A047891 -id:A047891 - cp's OEIS frontend

A054872 Number of (12345, 13245, 21345, 23145, 31245, 32145)-avoiding permutations.

1, 1, 2, 6, 24, 114, 600, 3372, 19824, 120426, 749976, 4762644, 30723792, 200778612, 1326360048, 8842981848, 59425117152, 402092408346, 2737156004376, 18732169337604, 128806616999184, 889479590046108, 6165939982059600, 42891532191557736, 299307319060137504

Offset: 0

Elisa Pergola (elisa(AT)dsi.unifi.it), May 26 2000

nonn

Hankel transform is A083667, the number of different antisymmetric relations on n labeled points. - Paul Barry, Jun 26 2008
Conjectured to be the number of permutations of length n+1 avoiding the partially ordered pattern (POP) {5>1, 1>2, 1>3, 1>4} of length 5. That is, conjectured to be the number of length n+1 permutations having no subsequences of length 5 in which the fifth element is the largest and the first element is the next largest - Sergey Kitaev, Dec 13 2020
This conjecture has been proven. There are six sets of permutations avoiding six size five permutations including the two sets discussed in this sequence that are known to match this sequence. A further two are conjectured to match this sequence. - Christian Bean, Jul 23 2024

			G.f. = 1 + x + 2*x^2 + 6*x^3 + 24*x^4 + 114*x^5 + 600*x^6 + 3372*x^7 + 19824*x^8 + ...

Alois P. Heinz, Table of n, a(n) for n = 0..1000 (terms n=1..200 from Vincenzo Librandi)
Michael H. Albert, Christian Bean, Anders Claesson, Émile Nadeau, Jay Pantone, and Henning Ulfarsson, PermPAL database.
Elena Barcucci, Alberto Del Lungo, Elisa Pergola, and Renzo Pinzani, Permutations avoiding an increasing number of length-increasing forbidden subsequences, Discrete Mathematics and Theoretical Computer Science 4, 2000, 31-44.
Christian Bean, Émile Nadeau, Jay Pantone, and Henning Ulfarsson, Permutations avoiding bipartite partially ordered patterns have a regular insertion encoding, The Electronic Journal of Combinatorics, Volume 31, Issue 3 (2024); arXiv preprint, arXiv:2312.07716 [math.CO], 2023.
Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, arXiv:1903.08946 [math.CO], 2019.
Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, The Electronic Journal of Combinatorics 26(3) (2019), P3.26.
Eric Weisstein's World of Mathematics, Legendre Polynomial.

Cf. A000108, A047891.

Set j=3 in the following: f := (x,j)->1-(j+1)*x- sqrt(1-2*(j+1)*x+(j-1)^2*x^2); t := (x,j)->sum(k!*x^k, k=1..(j-1)); s := (x,j)->x^(j-2)*(j-1)!*(f(x,j))/(2)+ t(x,j);

Table[SeriesCoefficient[x*(2-2*x-(1-8*x+4*x^2)^(1/2)),{x,0,n}],{n,1,20}] (* Vaclav Kotesovec, Oct 11 2012 *)
Table[2^(n-1) (LegendreP[n-1, 2] - LegendreP[n-3, 2])/(2n-3), {n, 1, 20}] (* Vladimir Reshetnikov, Nov 01 2015 *)

