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A084769 a(n) = P_n(9), where P_n is n-th Legendre polynomial; also, a(n) = central coefficient of (1 + 9*x + 20*x^2)^n.

%I A084769 #81 Aug 30 2025 10:13:05
%S A084769 1,9,121,1809,28401,458649,7544041,125700129,2114588641,35836273449,
%T A084769 610897146201,10463745263409,179939616743121,3104680678772409,
%U A084769 53721299280288201,931852905510160449,16198821321758152641
%N A084769 a(n) = P_n(9), where P_n is n-th Legendre polynomial; also, a(n) = central coefficient of (1 + 9*x + 20*x^2)^n.
%H A084769 Vincenzo Librandi, <a href="/A084769/b084769.txt">Table of n, a(n) for n = 0..200</a>
%H A084769 Hacène Belbachir and Abdelghani Mehdaoui, <a href="https://doi.org/10.2989/16073606.2020.1729269">Recurrence relation associated with the sums of square binomial coefficients</a>, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
%H A084769 Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Szalay/szalay42.html">Diagonal Sums in the Pascal Pyramid, II: Applications</a>, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
%H A084769 Vaclav Kotesovec, <a href="http://www.kotesovec.cz/math_articles/kotesovec_binomial_asymptotics.pdf">Asymptotic of a sums of powers of binomial coefficients * x^k</a>, 2012.
%H A084769 Tony D. Noe, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Noe/noe35.html">On the Divisibility of Generalized Central Trinomial Coefficients</a>, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.
%F A084769 G.f.: 1/sqrt(1-18*x+x^2).
%F A084769 Also a(n) = (n+1)-th term of the binomial transform of 1/(1-4x)^(n+1).
%F A084769 From _Paul Barry_, May 25 2005: (Start)
%F A084769 E.g.f.: exp(9*x) * Bessel_I(0, 2*sqrt(20)*x).
%F A084769 a(n) = Sum_{k=0..n} C(n, k)*C(n+k, k)4^k. (End)
%F A084769 D-finite with recurrence: n*a(n) + 9*(1-2*n)*a(n-1) + (n-1)*a(n-2) = 0. - _R. J. Mathar_, Sep 27 2012
%F A084769 a(n) = Sum_{k=0..n} binomial(n,k)^2 * 4^k * 5^(n-k). - _Paul D. Hanna_, Sep 29 2012
%F A084769 a(n) ~ sqrt(200 + 90*sqrt(5))*(9 + 4*sqrt(5))^n/(20*sqrt(Pi*n)) = (2 + sqrt(5))^(2*n+1)/(5^(1/4)*2*sqrt(2*Pi*n)). - _Vaclav Kotesovec_, Oct 14 2012
%F A084769 a(n) = hypergeom([-n, n+1], [1], -4). - _Peter Luschny_, May 23 2014
%F A084769 x*exp(Sum_{n >= 1} a(n)*x^n/n) = x + 9*x^2 + 101*x^3 + 1269*x^4 + ... is an integral power series, the o.g.f. for A269732. - _Peter Bala_, Jan 25 2018
%F A084769 a(n) = (-1)^n * Sum_{k=0..n} (1/18)^(n-2*k) * binomial(-1/2,k) * binomial(k,n-k). - _Seiichi Manyama_, Aug 28 2025
%F A084769 a(n) = Sum_{k=0..floor(n/2)} 20^k * 9^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). - _Seiichi Manyama_, Aug 30 2025
%t A084769 Table[SeriesCoefficient[1/Sqrt[1-18*x+x^2],{x,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 14 2012 *)
%t A084769 LegendreP[Range[0, 40], 9] (* _G. C. Greubel_, May 17 2023 *)
%t A084769 a[n_] := 4^n*Sum[(5/4)^k*Binomial[n, k]^2, {k, 0, n}];
%t A084769 Table[a[n], {n, 0, 16}]  (* _Detlef Meya_, May 22 2024 *)
%o A084769 (PARI) for(n=0,30,print1(subst(pollegendre(n),x,9)","))
%o A084769 (PARI) {a(n)=sum(k=0, n, binomial(n, k)^2*4^k*5^(n-k))} \\ _Paul D. Hanna_, Sep 29 2012
%o A084769 (Magma) [Evaluate(LegendrePolynomial(n), 9) : n in [0..40]]; // _G. C. Greubel_, May 17 2023
%o A084769 (SageMath) [gen_legendre_P(n,0,9) for n in range(41)] # _G. C. Greubel_, May 17 2023
%Y A084769 Column k=4 of A335333.
%Y A084769 Sequences of the form LegendreP(n, 2*m+1): A000012 (m=0), A001850 (m=1), A006442 (m=2), A084768 (m=3), this sequence (m=4).
%Y A084769 Cf. A243007 (a(n)^2), A269732.
%K A084769 nonn,easy,changed
%O A084769 0,2
%A A084769 _Paul D. Hanna_, Jun 03 2003