A101500 A Chebyshev transform of the central binomial numbers.

1, 2, 5, 16, 53, 178, 609, 2112, 7393, 26066, 92437, 329360, 1178149, 4228322, 15218305, 54907136, 198527617, 719170850, 2609577701, 9483269008, 34508808789, 125727351186, 458573578977, 1674270763584, 6118472289889, 22378379004146, 81913223571701

Offset: 0

Author

Paul Barry, Dec 04 2004

Keywords

• nonn
• easy

Comments

A Chebyshev transform of A000984. Under the Chebyshev transform, we map a g.f. g(x) to (1/(1+x^2))*g(x/(1+x^2)).

Also equal to the Riordan array (1/(1-x)^2,x/(1-x)^2) applied to aerated central binomial coefficients (with g.f. 1/sqrt(1-4x^2)). - Paul Barry, Jul 06 2009

Directed 2-D walks with n steps starting at (0,0) and ending on the X-axis using steps N,S,E,W and avoiding N followed by S. - David Scambler, Jun 24 2013

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• C. Banderier and P. Hitczenko, Enumeration and asymptotics of restricted compositions having the same number of parts, Disc. Appl. Math. 160 (18) (2012) 2542-2554 [The formula in Table 2 for A101500 is not correct], also arXiv:1201.6116
• P. Barry, On the Connection Coefficients of the Chebyshev-Boubaker polynomials, The Scientific World Journal, Volume 2013 (2013), Article ID 657806, 10 pages.
• Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.

Crossrefs

Cf. A000984.

Programs

Mathematica

CoefficientList[Series[1/Sqrt[(1+x^2)*(1-4*x+x^2)], {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 08 2014 *)
Table[1/2^n* Sum[(-1)^k*Binomial[2 k, k]* Sum[Binomial[n - 2 k, j]^2*3^j, {j, 0, n - 2 k}], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jun 30 2018 *)
Table[Sum[Binomial[n - k, k]*(-1)^k*Binomial[2 (n - 2 k), n - 2 k], {k, 0, Floor[n/2]}], {n, 0, 20}] (* Vaclav Kotesovec, Jun 30 2018 *)
a[ n_] := Sum[Binomial[n + k + 1, 2k + 1] Binomial[k, Quotient[k, 2]], {k, 0, n, 2}]; (* Michael Somos, Jun 30 2018 *)

PARI

A101500(maxx)={n=0;while(n<=maxx,z=sum(k=0,floor(n/2), binomial(n-k,k)*binomial(2*(n-2*k),n-2*k)*(-1)^k ) ;print1(z,", ");n+=1);} \\ Bill McEachen, Jan 02 2016

PARI

x='x+O('x^40); Vec(1/(sqrt(1+x^2)*sqrt(1-4*x+x^2))); \\ Michel Marcus, Jan 03 2016

Formula

G.f.: 1/(sqrt(1+x^2)*sqrt(1-4*x+x^2)).

a(n) = Sum_{k=0..floor(n/2)} C(n-k, k)*(-1)^k*C(2(n-2k), n-2k).

From Paul Barry, Jul 06 2009: (Start)

G.f.: 1/((1-x)^2-2*x^2/((1-x)^2-x^2/((1-x)^2-x^2/((1-x)^2-... (continued fraction);

a(n) = Sum_{k=0..n} C(n+k+1,n-k)*C(k,k/2)*(1+(-1)^k)/2. (End)

Conjecture: n*a(n) +2*(-2*n+1)*a(n-1) +2*(n-1)*a(n-2) +2*(-2*n+3)*a(n-3) +(n-2)*a(n-4)=0. - R. J. Mathar, Nov 16 2012

a(n) ~ sqrt(1/2 + 7/(8*sqrt(3))) * (2+sqrt(3))^n / sqrt(Pi*n). - Vaclav Kotesovec, Feb 08 2014

a(n) = Sum_{k=ceiling(n/2)..n} C(k,n-k)^2*2^(2*k-n). - Vladimir Kruchinin, Apr 09 2014