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
KEYWORD
easy,sign
AUTHOR
Paul Barry, Sep 03 2004
STATUS
approved