A098341 Expansion of 1/sqrt(1 - 6*x + 25*x^2).

1, 3, 1, -45, -255, -477, 2689, 25203, 82945, -90045, -2379519, -11581677, -12063999, 197669475, 1423716225, 3645266355, -12180238335, -156702949245, -626511576575, 51239061075, 15179398450945, 87687927568035, 151934475887745

Offset

0,2

Comments

Central coefficients of (1 + 3*x - 4*x^2)^n.

(-1)^n*a(n) is the sum of squares of coefficients of (1+2*i*x)^n where i=sqrt(-1) (see PARI code). - Joerg Arndt, Jul 06 2011

Binomial transform of A098337.

Second binomial transform of A098334.

Links

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.

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.

Formula

E.g.f.: exp(3*x)*BesselI(0, 4*I*x), I=sqrt(-1).

a(n) = (-1)^n*Sum_{k=0..n} binomial(n, k)^2*(-4)^k.

a(n) = (-1)^n*hypergeometric([-n,-n], [1], -4). - Peter Luschny, Sep 23 2014

D-finite with recurrence: n*a(n) +3*(-2*n+1)*a(n-1) +25*(n-1)*a(n-2)=0. - R. J. Mathar, Nov 27 2014

From Peter Bala, Nov 28 2021: (Start)

a(n) = (5^n)*P(n,3/5), where P(n,x) is the n-th Legendre polynomial.

a(n) = [x^n] ((1 - x)*(1 + 4*x))^n.

a(n) = 5^(2*n+1)*Sum_{k >= n} (-1)^(n+k)*binomial(k,n)^2*(1/4)^(k+1).

a(n) = (5/4)*(25/4)^n*hypergeom([n+1, n+1], [1], -1/4). (End)

Mathematica

Table[(-5)^n*LegendreP[n, -3/5], {n, 0, 20}] (* Vaclav Kotesovec, Jul 23 2013 *)
CoefficientList[Series[1/Sqrt[1-6x+25x^2], {x, 0, 30}], x] (* Harvey P. Dale, Aug 22 2014 *)

Prog

(PARI) a(n)={local(v=Vec((1+2*I*x)^n)); (-1)^n*sum(k=1, #v, v[k]^2); } /* Joerg Arndt, Jul 06 2011 */
(PARI) a(n)={local(v=Vec((1+2*I*x)^n)); sum(k=1, #v, real(v[k])^2-imag(v[k])^2); } /* Joerg Arndt, Jul 06 2011 */
(Sage)
A098341 = lambda n: (-1)^n*hypergeometric([-n, -n], [1], -4)
[Integer(A098341(n).n(100)) for n in (0..22)] # Peter Luschny, Sep 23 2014

Crossrefs

Cf. A098332, A012000, A116091, A126869.

Sequence in context: A344379 A155812 A158958 * A354391 A223173 A010292

Adjacent sequences: A098338 A098339 A098340 * A098342 A098343 A098344

Keyword

easy,sign

Author

Paul Barry, Sep 03 2004

Status

approved