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%I #33 Jun 10 2019 00:52:55
%S 1,9,58,330,1770,9198,46928,236736,1185645,5909805,29362806,145570230,
%T 720606705,3563543025,17610412600,86989143480,429579843435,
%U 2121099312195,10472653252550,51708363376950,255326054688320
%N a(n) = number of integer strings s(0),...,s(n) counted by array T in A026374 that have s(n)=4; also a(n) = T(2n,n-2).
%C Number of lattice paths from (0,0) to (2n,4), using steps U=(1,1), D=(1,-1) and at even levels(zero, positive and negative) also H=(2,0). Example: a(3)=9 because we have UUUUUD, UUUUDU, UUUDUU, UUDUUU, UDUUUU, DUUUUU, HUUUU, UUHUU and UUUUH. - _Emeric Deutsch_, Jan 30 2004
%H G. C. Greubel, <a href="/A026377/b026377.txt">Table of n, a(n) for n = 2..1000</a>
%F From _Emeric Deutsch_, Jan 30 2004: (Start)
%F a(n) = [t^(n+2)](1+3t+t^2)^n.
%F a(n) = Sum_{j=ceiling((n+2)/2)..n} (3^(2j-n-2)*binomial(n, j)*binomial(j, n+2-j)). (End)
%F From _Paul Barry_, Sep 20 2004: (Start)
%F E.g.f.: exp(3x) * BesselI(2, 2x);
%F a(n) = Sum_{k=0..n} binomial(n, k)*binomial(2k, k+2). (End)
%F Conjecture: n*(n+4)*a(n) - 3*(n+2)*(2*n+3)*a(n-1) + 5*(n+2)*(n+1)*a(n-2) = 0. - _R. J. Mathar_, Nov 24 2012
%F G.f.: (3*x - 1 + (1-6*x+7*x^2)/sqrt(5*x^2-6*x+1))/(2*x^2). - _Mark van Hoeij_, Apr 18 2013
%F a(n) ~ 5^(n+1/2)/(2*sqrt(Pi*n)). - _Vaclav Kotesovec_, Oct 07 2013
%F Assuming offset 0: a(n) = C(2*n+4,n)*hypergeom([-n,-n-4],[-3/2-n],-1/4). - _Peter Luschny_, May 09 2016
%p series( (3*x-1+(1-6*x+7*x^2)/sqrt(5*x^2-6*x+1))/(2*x^2), x=0, 30); # _Mark van Hoeij_, Apr 18 2013
%t CoefficientList[Series[(3*x-1+(1-6*x+7*x^2)/Sqrt[5*x^2-6*x+1])/(2*x^2), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Oct 07 2013 *)
%o (PARI) x='x+O('x^66); Vec((3*x-1+(1-6*x+7*x^2)/sqrt(5*x^2-6*x+1))/(2*x^2)) /* _Joerg Arndt_, Apr 19 2013 */
%Y Cf. A026374.
%K nonn
%O 2,2
%A _Clark Kimberling_
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