A173992 Sequence whose Hankel transform is the Somos (4) sequence.
1, 1, 3, 6, 15, 34, 83, 198, 488, 1202, 3015, 7608, 19432, 49994, 129779, 339176, 892600, 2362634, 6288156, 16816232, 45170466, 121812152, 329679487, 895171236, 2437885058, 6657311202, 18224979884, 50006899724, 137502724754
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..2000
- Ricardo Gómez Aíza, Trees with flowers: A catalog of integer partition and integer composition trees with their asymptotic analysis, arXiv:2402.16111 [math.CO], 2024.
Programs
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Magma
m:=50; R
:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1-2*x-Sqrt((1-2*x)*(1-2*x-4*x^2+4*x^3)))/(2*x^2*(1-2*x)))); // G. C. Greubel, Sep 25 2018 -
Maple
with(LREtools): with(FormalPowerSeries): # requires Maple 2022 ogf:=(1-2*x-sqrt((1-2*x)*(1-2*x-4*x^2+4*x^3)))/(2*x^2*(1-2*x)): req1:= FindRE(ogf,x,u(n)); inits:= {seq(u(i-1)=[1, 1, 3, 6, 15, 34][i],i=1..6)}: req2:= subs(n=n-4, MinimalRecurrence(req1,u(n),inits)[1]); # Mathar's recurrence a:= gfun:-rectoproc({req2} union inits, u(n), remember): seq(a(n),n=0..28); # Georg Fischer, Nov 03 2022
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Mathematica
A173992[n_] := Sum[CatalanNumber[k] Sum[Binomial[k + 1, i] Binomial[n - k - i, n - 2 k - i] (-1)^i Floor[2^(n - 2 k - i)], {i, 0, k + 1}], {k, 0, Floor[n/2]}] (* Eric Rowland, May 15 2017 *) CoefficientList[Series[(1-2*x -Sqrt[(1-2*x)*(1-2*x-4*x^2+4*x^3)])/(2*x^2* (1-2*x)), {x, 0, 50}], x] (* G. C. Greubel, Sep 25 2018 *)
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PARI
a(n) = sum(k=0, n\2, binomial(2*k,k)/(k+1)*sum(i=0, k+1, binomial(k+1,i)*binomial(n-k-i,n-2*k-i)*(-1)^i*2^(n-2*k-i))); \\ Michel Marcus, May 15 2017
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PARI
x='x+O('x^50); Vec((1-2*x-((1-2*x)*(1-2*x-4*x^2+4*x^3))^(1/2))/(2*x^2*(1-2*x))) \\ Altug Alkan, Sep 25 2018
Formula
G.f.: ((1-x)/(1-2*x)) * c(x^2*(1-x)/(1-2*x)) = (1-2*x-sqrt((1-2x)*(1-2*x-4*x^2+4*x^3)))/(2*x^2*(1-2*x)), c(x) the g.f. of A000108;
a(n) = Sum_{k=0..floor(n/2), A000108(k)*Sum_{i=0..k+1, C(k+1,i)*C(n-k-i,n-2k-i)*(-1)^i*2^(n-2k-i)}}.
D-finite with recurrence: (n+2)*a(n) -4*(n+1)*a(n-1) +4*a(n-2) +2*(6n-11)*a(n-3) +8*(3-n)*a(n-4)=0. - R. J. Mathar, Nov 17 2011
a(n) ~ sqrt(2-5*c+4*c^2)/(2*c*(1-2*c)*sqrt(Pi*n^3))*(1/c)^n where c=(4+(1+i*sqrt(3))*(1+3*i*sqrt(111))^(1/3)+80/((sqrt(3)+i)^2*(1+3*i*sqrt(111))^(1/3)))/12. - Ricardo Gómez Aíza, Feb 26 2024
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