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%I #65 Feb 24 2023 11:24:48
%S 1,4,15,55,200,725,2625,9500,34375,124375,450000,1628125,5890625,
%T 21312500,77109375,278984375,1009375000,3651953125,13212890625,
%U 47804687500,172958984375,625771484375,2264062500000,8191455078125
%N Row sums of convolution triangle A030523.
%C Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 3, s(2n) = 5.
%C With offset 0 = INVERT transform of A001792: (1, 3, 8, 20, 48, 112, ...). - _Gary W. Adamson_, Oct 26 2010
%C From _Tom Copeland_, Nov 09 2014: (Start)
%C The array belongs to a family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the o.g.f. (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse x*(1-x)/(1 + (t-1)*x*(1-x)). See A091867 for more info on this family. Here t = -4 (mod signs in the results).
%C Let C(x) = (1 - sqrt(1-4x))/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1+t*x) with inverse P(x,-t).
%C O.g.f.: G(x) = x*(1-x)/(1 - 5x*(1-x)) = P(Cinv(x),-5).
%C Inverse O.g.f.: Ginv(x) = (1 - sqrt(1 - 4*x/(1+5x)))/2 = C(P(x,5)) (signed A026378). Cf. A030528. (End)
%C p-INVERT of (2^n), where p(s) = 1 - s - s^2; see A289780. - _Clark Kimberling_, Aug 10 2017
%H Michel Marcus, <a href="/A039717/b039717.txt">Table of n, a(n) for n = 1..1000</a>
%H Wolfdieter Lang, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL3/LANG/lang.html">On generalizations of Stirling number triangles</a>, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,-5).
%F G.f.: x*(1-x)/(1-5*x+5*x^2) = g1(3, x)/(1-g1(3, x)), g1(3, x) := x*(1-x)/(1-2*x)^2 (g.f. first column of A030523).
%F From _Paul Barry_, Apr 16 2004: (Start)
%F Binomial transform of Fibonacci(2n+2).
%F a(n) = (sqrt(5)/2 + 5/2)^n*(3*sqrt(5)/10 + 1/2) - (5/2 - sqrt(5)/2)^n*(3*sqrt(5)/10 - 1/2). (End)
%F a(n) = (1/5)*Sum_{r=1..9} sin(3*r*Pi/10)*sin(r*Pi/2)*(2*cos(r*Pi/10))^(2n)).
%F a(n) = 5*a(n-1) - 5*a(n-2).
%F a(n) = Sum_{k=0..n} Sum_{i=0..n} binomial(n, i)*binomial(k+i+1, 2k+1). - _Paul Barry_, Jun 22 2004
%F From _Johannes W. Meijer_, Jul 01 2010: (Start)
%F Limit_{k->infinity} a(n+k)/a(k) = (A020876(n) + A093131(n)*sqrt(5))/2.
%F Limit_{n->infinity} A020876(n)/A093131(n) = sqrt(5).
%F (End)
%F From _Benito van der Zander_, Nov 19 2015: (Start)
%F Limit_{k->infinity} a(k+1)/a(k) = 1 + phi^2 = (5 + sqrt(5)) / 2.
%F a(n) = a(n-1) * 3 + A081567(n-2) (not proved).
%F (End)
%t CoefficientList[Series[(1 - x) / (1 - 5 x + 5 x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Nov 10 2014 *)
%o (PARI) Vec(x*(1-x)/(1-5*x+5*x^2) + O(x^40)) \\ _Altug Alkan_, Nov 20 2015
%Y Cf. A000045.
%Y Appears in A109106. - _Johannes W. Meijer_, Jul 01 2010
%Y Cf. A001792. - _Gary W. Adamson_, Oct 26 2010
%Y Cf. A000108, A005043, A091867, A026378, A030528.
%K easy,nonn
%O 1,2
%A _Wolfdieter Lang_
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