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A054391
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Number of permutations with certain forbidden subsequences.
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12
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1, 1, 2, 5, 14, 41, 123, 374, 1147, 3538, 10958, 34042, 105997, 330632, 1032781, 3229714, 10109310, 31667245, 99260192, 311294876, 976709394, 3065676758, 9625674442, 30231524869, 94972205349, 298419158008, 937861780439, 2947969125284
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OFFSET
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0,3
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COMMENTS
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The inverse Motzkin transform apparently yields 1 followed by A000930, which implies a generating function g(x)=1+z/(1-z-z^3) where z=x*A001006(x). - R. J. Mathar, Jul 07 2009
It appears that the infinite set of interpolated sequences between the Motzkin and the Catalan can be generated with a succession of INVERT transforms, given each sequence has two leading 1's. Also, the N-th sequence in the set starting with (N=1, A001006) can be generated from a production matrix of the form "M" in the formula section, such that the main diagonal is (N leading 1's, 0, 0, 0, ...). M with a diagonal of (1, 0, 0, 0, ...) generates A001006, while M with a main diagonal of all 1's is the production matrix for A000108. - Gary W. Adamson, Jul 29 2011
Conjecture: Also the number of non-capturing set partitions of {1..n} (A326254). A set partition is capturing if it has two blocks of the form {...x...y...} and {...z...t...} where x < z and y > t or x > z and y < t. This is a weaker condition than nesting, so for example {{1,3,5},{2,4}} is capturing but not nesting. The a(0) = 1 through a(4) = 14 non-capturing set partitions are:
{} {{1}} {{1,2}} {{1,2,3}} {{1,2,3,4}}
{{1},{2}} {{1},{2,3}} {{1},{2,3,4}}
{{1,2},{3}} {{1,2},{3,4}}
{{1,3},{2}} {{1,2,3},{4}}
{{1},{2},{3}} {{1,2,4},{3}}
{{1,3},{2,4}}
{{1,3,4},{2}}
{{1},{2},{3,4}}
{{1},{2,3},{4}}
{{1,2},{3},{4}}
{{1},{2,4},{3}}
{{1,3},{2},{4}}
{{1,4},{2},{3}}
{{1},{2},{3},{4}}
(End)
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LINKS
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FORMULA
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G.f.: 1 - 2*x^2 / (2*x^2 - 3*x + 1 - sqrt(1 - 2*x - 3*x^2)). - Mansour and Shattuck
G.f.: 1/(1-x-x^2/(1-2x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction) (conjecture). - Paul Barry, Jan 19 2009
a(n) = upper left term of M^n, a(n+1) = sum of top row terms of M^n; M = an infinite square production matrix as follows with a main diagonal of (1, 1, 1, 0, 0, 0, ...):
1, 1, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
1, 1, 1, 1, 0, 0, ...
1, 1, 1, 0, 1, 0, ...
1, 1, 1, 1, 0, 1, ...
1, 1, 1, 1, 1, 0, ...
... (End)
a(n) = Sum_{k=1..n-1} (sum(l=1..k, (binomial(k,l)*l*sum(j=0..n+l-k-1, binomial(j,1-n-2*l+k+2*j)*binomial(n-1+l-k,j)))/(n+l-k-1))) + 1. - Vladimir Kruchinin, Oct 31 2011
D-finite with recurrence (-n+1)*a(n) + 3*(2*n-3)*a(n-1) + (-8*n+11)*a(n-2) + (-5*n+32)*a(n-3) + (7*n-31)*a(n-4) + 3*(-n+4)*a(n-5)= 0. - R. J. Mathar, Nov 26 2012
G.f.: 1 - x*(2*x^2-3*x+1 + 1/G(0))/(2*(x^3-3*x^2+4*x-1)), where G(k)= 1 + x*(2+3*x)*(4*k+1)/( 4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jun 29 2013
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EXAMPLE
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a(4) = 14, a(5) = 41 since the top row of M^4 = (14, 14, 9, 3, 1), with 41 = (14 + 14 + 9 + 3 + 1).
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MAPLE
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c := x->(1-sqrt(1-4*x))/(2*x); a := (x, j)->(x)/((1-4*x)*(c(x))^2*(1-c(x))^(j))*(-x^2*(c(x))^2*(1-c(x))*(x^2*(c(x))^4)^(j)-(1-3*x-2*x^2)*(c(x))^2*(x*(c(x))^2)^(j)+x);
b := (x, j)->1+(1)/((1-4*x)*c(x)*(1-c(x))^(j))*(-2*x^3*(c(x))^2*(x^2*(c(x))^4)^(j)+(1-3*x-2*x^2)*c(x)*(x*(c(x))^2)^(j)-2*x^2);
co := (x, j)->(1)/((1-4*x)*(1-c(x))^(j))*(x^2*(x^2*(c(x))^4)^(j)-(1-3*x-2*x^2)*(x*(c(x))^2)^(j)+x^2);
s := (x, j)->(1-b(x, j)+(-1)^j*sqrt((1-b(x, j))^2-4*a(x, j)*co(x, j)))/(2*a(x, j)); j := 3; series(s(x, j), x=0..60); od; # j=1, 2, 3, ... inf gives A001006, A005773, A054391, A054392, ..., A000108
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MATHEMATICA
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CoefficientList[Series[1 - 2*x^2/(2*x^2 - 3*x + 1 - Sqrt[1 - 2*x - 3*x^2]), {x, 0, 50}], x] (* G. C. Greubel, Apr 27 2017 *)
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PROG
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(Maxima) a(n):=sum((sum((binomial(k, l)*l*sum(binomial(j, 1-n-2*l+k+2*j)*binomial(n-1+l-k, j), j, 0, n+l-k-1))/(n+l-k-1), l, 1, k)), k, 1, n-1)+1; \\ Vladimir Kruchinin, Oct 31 2011
(PARI) x='x+O('x^66); gf=1-2*x^2/(2*x^2-3*x+1-sqrt(1-2*x-3*x^2)); Vec(gf) \\ Joerg Arndt, Jun 29 2013
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CROSSREFS
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Conjectured to be equal to A326254.
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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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