x='x+O('x^50); Vec(x*(2-2*x-(1-8*x+4*x^2)^(1/2))) \\ Altug Alkan, Nov 02 2015

G.f.: 1 + x*(2 - 2*x - (1 - 8*x + 4*x^2)^(1/2)). - corrected by Vaclav Kotesovec, Oct 11 2012
a(n) = 2*A047891(n-1), n>=2. - Philippe Deléham, Aug 17 2007
Recurrence: (n-1)*a(n) = 4*(2*n-5)*a(n-1) - 4*(n-4)*a(n-2). - Vaclav Kotesovec, Oct 11 2012
a(n) ~ sqrt(26*sqrt(3)-45)*(4+2*sqrt(3))^n/(sqrt(8*Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 11 2012
From Vladimir Reshetnikov, Nov 01 2015: (Start)
a(n) = 2^(n-1)*(LegendreP_{n-1}(2) - LegendreP_{n-3}(2))/(2*n-3).
For n > 2, a(n) = 6*hypergeom([2-n,3-n], [2], 3).
(End)
G.f. satisfies: A(x) = x * Sum_{n>=0} ( A(x)/x + 4*x + x/A(x) )^n / (2*4^n). - Paul D. Hanna, Mar 24 2016
G.f. satisfies: A(x) = x * Sum_{n>=0} ( A(x)/x + 4*x - x/A(x) )^n / 4^n. - Paul D. Hanna, Mar 24 2016

a(0)=1 prepended by Alois P. Heinz, Dec 13 2020

A082302 Expansion of g.f.: (1 - 5x - sqrt(25x^2 - 14x + 1))/(2x).

Original entry on oeis.org

1, 6, 42, 330, 2814, 25422, 239442, 2326434, 23151030, 234784662, 2417832186, 25216231866, 265796560302, 2827138163550, 30306009654690, 327081253546770, 3551148743559270, 38758882760119590, 425024567305557450

Offset: 0

Benoit Cloitre, May 10 2003

nonn

More generally coefficients of (1 - m*x - sqrt(m^2*x^2 - (2*m + 4)*x + 1))/(2*x) are given by a(0)=1 and, for n > 0, a(n) = (1/n)*Sum_{k=0..n} (m+1)^k*C(n,k)*C(n,k-1).

Hankel transform is 6^C(n+1,2). - Philippe Deléham, Feb 11 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
Paul Barry, Generalized Catalan Numbers Associated with a Family of Pascal-like Triangles, J. Int. Seq., Vol. 22 (2019), Article 19.5.8.

Cf. A006318, A047891.

Concatenation([1],List([1..20],n->(1/n)*Sum([0..n],k->6^k*Binomial(n,k)*Binomial(n,k-1)))); # Muniru A Asiru, Apr 05 2018

m:=50; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1-5*x-Sqrt(25*x^2-14*x+1))/(2*x))); // G. C. Greubel, Aug 16 2018

A082302_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
for w from 1 to n do a[w] := 6*a[w-1]+add(a[j]*a[w-j-1], j=1..w-1) od; convert(a,list)end: A082302_list(18); # Peter Luschny, May 19 2011
a := n -> `if`(n=0, 1, 6*hypergeom([1 - n, -n], [2], 6)):
seq(simplify(a(n)), n=0..18); # Peter Luschny, May 22 2017

Table[SeriesCoefficient[(1-5*x-Sqrt[25*x^2-14*x+1])/(2*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)

a(n)=if(n<1,1,sum(k=0,n,6^k*binomial(n,k)*binomial(n,k-1))/n)

x='x+O('x^99); Vec((1-5*x-(25*x^2-14*x+1)^(1/2))/(2*x)) \\ Altug Alkan, Apr 04 2018

Equals 6*A078018(n) for n > 0.
a(0)=1; for n > 0, a(n) = (1/n)*Sum_{k=0..n} 6^k*C(n, k)*C(n, k-1).
D-finite with recurrence: (n+1)*a(n) + 7*(1-2n)*a(n-1) + 25*(n-2)*a(n-2) = 0. - R. J. Mathar, Dec 08 2011
a(n) ~ sqrt(12 + 7*sqrt(6))*(7 + 2*sqrt(6))^n/(2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2012
a(n) = 6*hypergeom([1 - n, -n], [2], 6) for n > 0. - Peter Luschny, May 22 2017
G.f.: 1/(1 - 5*x - x/(1 - 5*x - x/(1 - 5*x - x/(1 - 5*x - x/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Apr 04 2018

A082366 G.f.: (1 - 7x - sqrt(49x^2 - 18x + 1))/(2x).

