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A218225
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G.f. A(x) satisfies: (1 - x*A(x)) / (1 - x^2*A(x)^2)^2 = 1 - x.
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4
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1, 2, 6, 23, 101, 480, 2400, 12434, 66142, 359112, 1981904, 11085198, 62696874, 357970472, 2060459256, 11943445311, 69656978837, 408466559630, 2406825745010, 14243262687023, 84618295006269, 504485687485408, 3017344000161296, 18099717207764928
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OFFSET
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0,2
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COMMENTS
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Conjecture: a(n) is the number of permutations of [1..n+1] that avoid one of the following sets of patterns: (2134, 42153, 24153), (3124, 42153, 24153), (2143, 42135, 24135). - Alexander Burstein, Dec 20 2017
Chern et al. (see links) proves the above conjecture for (3124, 42153, 24153) as well as the following conjecture of Hong and Li:
a(n) is the number of inversion sequences of length n+1 avoiding pattern 0021. (End)
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LINKS
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FORMULA
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G.f. A(x) satisfies:
(1) A(1 - (1-x)/(1-x^2)^2) = x + 1/(1-x-x^2).
(2) A(x) = (1/x) * Series_Reversion( x*(1-x-x^2)/((1-x)*(1+x)^2) ).
(3) A(x) = (1 - x*A(x)) * (1 + x*A(x))^2 / (1 - x*A(x) - x^2*A(x)^2).
(4) A(x) = exp( Sum_{n>=1} (x^n/n) * Sum_{k=0..n} binomial(n,k)^2 * (1-x)^k * A(x)^k ).
Recurrence: 5*n*(n+1)*a(n) = 21*n*(2*n-1)*a(n-1) - 3*(23*n^2-46*n+24)*a(n-2) + 16*(n-2)*(2*n-3)*a(n-3). - Vaclav Kotesovec, May 22 2013
a(n) ~ 2^(5*n+6)/(27*sqrt(Pi)*5^(n+1/2)*n^(3/2)). - Vaclav Kotesovec, May 22 2013
G.f.: (-1+cos(2/3*(arccot(3*sqrt(3/5))-arccot((3*sqrt(3))/sqrt(5-32*x))))+sqrt(15)*sin(2/3*(arccot(3*sqrt(3/5))-arccot((3*sqrt(3))/sqrt(5-32*x)))))/(3*x). - Vaclav Kotesovec, Jul 06 2013
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EXAMPLE
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G.f.: A(x) = 1 + 2*x + 6*x^2 + 23*x^3 + 101*x^4 + 480*x^5 + 2400*x^6 + ...
The series reversion of x*A(x) begins:
x - 2*x^2 + 2*x^3 - 3*x^4 + 3*x^5 - 4*x^6 + 4*x^7 - 5*x^8 + 5*x^9 + ...
so A(1 - (1-x)/(1-x^2)^2) = x + 1/(1-x-x^2).
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + (1-x)*A(x))*x +
(1 + 2^2*(1-x)*A(x) + (1-x)^2*A(x)^2)*x^2/2 +
(1 + 3^2*(1-x)*A(x) + 3^2*(1-x)^2*A(x)^2 + (1-x)^3*A(x)^3)*x^3/3 +
(1 + 4^2*(1-x)*A(x) + 6^2*(1-x)^2*A(x)^2 + 4^2*(1-x)^3*A(x)^3 + (1-x)^4*A(x)^4)*x^4/4 +
(1 + 5^2*(1-x)*A(x) + 10^2*(1-x)^2*A(x)^2 + 10^2*(1-x)^3*A(x)^3 + 5^2*(1-x)^4*A(x)^4 + (1-x)^5*A(x)^5)*x^5/5 + ...
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MATHEMATICA
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Table[1/(n+1)*SeriesCoefficient[(((x-1)*(x+1)^2)/(x^2+x-1))^(n+1), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 22 2013 *)
Flatten[{1, Table[FullSimplify[SeriesCoefficient[(2*(1-x)^(1/3)-2^(2/3)*(-11-16*x-3*Sqrt[-15+96*x])^(1/3)+2^(2/3)*(11+16*x-3*Sqrt[-15+96*x])^(1/3))/(6*(-1+x)^(1/3)*x), {x, 0, n}]], {n, 1, 10}]}] (* Vaclav Kotesovec, Jul 06 2013 *)
CoefficientList[Series[(-1+Cos[2/3*(ArcCot[3*Sqrt[3/5]]-ArcCot[(3*Sqrt[3])/Sqrt[5-32*x]])]+Sqrt[15]*Sin[2/3*(ArcCot[3*Sqrt[3/5]]-ArcCot[(3*Sqrt[3])/Sqrt[5-32*x]])])/(3*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Jul 06 2013 *)
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PROG
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(PARI) {a(n)=polcoeff((1/x)*serreverse(x*(1-x-x^2)/((1-x)*(1+x)^2 +x*O(x^n))), n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) {a(n)=local(A=1); for(i=1, n, A=exp(sum(m=1, n, x^m/m*sum(k=0, m, binomial(m, k)^2*(1-x)^k*A^k)+x*O(x^n)))); polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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