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Let a(0) = a(1) = 1, and n*a(n) = 2*(-7+5*n)*a(n-1) + 9*(2-n)*a(n-2) for n >= 2.
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%I #86 Sep 03 2024 01:41:31

%S 1,1,3,13,71,441,2955,20805,151695,1135345,8671763,67320573,529626839,

%T 4213228969,33833367963,273892683573,2232832964895,18314495896545,

%U 151037687326755,1251606057754605,10416531069771111,87029307323766681

%N Let a(0) = a(1) = 1, and n*a(n) = 2*(-7+5*n)*a(n-1) + 9*(2-n)*a(n-2) for n >= 2.

%C Let y = y(x) be implicitly defined by g(x,y(x)) = 0, with dg/dy not identically zero. For n >= 1, the sequence a(n) is the number of terms in the expansion of the divided difference [x0,...,xn]y in terms of bivariate divided differences of g.

%C (1 + 3*x + 13*x^2 + 71*x^3 + ...) = (1 + 4*x + 20*x^2 + 116*x^3 + ...) * 1/(1 + x + 4*x^2 + 20*x^3 + 116*x^4 + ...); where A082298 = (1, 4, 20, 116, 740, ...). - _Gary W. Adamson_, Nov 17 2011

%C The shifted sequence 1,3,13,71,... is the binomial transform of A151374. - _Georg Muntingh_, Jul 19 2012

%C a(n+1) is the number of Schröder paths of semilength n in which the (2,0)-steps come in 3 colors and with no peaks at level 1. - _José Luis Ramírez Ramírez_, Mar 31 2013

%C Define an infinite triangle by T(n,0)=1 and the other cells by T(n,k) = Sum_{c=0..k-1} T(n,c) + Sum_{r=k..n-1} T(r,k), the sum of the cells to the left and above a cell. The column k=1 contains A000079, the column k=2 essentially A001792. Then T(n,n)=a(n) on the diagonal. - _J. M. Bergot_, May 22 2013

%H Vincenzo Librandi, <a href="/A162326/b162326.txt">Table of n, a(n) for n = 0..300</a>

%H G. Muntingh, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL15/Muntingh/muntingh2.html">Implicit Divided Differences, Little Schroeder Numbers, and Catalan Numbers</a>, J. Integ. Seqs., Vol. 15 (2012), Article 12.6.5.

%F Let E = N x N \ {(0,0), (0,1)} be a set of pairs of natural numbers. The number of terms a(n) is the coefficient of x^n*y^{n-1} of the generating function 1 - log(1 - Sum_{(s,t) in E} x^s*y^{s+t-1}) = 1 + Sum_{q >= 1} (Sum_{(s,t) in E} x^s*y^{s+t-1})^q / q.

%F From _Georg Muntingh_, Jul 19 2012: (Start)

%F a(n) = 2F1(1/2,1-n;2;-8), where 2F1 is the Gauss hypergeometric series.

%F G.f.: (5 - sqrt( (1-9*x)/(1-x) ))/4.

%F Quadratic recurrence relation: a(n) = 1 + 2*Sum_{m=1..n-1} a(m)*a(n-m).

%F (End)

%F a(n) ~ 3^(2*n+1)/(16*sqrt(2*Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 20 2012

%F a(n) = Sum_{k=0..n} (binomial(n,k)*2^(n-k-1)*binomial(2*n-2*k-2,n-k-1))/n, n>0, a(0)=1. - _Vladimir Kruchinin_, Mar 13 2016

%F From _Peter Bala_, Jan 19 2020: (Start)

%F a(n+1) = Sum_{k = 0..n} 2^k*C(n,k)*Catalan(k).

%F a(n+1) = (2/Pi) * Integral_{x = -1..1} (1 + 8*x^2)^n*sqrt(1 - x^2) dx.

%F O.g.f.: 1 + x/(1 - x)*c(2*x/(1-x)), where c(x) is the o.g.f. for A000108. (End)

%F Conjecture: a(n) = t_n for n > 0 with a(0) = 1 where we start with vector v of fixed length m with elements v_i = 1, then set t = v and for i=1..m-1, for j=i+1..m apply [v_i, v_j] := [v_i + 2*v_j, 2*v_i + v_j] (here square brackets mean that instead of sequentially assigning v_i and then v_j, we reserve their values (for example, as A = v_i, B = v_j) and then assign them in any order) and t_{i+1} := v_{i+1} (after ending each cycle for j). It also looks like that if we change 2*v_i to z*v_i it gives us a(n+1) = Sum_{k=0..n} A090981(n, k)*2^(n-k) for n >= 0. - _Mikhail Kurkov_, Aug 14 2024

%e Write [0...n]y for [x0,...,xn]y and [0...s,0...t]g for [x0,...,xs;y0,...,yt]g.

%e For n = 1 one finds 1 term, [01]y = -[01;1]g/[0;01]g.

%e For n = 2 one finds 3 terms, [012]y = -[012;2]g/[0;02]g + ([01;12]g[12;2]g)/([0;02]g[1;12]g) - ([0;012]g[01;1]g[12;2]g)/([0;02]g[0;01]g[1;12]g).

%t CoefficientList[Series[(5-Sqrt[(1-9*x)/(1-x)])/4, {x, 0, 20}], x] (* _Vaclav Kotesovec_, Oct 20 2012 *)

%o (Python)

%o L = [1, 1]

%o for n in range(2,22):

%o L.append( ((-14 + 10*n)*L[-1] + (18-9*n)*L[-2])//n )

%o print(L)

%o # _Georg Muntingh_, Jul 19 2012

%o (PARI) a(n) = if(n<2, 1, (2*(-7+5*n)*a(n-1) + 9*(2-n)*a(n-2))/n);

%o vector(25, n, a(n-1)) \\ _Altug Alkan_, Oct 06 2015

%o (PARI) my(x='x+O('x^20)); Vec((5-sqrt((1-9*x)/(1-x)))/4) \\ _G. C. Greubel_, Feb 07 2019

%o (Maxima)

%o a(n):=if n=0 then 1 else sum(binomial(n,k)*2^(n-k-1)*binomial(2*n-2*k-2,n-k-1),k,0,n)/n; /* _Vladimir Kruchinin_, Mar 13 2016 */

%o (Magma) m:=20; R<x>:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (5-Sqrt((1-9*x)/(1-x)))/4 )); // _G. C. Greubel_, Feb 07 2019

%o (Magma) a:=[1,3]; for n in [3..21] do Append(~a,(2*(-7+5*n)*a[n-1] + 9*(2-n)*a[n-2]) div n); end for ; [1] cat a; // _Marius A. Burtea_, Jan 20 2020

%o (Sage) ((5-sqrt((1-9*x)/(1-x)))/4).series(x, 20).coefficients(x, sparse=False) # _G. C. Greubel_, Feb 07 2019

%Y Cf. A172003, which is a generalization to bivariate implicit functions.

%Y Cf. A003262, which is the analogous sequence for implicit derivatives, and A172004 for its generalization to bivariate implicit functions.

%Y Cf. A082298, A000108, A151374.

%K nonn

%O 0,3

%A _Georg Muntingh_, Jul 01 2009

%E Edited by _Georg Muntingh_, Jan 22 2010