A090366 -id:A090366 - cp's OEIS frontend

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

Paul D. Hanna, Nov 26 2003

nonn

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

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

Cf. A090366, A090367.
Cf. A074664.
Similar recurrences: A006014, A124758, A217924, A243499, A284005, A290615, A329369, A341392, A347204, A347205.

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

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 *)

{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))}

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

A090367 Shifts 1 place left under the INVERT transform of the BINOMIAL transform of the self-convolution cube of this sequence.

Original entry on oeis.org

1, 1, 5, 34, 276, 2509, 24739, 259815, 2873376, 33207790, 398897289, 4960652325, 63676368387, 841741913795, 11438028248093, 159536511439266, 2281321298635427, 33411684617642665, 500761214428795093, 7674842860939188928, 120209960716130232745

Offset: 0

Paul D. Hanna, Nov 26 2003

nonn

Alois P. Heinz, Table of n, a(n) for n = 0..200

Cf. A090365, A090366.

bintr:= proc(p) local b; b:= proc(n) option remember;
           add(p(k) *binomial(n,k), k=0..n) end
        end:
invtr:= proc(p) local b; b:= proc(n) option remember;
           `if`(n<1, 1, add(b(n-i) *p(i-1), i=1..n+1)) end
        end:
s:= proc(n) option remember; add(a(i)*a(n-i), i=0..n) end:
b:= invtr(bintr(n-> add(s(i)*a(n-i), i=0..n))):
a:= n-> `if`(n<0, 0, b(n-1)):
seq(a(n), n=0..25);  # Alois P. Heinz, Jun 28 2012

m = 25; A[] = 1; Do[A[x] = 1/(1 - A[x/(1-x)]^3*(x/(1-x))) + O[x]^m // Normal, {m}]; CoefficientList[A[x], x] (* Jean-François Alcover, Jun 04 2018 *)

{a(n)=local(A); if(n<0,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A^3,x,x/(1-x))/(1-x)+x*O(x^n); A=1+x*A*B);polcoeff(A,n,x))}

G.f.: A(x) = 1/(1 - A(x/(1-x))^3*x/(1-x) ).

cp's OEIS Frontend

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

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