A090365 Shifts 1 place left under the INVERT transform of the BINOMIAL transform of this sequence.

1, 1, 3, 11, 47, 225, 1177, 6625, 39723, 251939, 1681535, 11764185, 86002177, 655305697, 5193232611, 42726002123, 364338045647, 3215471252769, 29331858429241, 276224445794785, 2682395337435723, 26832698102762435, 276221586866499839, 2923468922184615897

Offset: 0

Author

Paul D. Hanna, Nov 26 2003

Keywords

• nonn

Comments

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

The Hankel transform of this sequence without the a(0)=1 term is also A000178(n+1). - Michael Somos, Oct 02 2024

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..574

Crossrefs

Cf. A090366, A090367.

Cf. A074664.

Similar recurrences: A006014, A124758, A217924, A243499, A284005, A290615, A329369, A341392, A347204, A347205.

Programs

Maple

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)):
seq(a(n), n=0..30);  # Alois P. Heinz, Jun 28 2012

Mathematica

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 *)

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))}

Formula

G.f.: A(x) satisfies A(x) = 1/(1 - A(x/(1-x))*x/(1-x) ).

a(n) = Sum_{k = 0..n} A085838(n, k). - Philippe Deléham, Jun 04 2004

G.f.: 1/x-1-1/(B(x)-1) where B(x) = g.f. for A000110 the Bell numbers. - Vladeta Jovovic, Aug 08 2004

a(n) = Sum_{k=0..n} A094456(n,k). - Philippe Deléham, Nov 07 2007

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

From Sergei N. Gladkovskii, Jan 06 2012 - May 12 2013: (Start)

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