A103885 a(n) = [x^(2*n)] ((1 + x)/(1 - x))^n.

1, 2, 16, 146, 1408, 14002, 142000, 1459810, 15158272, 158611106, 1669752016, 17664712562, 187641279616, 2000029880786, 21380213588848, 229129634462146, 2460955893981184, 26482855453375042, 285475524009208720, 3082024598888203090, 33319523640218177408

Offset: 0

Author

Ralf Stephan, Feb 20 2005

Keywords

• nonn
• easy

Comments

From Peter Bala, Mar 01 2020: (Start)

The recurrence given below can be rewritten in the form

(2*n+1)*(2*n+2)*P(2,n)*a(n+1) - (2*n-1)*(2*n-2)*P(2,-n)*a(n-1) = Q(2,n^2)*a(n), where the polynomial Q(2,n) = 4*(55*n^2 - 34*n + 3) and the polynomial P(2,n) = 5*n^2 - 5*n + 1 satisfies the symmetry condition P(2,n) = P(2,1-n) and has real zeros.

More generally, for fixed m = 1,2,3,..., we conjecture that the sequence b(n) := a(m*n) satisfies a recurrence of the form ( Product_{k = 1..2*m} (2*m*n + k) ) * P(2*m,n)*b(n+1) + (-1)^m*( Product_{k = 1..2*m} (2*m*n - k) ) * P(2*m,-n)*b(n-1) = Q(2*m,n^2)*b(n), where the polynomials P(2*m,n) and Q(2*m,n) have degree 2*m. Conjecturally, the polynomial P(2*m,n) = P(2*m,1-n) and has real zeros in the interval [0, 1]. The 4*m zeros of the polynomial Q(2*m,n^2) seem to belong to the interval [-1, 1] and 4*m - 2 of these zeros appear to be approximated by the rational numbers +- k/(3*m), where 1 <= k <= 3*m - 2, k not a multiple of 3. (End)

Links

• G. C. Greubel, Table of n, a(n) for n = 0..950 [a(0) = 1 inserted by _Georg Fischer_, Apr 03 2020]
• Peter Bala, Notes on A103885
• V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

Crossrefs

Cf. A002003, A006318, A027307, A103882, A103884, A123164, A144097, A156894, A266213, A370102.

Programs

Magma

A103885:= func< n | n eq 0 select 1 else (&+[ Binomial(n, k)*Binomial(2*n+k-1, n-1): k in [0..n]]) >;
[A103885(n): n in [0..40]]; // G. C. Greubel, Oct 27 2024

Maple

a := n -> `if`(n=0, 1, 2*n*hypergeom([1 - 2*n, 1 - n], [2], 2)):
seq(simplify(a(n)), n=0..17); # Peter Luschny, Dec 30 2019
# Alternative (after Peter Bala ):
gf := n -> ( (1 + x)/(1 - x) )^n: ser := n -> series(gf(n), x, 40):
seq(coeff(ser(n), x, 2*n), n=0..17); # Peter Luschny, Mar 20 2020

Mathematica

Prepend[Table[Sum[2^i Binomial[n, i] Binomial[2n-1, i-1], {i, 1, 2n}], {n,1,20}], 1] (* Vaclav Kotesovec, Jul 01 2015 *)

PARI

a(n) = if (n==0, 1, sum(i=0, n, 2^i * binomial(n, i) * binomial(2*n-1, i-1))); \\ Michel Marcus, Mar 21 2020

SageMath

def A103885(n): return 1 if n==0 else sum(binomial(n, k)*binomial(2*n+k-1, n-1) for k in range(n+1))
[A103885(n) for n in range(41)] # G. C. Greubel, Oct 27 2024

Formula

a(n) = Sum_{i=0..n} 2^i * binomial(n,i) * binomial(2*n-1,i-1). [Original definition, with summation range {i=1..n}.]

a(n) = A103884(n, n).

G.f.: A(x) = x*B(x)'/B(x), where B(x) is g.f. of A027307. - Vladimir Kruchinin, Jun 30 2015

From Vaclav Kotesovec, Jul 01 2015: (Start)

Recurrence: n*(2*n-1)*(5*n^2 - 15*n + 11)*a(n) = 2*(55*n^4 - 220*n^3 + 296*n^2 - 152*n + 24)*a(n-1) + (n-2)*(2*n-3)*(5*n^2 - 5*n + 1)*a(n-2).

a(n) ~ ((11 + 5*sqrt(5))/2)^n / (2 * 5^(1/4) * sqrt(Pi*n)). (End)

a(n) = [x^n] (1/(1 - x - x/(1 - x - x/(1 - x - x/(1 - x - x/(1 - ...))))))^n, a continued fraction. - Ilya Gutkovskiy, Sep 29 2017

a(n) = 2*n*hypergeom([1 - 2*n, 1 - n], [2], 2) for n >= 1. - Peter Luschny, Dec 30 2019

From Peter Bala, Mar 01 2020: (Start)

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

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

a(n) = (1/2)*Sum_{k = 0..n} C(2*n, n-k)*C(2*n+k-1, k). Cf. A156894.

a(n) = [x^n] S(x)^n, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318.

a(n) = (1/2) * [x^(n)] ( (1 + x)/(1 - x) )^(2*n). Cf. A002003(n) = [x^n] ( (1 + x)/(1 - x) )^n.

Conjecture: a(n) = - [x^n] G(x)^(-n), where G(x) = 1 + 2*x + 14*x^2 + 134*x^3 + 1482*x^4 + ... is the o.g.f. of A144097.

a(p) == 2 ( mod p^3 ) for prime p >= 5. (End)

From Peter Bala, Sep 22 2021: (Start)

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

Equivalently, a(n) = [x^n] T(n,(1+x)/(1-x)), where T(n,x) is the n-th Chebyshev polynomial of the first kind. Cf. A103882. (End)

For n>0, a(n) = (1/3) * [x^n] (1/S(-x))^(3*n), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318. Cf. A370102. - Peter Bala, Jul 29 2024

Extensions

a(0) = 1 added and new name by Peter Bala, Mar 01 2020