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A090365
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Shifts 1 place left under the INVERT transform of the BINOMIAL transform of this sequence.
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7
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1, 1, 3, 11, 47, 225, 1177, 6625, 39723, 251939, 1681535, 11764185, 86002177, 655305697, 5193232611, 42726002123, 364338045647, 3215471252769, 29331858429241, 276224445794785, 2682395337435723, 26832698102762435, 276221586866499839, 2923468922184615897
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OFFSET
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0,3
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COMMENTS
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The Hankel transform of this sequence is A000178(n+1); example: det([1,1,3; 1,3,11; 3,11,47]) = 12. - Philippe Deléham, Mar 02 2005
a(n) appears to be the number of indecomposable permutations (A003319) of [n+1] that avoid both of the dashed patterns 32-41 and 41-32. - David Callan, Aug 27 2014
This is true: A nonempty permutation avoids 32-41 and 41-32 if and only if all its components do so. So if A(x) denotes the g.f. for indecomposable {32-41,41-32}-avoiders, then F(x):=1/(1-A(x)) is the g.f. for all {32-41,41-32}-avoiders. From A074664, F(x)=1/x(1-1/B(x)) where B(x) is the o.g.f. for the Bell numbers. Solve for A(x). - David Callan, Jul 21 2017
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LINKS
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FORMULA
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G.f.: A(x) satisfies A(x) = 1/(1 - A(x/(1-x))*x/(1-x) ).
G.f.: 1/(1-x/(1-2x/(1-x/(1-3x/(1-x/(1-4x/(1-x/(1-5x/(1-... (continued fraction). - Paul Barry, Feb 25 2010
Continued fractions:
G.f.: 1 - x/(G(0)+x); G(k) = x - 1 + x*k + x*(x-1+x*k)/G(k+1).
G.f.: 1/x - 1/2 + (x^2-4)/(4*U(0)-2*x^2+8) where U(k) = k*(2*k+3)*x^2 + x - 2 - (2-x+2*k*x)*(2+3*x+2*k*x)*(k+1)*x^2/U(k+1).
G.f.: 1/x+1/(U(0)-1) where U(k) = -x*k + 1 - x - x^2*(k+1)/U(k+1).
G.f.: (1 - U(0))/x - 1 where U(k) = 1 - x*(k+2) - x^2*(k+1)/U(k+1).
G.f.: (1 - U(0))/x where U(k) = 1 - x*(k+1)/(1-x/U(k+1)).
G.f.: 1/x + 1/( G(0)-1) where G(k) = 1 - x/(1 - x*(2*k+1)/(1 - x/(1 - x*(2*k+2)/ G(k+1) ))).
G.f.:1/x + 1/( G(0) - 1 ) where G(k) = 1 - x/(1 - x*(k+1)/G(k+1) ).
G.f.: (1 - Q(0))/x where Q(k) = 1 + x/(x*k - 1 )/Q(k+1).
G.f.: 1/x - 1/x/Q(0), where Q(k) = 1 + x/(1 - x + x*(k+1)/(x - 1/Q(k+1))).
(End)
Conjecture: a(n) = b(2^(n-1) - 1) for n > 0 with a(0) = 1 where b(n) = b((n - 2^f(n))/2) + b(floor((2n - 2^f(n))/2)) + b(A025480(n-1)) for n > 0 with b(0) = 1 and where f(n) = A007814(n). - Mikhail Kurkov, Jan 11 2022
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MAPLE
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bintr:= proc(p) proc(n) add(p(k) *binomial(n, k), k=0..n) end end:
invtr:= proc(p) local b;
b:= proc(n) option remember; local i;
`if`(n<1, 1, add(b(n-i) *p(i-1), i=1..n+1))
end;
end:
b:= invtr(bintr(a)):
a:= n-> `if`(n<0, 0, b(n-1)):
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MATHEMATICA
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a[n_] := Module[{A, B}, A = 1+x; For[k=1, k <= n, k++, B = (A /. x -> x/(1 - x))/(1-x) + O[x]^n // Normal; A = 1 + x*A*B]; SeriesCoefficient[A, {x, 0, n}]]; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Oct 23 2016, adapted from PARI *)
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PROG
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(PARI) {a(n)=local(A); if(n<0, 0, A=1+x+x*O(x^n); for(k=1, n, B=subst(A, x, x/(1-x))/(1-x)+x*O(x^n); A=1+x*A*B); polcoeff(A, n, x))}
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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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