login
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A084605 G.f.: 1/(1-2x-15x^2)^(1/2); also, a(n) is the central coefficient of (1+x+4x^2)^n. 14

%I

%S 1,1,9,25,145,561,2841,12489,60705,281185,1353769,6418809,30917041,

%T 148331665,716698425,3462260265,16786700865,81464917185,396215601225,

%U 1929237099225,9408084660945,45928695279345,224476389327705

%N G.f.: 1/(1-2x-15x^2)^(1/2); also, a(n) is the central coefficient of (1+x+4x^2)^n.

%C Also number of paths from (0,0) to (n,0) using steps U=(1,1), H=(1,0) and D=(1,-1), the U (or D) steps come in four colors. - _N-E. Fahssi_, Mar 30 2008

%C Ignoring initial term, equals the logarithmic derivative of A091147. - _Paul D. Hanna_, Dec 08 2018

%H Seiichi Manyama, <a href="/A084605/b084605.txt">Table of n, a(n) for n = 0..1433</a> (terms 0..200 from Vincenzo Librandi)

%H Hacène Belbachir, Abdelghani Mehdaoui, László Szalay, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Szalay/szalay42.html">Diagonal Sums in the Pascal Pyramid, II: Applications</a>, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.

%H Tony D. Noe, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Noe/noe35.html">On the Divisibility of Generalized Central Trinomial Coefficients</a>, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

%H Sheng-Liang Yang, Yan-Ni Dong, and Tian-Xiao He, <a href="https://doi.org/10.1016/j.disc.2017.07.006">Some matrix identities on colored Motzkin paths</a>, Discrete Mathematics 340.12 (2017): 3081-3091.

%F E.g.f.: exp(x)*BesselI(0, 4*x). - _Vladeta Jovovic_, Aug 20 2003

%F a(n) is also the central coefficient of (4+x+x^2)^n; a(n) = Sum_{k=0..n} 3^(n-k) C(n,k) T(k,n), where T(k,n) is the triangle of trinomial coefficients = Coefficient of x^n of (1+x+x^2)^k : A027907. - _N-E. Fahssi_, Mar 30 2008

%F a(n) = (1/Pi)*integral(x=-2..2, (2*x+1)^n/sqrt((2-x)*(2+x))). - _Peter Luschny_, Sep 12 2011

%F D-finite with recurrence a(n+2) = ((2*n+3)*a(n+1) + 15*(n+1)*a(n))/(n+2); a(0)=a(1)=1 - _Sergei N. Gladkovskii_, Aug 01 2012

%F a(n) ~ 5^(n+1/2)/(2*sqrt(2*Pi*n)). - _Vaclav Kotesovec_, Oct 14 2012

%F a(n) = 2^n*GegenbauerC(n, -n, -1/4). - _Peter Luschny_, May 08 2016

%F a(n) = hypergeom([1/2 - n/2, -n/2], [1], 16). - _Peter Luschny_, Mar 18 2018

%F a(n) = Sum_{k=0..n} (-3)^(n-k) * 2^k * binomial(n,k)*binomial(2*k,k). - _Paul D. Hanna_, Dec 09 2018

%F a(n) = Sum_{k=0..n} 5^(n-k) * (-2)^k * binomial(n,k)*binomial(2*k,k). - _Seiichi Manyama_, May 01 2019

%p a := n -> simplify(2^n*GegenbauerC(n,-n, -1/4)):

%p seq(a(n), n=0..22); # _Peter Luschny_, May 08 2016

%t Table[n!*SeriesCoefficient[E^x*BesselI[0,4*x],{x,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 14 2012 *)

%t a[n_] := Hypergeometric2F1[1/2 - n/2, -n/2, 1, 16];

%t Table[a[n], {n, 0, 22}] (* _Peter Luschny_, Mar 18 2018 *)

%o (PARI) for(n=0,30,t=polcoeff((1+x+4*x^2)^n,n,x); print1(t","))

%o for(n=0,20,print1(a(n),", "))

%o (PARI) {a(n) = sum(k=0,n, (-3)^(n-k)*2^k*binomial(n,k)*binomial(2*k,k))}

%o for(n=0,20,print1(a(n),", ")) \\ _Paul D. Hanna_, Dec 09 2018

%Y Cf. A002426, A084600-A084604, A084606-A084615.

%Y Cf. A322240 (a(n)^2), A091147.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jun 01 2003

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified July 24 05:05 EDT 2021. Contains 346273 sequences. (Running on oeis4.)