A098455 -id:A098455 - cp's OEIS frontend

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

1, 2, 14, 68, 406, 2332, 13964, 83848, 509926, 3118892, 19194724, 118654648, 736365436, 4584612632, 28623792344, 179142212368, 1123532958086, 7059622447052, 44431918660724, 280059644507608, 1767597777222676

Offset: 0

Paul D. Hanna, Jun 10 2003

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

			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 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
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.

Column k=2 of A387466.

Cf. A000984, A006139, A098455, A098456.

[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

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 *)

for(n=0,30,t=polcoeff((1+2*x+5*x^2)^n,n,x); print1(t","))

a(n) = 4^n*abs(pollegendre(n, I/2)) \\ after 2nd Mathematica; Michel Marcus, Oct 22 2015

{a(n) = (4/I)^n * pollegendre(n, I/2)}; /* Michael Somos, Sep 30 2017 */

[(-4*i)^n*gen_legendre_P(n, 0, i/2) for n in range(41)] # G. C. Greubel, May 30 2023

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
From Seiichi Manyama, Aug 30 2025: (Start)
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.
a(n) = Sum_{k=0..floor(n/2)} 5^k * 2^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). (End)

A387466 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 4x - (2k*x)^2).

Original entry on oeis.org

1, 1, 2, 1, 2, 6, 1, 2, 8, 20, 1, 2, 14, 32, 70, 1, 2, 24, 68, 136, 252, 1, 2, 38, 128, 406, 592, 924, 1, 2, 56, 212, 1096, 2332, 2624, 3432, 1, 2, 78, 320, 2566, 7632, 13964, 11776, 12870, 1, 2, 104, 452, 5320, 20092, 60864, 83848, 53344, 48620

Offset: 0

Seiichi Manyama, Aug 29 2025

			Square array begins:
    1,    1,     1,     1,      1,      1,       1, ...
    2,    2,     2,     2,      2,      2,       2, ...
    6,    8,    14,    24,     38,     56,      78, ...
   20,   32,    68,   128,    212,    320,     452, ...
   70,  136,   406,  1096,   2566,   5320,   10006, ...
  252,  592,  2332,  7632,  20092,  44752,   88092, ...
  924, 2624, 13964, 60864, 210524, 607424, 1523724, ...

Columns k=0..4 give A000984, A006139, A084770, A098455, A098456.
Main diagonal gives A387467.
Cf. A386621.

a(n, k) = sum(j=0, n\2, (k^2+1)^j*2^(n-2*j)*binomial(n, 2*j)*binomial(2*j, j));

A(n,k) = Sum_{j=0..n} (1-k*i)^j * (1+k*i)^(n-j) * binomial(n,j)^2, where i is the imaginary unit.
A(n,k) = Sum_{j=0..floor(n/2)} k^(2*j) * binomial(2*(n-j),n-j) * binomial(n-j,j).
n*A(n,k) = 2*(2*n-1)*A(n-1,k) + 4*k^2*(n-1)*A(n-2,k) for n > 1.
A(n,k) = Sum_{j=0..floor(n/2)} (k^2+1)^j * 2^(n-2*j) * binomial(n,2*j) * binomial(2*j,j).
A(n,k) = [x^n] (1 + 2*x + (k^2+1)*x^2)^n.
E.g.f. of column k: exp(2*x) * BesselI(0, 2*sqrt(k^2+1)*x).

A098456 Expansion of 1/sqrt(1-4x-64x^2).

Original entry on oeis.org

1, 2, 38, 212, 2566, 20092, 210524, 1884136, 18854854, 178415852, 1764019828, 17115907096, 169100140444, 1661540282456, 16458178007288, 162887627833552, 1618680238292294, 16095872154638156, 160435286115927044, 1600771362880092472, 15997473711080724916

Offset: 0

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)}. Then a(n)=4^n*Q(n,1/4). Central coefficients of (1+2*x+17*x^2)^n.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Hacène Belbachir, Abdelghani Mehdaoui, László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.

Cf. A084770, A098455.

CoefficientList[Series[1/Sqrt[1-4x-64x^2],{x,0,30}],x] (* Harvey P. Dale, Dec 27 2011 *)

x='x+O('x^66); Vec(1/sqrt(1-4*x-64*x^2)) \\ Joerg Arndt, May 11 2013

E.g.f.: exp(2x) * BesselI(0, 2*sqrt(17)*x).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(2(n-k), n)*16^k.
D-finite with recurrence: n*a(n) +2*(1-2*n)*a(n-1) +64*(1-n)*a(n-2)=0. - R. J. Mathar, Sep 26 2012
a(n) ~ sqrt(578+34*sqrt(17))*(2+2*sqrt(17))^n/(34*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 15 2012
From Seiichi Manyama, Aug 30 2025: (Start)
a(n) = Sum_{k=0..n} (1-4*i)^k * (1+4*i)^(n-k) * binomial(n,k)^2, where i is the imaginary unit.
a(n) = Sum_{k=0..floor(n/2)} 17^k * 2^(n-2*k) * binomial(n,2*k) * binomial(2*k,k). (End)

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

PARI

SageMath

Formula

A387466 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 4x - (2k*x)^2).

Views

Author

Keywords

Examples

Crossrefs

Programs

PARI

Formula

A098456 Expansion of 1/sqrt(1-4x-64x^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

PARI

SageMath

Formula

A387466 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 4*x - (2*k*x)^2).

Views

Author

Keywords

Examples

Crossrefs

Programs

PARI

Formula

A098456 Expansion of 1/sqrt(1-4*x-64*x^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A387466 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 4x - (2k*x)^2).

A098456 Expansion of 1/sqrt(1-4x-64x^2).