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A084770 Coefficients of 1/(1-4x-16x^2)^(1/2); also, a(n) is the central coefficient of (1+2x+5x^2)^n.

%I A084770 #63 Aug 30 2025 10:13:13
%S A084770 1,2,14,68,406,2332,13964,83848,509926,3118892,19194724,118654648,
%T A084770 736365436,4584612632,28623792344,179142212368,1123532958086,
%U A084770 7059622447052,44431918660724,280059644507608,1767597777222676
%N A084770 Coefficients of 1/(1-4x-16x^2)^(1/2); also, a(n) is the central coefficient of (1+2x+5x^2)^n.
%C A084770 Also number of paths from (0,0) to (n,0) using steps U=(1,1), H=(1,0) and D=(1,-1), U can have 5 colors and H can have 2 colors. - _N-E. Fahssi_, Mar 30 2008
%C A084770 Self-convolution of a(n)/4^n gives Fibonacci numbers A000045(n+1). - _Vladimir Reshetnikov_, Oct 09 2016
%C A084770 Let A(x) be the g.f. and f(x) := x * A(-x/4) = x / sqrt(1 + x - x^2) = x - x^2*1/2 + x^3*7/8 - ..., then f() maps the unit interval to itself monotonically with 0 an attractive fixed point. Let b(n, t) := 1/(n/2 + (t-c_0) - 5/4*log(n + 2*(t-c_1) - 5/2*log(n + 2*(t-c_2) - 5/2*log(n + 2*t ...)))), where c_0=0, c_1=1, c_2=121/120, ..., then b(n+1, t) = f(b(n, t)). - _Michael Somos_, Sep 30 2017
%H A084770 Vincenzo Librandi, <a href="/A084770/b084770.txt">Table of n, a(n) for n = 0..200</a>
%H A084770 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 A084770 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.
%H A084770 Michael Z. Spivey and Laura L. Steil, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Spivey/spivey7.html">The k-Binomial Transforms and the Hankel Transform</a>, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
%F A084770 E.g.f.: exp(2*x) * BesselI(0, 2*sqrt(5)*x). More generally, e.g.f.: exp(b*x) * BesselI(0, 2*sqrt(c)*x) yields central coefficients of (1+b*x+c*x^2)^n. - _Vladeta Jovovic_, Mar 21 2004
%F A084770 a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(2(n-k), n)*4^k. - _Paul Barry_, Sep 08 2004
%F A084770 Define Q(n, x) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(2(n-k), n)*x^(n-2k). A084770(n) is 2^n*Q(n, 1/2). - _Paul Barry_, Sep 08 2004
%F A084770 Recurrence: n*a(n) = 2*(2*n-1)*a(n-1) + 16*(n-1)*a(n-2). - _Vaclav Kotesovec_, Oct 14 2012
%F A084770 a(n) ~ sqrt(50+10*sqrt(5))*(2+2*sqrt(5))^n/(10*sqrt(Pi*n)). - _Vaclav Kotesovec_, Oct 14 2012
%F A084770 G.f.: G(0), where G(k)= 1 + 4*x*(1+4*x)*(4*k+1)/(4*k+2 - 4*x*(1+4*x)*(4*k+2)*(4*k+3)/(4*x*(1+4*x)*(4*k+3) + 4*(k+1)/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Jun 17 2013
%F A084770 a(n) = 2^n * hypergeom([(1-n)/2, -n/2], [1], 5). - _Vladimir Reshetnikov_, Oct 10 2016
%F A084770 a(n) = (4/i)^(2*n+1) * a(-1-n), and 0 = a(n)*(+256*a(n+1) + 96*a(n+2) - 32*a(n+3)) + a(n+1)*(+32*a(n+1) + 16*a(n+2) - 6*a(n+3)) + a(n+2)*(-2*a(n+2) + a(n+3)) for all n in Z. - _Michael Somos_, Sep 30 2017
%F A084770 From _Seiichi Manyama_, Aug 30 2025: (Start)
%F A084770 a(n) = Sum_{k=0..n} (1-2*i)^k * (1+2*i)^(n-k) * binomial(n,k)^2, where i is the imaginary unit.
%F A084770 a(n) = Sum_{k=0..floor(n/2)} 5^k * 2^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). (End)
%e A084770 G.f.: 1/sqrt(1-2*b*x+(b^2-4*c)*x^2) yields central coefficients of (1+b*x+c*x^2)^n.
%e A084770 G.f. = 1 + 2*x + 14*x^2 + 68*x^3 + 406*x^4 + 2332*x^5 + 13964*x^6 + 83848*x^7 + ...
%t A084770 Table[n!*SeriesCoefficient[E^(2*x)*BesselI[0,2*Sqrt[5]*x],{x,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 14 2012 *)
%t A084770 Table[Abs[LegendreP[n, I/2]] 4^n, {n, 0, 20}] (* _Vladimir Reshetnikov_, Oct 22 2015 *)
%t A084770 a[n_]:= (4/I)^n LegendreP[n, I/2]; (* _Michael Somos_, Sep 30 2017 *)
%o A084770 (PARI) for(n=0,30,t=polcoeff((1+2*x+5*x^2)^n,n,x); print1(t","))
%o A084770 (PARI) a(n) = 4^n*abs(pollegendre(n, I/2)) \\ after 2nd Mathematica; _Michel Marcus_, Oct 22 2015
%o A084770 (PARI) {a(n) = (4/I)^n * pollegendre(n, I/2)}; /* _Michael Somos_, Sep 30 2017 */
%o A084770 (Magma) [n le 2 select 2^(n-1) else (2*(2*n-3)*Self(n-1) + 16*(n-2)*Self(n-2))/(n-1): n in [1..30]]; // _G. C. Greubel_, May 30 2023
%o A084770 (SageMath) [(-4*i)^n*gen_legendre_P(n, 0, i/2) for n in range(41)] # _G. C. Greubel_, May 30 2023
%Y A084770 Cf. A000984, A006139, A098455, A098456.
%K A084770 nonn,changed
%O A084770 0,2
%A A084770 _Paul D. Hanna_, Jun 10 2003