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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. 7
 1, 2, 14, 68, 406, 2332, 13964, 83848, 509926, 3118892, 19194724, 118654648, 736365436, 4584612632, 28623792344, 179142212368, 1123532958086, 7059622447052, 44431918660724, 280059644507608, 1767597777222676 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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 Self-convolution of a(n)/4^n gives Fibonacci numbers A000045(n+1). - Vladimir Reshetnikov, Oct 09 2016 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 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..200 Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5. Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7. Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1. FORMULA 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 a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(2(n-k), n)*4^k. - Paul Barry, Sep 08 2004 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 Recurrence: n*a(n) = 2*(2*n-1)*a(n-1) + 16*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 14 2012 a(n) ~ sqrt(50+10*sqrt(5))*(2+2*sqrt(5))^n/(10*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012 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 a(n) = 2^n * hypergeom([(1-n)/2, -n/2], [1], 5). - Vladimir Reshetnikov, Oct 10 2016 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 EXAMPLE 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. G.f. = 1 + 2*x + 14*x^2 + 68*x^3 + 406*x^4 + 2332*x^5 + 13964*x^6 + 83848*x^7 + ... MATHEMATICA Table[n!*SeriesCoefficient[E^(2*x)*BesselI[0, 2*Sqrt[5]*x], {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 14 2012 *) Table[Abs[LegendreP[n, I/2]] 4^n, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 22 2015 *) a[n_]:= (4/I)^n LegendreP[n, I/2]; (* Michael Somos, Sep 30 2017 *) PROG (PARI) for(n=0, 30, t=polcoeff((1+2*x+5*x^2)^n, n, x); print1(t", ")) (PARI) a(n) = 4^n*abs(pollegendre(n, I/2)) \\ after 2nd Mathematica; Michel Marcus, Oct 22 2015 (PARI) {a(n) = (4/I)^n * pollegendre(n, I/2)}; /* Michael Somos, Sep 30 2017 */ (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 (SageMath) [(-4*i)^n*gen_legendre_P(n, 0, i/2) for n in range(41)] # G. C. Greubel, May 30 2023 CROSSREFS Sequence in context: A325925 A084132 A271235 * A086243 A258138 A206947 Adjacent sequences: A084767 A084768 A084769 * A084771 A084772 A084773 KEYWORD nonn AUTHOR Paul D. Hanna, Jun 10 2003 STATUS approved

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Last modified September 13 04:25 EDT 2024. Contains 375859 sequences. (Running on oeis4.)