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A098341
Expansion of 1/sqrt(1 - 6*x + 25*x^2).
3
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
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