o A352654     if n == 0: return 0
%o A352654     m, g = 1, 0
%o A352654     for k in range(n+1):
%o A352654         g += -m*n//(n+k) if (n+k)&1 else m*n//(n+k)
%o A352654         m *= (n+k+1)*(n-k)*(n+k)
%o A352654         m //= (k+1)**3
%o A352654     return g # _Chai Wah Wu_, Oct 03 2022
%o A352654 (SageMath)
%o A352654 def a(n): return n * (n-1) * hypergeometric([2-n, 1-n, 1+n], [2, 2], 1)
%o A352654 print([simplify(a(n)) for n in range(22)]) # _Peter Luschny_, Jan 03 2024
%Y A352654 Cf. A005258, A103882, A108628, A208675.
%K A352654 nonn,easy
%O A352654 0,3
%A A352654 _Peter Bala_, Apr 14 2022