|
%I #74 Oct 06 2021 13:17:33
%S 1,2,16,146,1408,14002,142000,1459810,15158272,158611106,1669752016,
%T 17664712562,187641279616,2000029880786,21380213588848,
%U 229129634462146,2460955893981184,26482855453375042,285475524009208720,3082024598888203090,33319523640218177408
%N a(n) = [x^(2*n)] ((1 + x)/(1 - x))^n.
%C From _Peter Bala_, Mar 01 2020: (Start)
%C The recurrence given below can be rewritten in the form
%C (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.
%C 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)
%H G. C. Greubel, <a href="/A103885/b103885.txt">Table of n, a(n) for n = 0..950</a> [a(0) = 1 inserted by _Georg Fischer_, Apr 03 2020]
%H Peter Bala, <a href="/A103885/a103885.pdf">Notes on A103885</a>
%H V. V. Kruchinin and D. V. Kruchinin, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Kruchinin/kruch9.html">A Generating Function for the Diagonal T_{2n,n} in Triangles</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.
%F 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}.]
%F a(n) = A103884(n, n).
%F G.f.: A(x)=x*B(x)'/B(x), where B(x) is g.f. of A027307. - _Vladimir Kruchinin_, Jun 30 2015
%F From _Vaclav Kotesovec_, Jul 01 2015: (Start)
%F 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).
%F a(n) ~ ((11 + 5*sqrt(5))/2)^n / (2 * 5^(1/4) * sqrt(Pi*n)). (End)
%F 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
%F a(n) = 2*n*hypergeom([1 - 2*n, 1 - n], [2], 2) for n >= 1. - _Peter Luschny_, Dec 30 2019
%F From _Peter Bala_, Mar 01 2020: (Start)
%F a(n) = Sum_{k = 0..n} C(n, k)*C(2*n+k-1, n-1).
%F a(n) = Sum_{k = 0..n} C(2*n, 2*k)*C(2*n-k-1, n-1).
%F a(n) = (1/2)*Sum_{k = 0..n} C(2*n, n-k)*C(2*n+k-1, k). Cf. A156894.
%F 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.
%F a(n) = (1/2) * [x^(n)] ( (1 + x)/(1 - x) )^(2*n). Cf. A002003(n) = [x^n] ( (1 + x)/(1 - x) )^n.
%F 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.
%F a(p) == 2 ( mod p^3 ) for prime p >= 5. (End)
%F From _Peter Bala_, Sep 22 2021: (Start)
%F a(n) = Sum_{k = 0..n} 4^k*binomial(n+k-1,n)*binomial(n,k)^2 / binomial(2*k,k).
%F 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)
%p a := n -> `if`(n=0, 1, 2*n*hypergeom([1 - 2*n, 1 - n], [2], 2)):
%p seq(simplify(a(n)), n=0..17); # _Peter Luschny_, Dec 30 2019
%p # Alternative (after _Peter Bala_ ):
%p gf := n -> ( (1 + x)/(1 - x) )^n: ser := n -> series(gf(n), x, 40):
%p seq(coeff(ser(n), x, 2*n), n=0..17); # _Peter Luschny_, Mar 20 2020
%t 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 *)
%o (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
%Y Cf. A002003, A123164, A266213, A103882, A103884, A027307, A006318, A144097, A156894.
%K nonn
%O 0,2
%A _Ralf Stephan_, Feb 20 2005
%E a(0) = 1 added and new name by _Peter Bala_, Mar 01 2020
|
