cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A101786 G.f. satisfies: A(x) = 1 + x*A(x)/(1 - 2*x^2*A(x)^2).

Original entry on oeis.org

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
Offset: 0

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Author

Paul D. Hanna, Dec 16 2004

Keywords

Comments

Formula may be derived using the Lagrange Inversion theorem (cf. A049124).

Examples

			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).
		

Crossrefs

Programs

  • 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 *)
  • 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)))}

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