This site is supported by donations to The OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A101786 G.f. satisfies: A(x) = 1 + x*A(x)/(1 - 2*x^2*A(x)^2). 1
 1, 1, 1, 3, 9, 25, 77, 247, 801, 2657, 8969, 30635, 105785, 368745, 1295493, 4582767, 16309953, 58357313, 209798289, 757461011, 2745281705, 9984464761, 36428252541, 133293594343, 489028250465, 1798543861537, 6629635284505 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS Formula may be derived using the Lagrange Inversion theorem (cf. A049124). LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 FORMULA a(n) = Sum_{k=0..[(n-1)/2]} C(n-k-1, k)*C(n, 2*k)*2^k/(2*k+1) for n>0, with a(0)=1. G.f.: A(x) = (1/x)*Series_Reversion( x*(1 - 2*x^2)/(1+x - 2*x^2) ). Recurrence: 2*n*(n+1)*(31*n^2 - 127*n + 120)*a(n) = 3*n*(62*n^3 - 285*n^2 + 359*n - 88)*a(n-1) + (62*n^4 - 378*n^3 + 1009*n^2 - 1425*n + 792)*a(n-2) + (n-3)*(682*n^3 - 3135*n^2 + 4133*n - 1272)*a(n-3) - 9*(n-4)*(n-3)*(31*n^2 - 65*n + 24)*a(n-4). - Vaclav Kotesovec, Sep 17 2013 a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 3/4 + 1/(4*sqrt(3/(35 - (176*2^(2/3))/(9959 + 465*sqrt(465))^(1/3) + 2*(19918 + 930*sqrt(465))^(1/3)))) + 1/2*sqrt(35/6 + (44*2^(2/3))/(3*(9959 + 465*sqrt(465))^(1/3)) - (9959 + 465*sqrt(465))^(1/3)/(3*2^(2/3)) + 127/2*sqrt(3/(35 - (176*2^(2/3))/ (9959 + 465*sqrt(465))^(1/3) + 2*(19918 + 930*sqrt(465))^(1/3)))) = 3.9027270552404829297969 = ... is the root of the equation 9 - 22*d - 2*d^2 - 6*d^3 + 2*d^4 = 0 and c = 0.68546565145612597016100560323891887595749... - Vaclav Kotesovec, Sep 17 2013 EXAMPLE Generated from Fibonacci polynomials (A011973) and coefficients of odd powers of 1/(1-x): a(1) = 1*1/1 a(2) = 1*1/1 + 0*1*2/3 a(3) = 1*1/1 + 1*3*2/3 a(4) = 1*1/1 + 2*6*2/3 + 0*1*2^2/5 a(5) = 1*1/1 + 3*10*2/3 + 1*5*2^2/5 a(6) = 1*1/1 + 4*15*2/3 + 3*15*2^2/5 + 0*1*2^3/7 a(7) = 1*1/1 + 5*21*2/3 + 6*35*2^2/5 + 1*7*2^3/7 a(8) = 1*1/1 + 6*28*2/3 + 10*70*2^2/5 + 4*28*2^3/7 + 0*1*2^4/9 This process is equivalent to the formula: a(n) = Sum_{k=0..[(n-1)/2]} C(n-k-1,k)*C(n,2*k)*2^k/(2*k+1). MATHEMATICA Flatten[{1, Table[Sum[Binomial[n-k-1, k]*Binomial[n, 2*k]*2^k/(2*k+1), {k, 0, Floor[(n-1)/2]}], {n, 1, 20}]}] (* Vaclav Kotesovec, Sep 17 2013 *) ShiftedReversion[ser_, n_, sgn_] := CoefficientList[(sgn/x)InverseSeries[Series[x sgn ser, {x, 0, n}]], x]; Jacobsthal := (2x^2 - 1)/((x + 1)(2x - 1)); (* with A001045(0) = 1 *) ShiftedReversion[Jacobsthal, 27, -1] (* Peter Luschny, Jan 10 2019 *) PROG (PARI) {a(n)=if(n==0, 1, sum(k=0, (n-1)\2, binomial(n-k-1, k)*binomial(n, 2*k)*2^k/ (2*k+1)))} CROSSREFS Cf. A011973, A001045, A049124, A101785. Sequence in context: A001189 A212352 A198180 * A217995 A246653 A192371 Adjacent sequences:  A101783 A101784 A101785 * A101787 A101788 A101789 KEYWORD nonn AUTHOR Paul D. Hanna, Dec 16 2004 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified August 20 20:50 EDT 2019. Contains 326155 sequences. (Running on oeis4.)