Original entry on oeis.org

1, 8, 72, 712, 7560, 84616, 985032, 11814728, 145043208, 1813915912, 23029334856, 296050614216, 3846007927944, 50412893051784, 665925356663496, 8855844075949128, 118467982501096968, 1593108078166843912

Offset: 0

Benoit Cloitre, May 10 2003

nonn

More generally coefficients of (1 - m*x - sqrt(m^2*x^2 - (2*m+4)*x + 1))/(2*x) are given by a(0)=1 and a(n) = (1/n)*Sum_{k=0..n} (m+1)^k*C(n,k)*C(n,k-1) for n > 0.

Hankel transform is 8^C(n+1,2). - Philippe Deléham, Feb 11 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

Cf. A006318, A047891.

Concatenation([1],List([1..20],n->(1/n)*Sum([0..n],k->8^k*Binomial(n,k)*Binomial(n,k-1)))); # Muniru A Asiru, Apr 05 2018

m:=25; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1-7*x-Sqrt(49*x^2-18*x+1))/(2*x))); // G. C. Greubel, Sep 16 2018

CoefficientList[Series[(1-7x-Sqrt[49x^2-18x+1])/(2x),{x,0,20}],x]  (* Harvey P. Dale, Feb 22 2011 *)

a(n)=if(n<1,1,sum(k=0,n,8^k*binomial(n,k)*binomial(n,k-1))/n)

x='x+O('x^99); Vec((1-7*x-(49*x^2-18*x+1)^(1/2))/(2*x)) \\ Altug Alkan, Apr 04 2018

a(0)=1; a(n) = (1/n)*Sum_{k=0..n} 8^k*C(n, k)*C(n, k-1) for n > 0.
D-finite with recurrence: (n+1)*a(n) + 9*(1-2n)*a(n-1) + 49*(n-2)*a(n-2) = 0. - R. J. Mathar, Dec 08 2011
a(n) ~ sqrt(16+18*sqrt(2))*(9+4*sqrt(2))^n/(2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2012
G.f.: 1/(1 - 7*x - x/(1 - 7*x - x/(1 - 7*x - x/(1 - 7*x - x/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Apr 04 2018

A082367 G.f.: (1-8x-sqrt(64x^2-20x+1))/(2x).

Original entry on oeis.org

1, 9, 90, 981, 11430, 140058, 1782900, 23369805, 313426350, 4281280686, 59360821740, 833312907522, 11820849447420, 169182862497108, 2440064033240040, 35428651752626109, 517446157031236350

Offset: 0

Benoit Cloitre, May 10 2003

nonn

More generally coefficients of (1-m*x-sqrt(m^2*x^2-(2*m+4)*x+1))/(2*x) are given by a(0)=1 and n>0 a(n)=(1/n)*Sum_{k=0..n} (m+1)^k*C(n,k)*C(n,k-1).

Hankel transform is 9^C(n+1,2). - Philippe Deléham, Feb 11 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

Cf. A006318, A047891.

m:=25; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1-8*x-Sqrt(64*x^2-20*x+1))/(2*x))); // G. C. Greubel, Sep 16 2018

f:= gfun:-rectoproc({64*n*a(n)+(-30-20*n)*a(1+n)+(3+n)*a(n+2), a(0) = 1, a(1) = 9}, a(n), remember):
map(f, [$0..30]); # Robert Israel, Mar 16 2018

Table[SeriesCoefficient[(1-8*x-Sqrt[64*x^2-20*x+1])/(2*x),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)

a(n)=if(n<1,1,sum(k=0,n,9^k*binomial(n,k)*binomial(n,k-1))/n)

x='x+O('x^99); Vec((1-8*x-(64*x^2-20*x+1)^(1/2))/(2*x)) \\ Altug Alkan, Apr 04 2018

a(0)=1; for n > 0, a(n) = (1/n)*Sum_{k=0..n} 9^k*C(n, k)*C(n, k-1).
D-finite with recurrence: (n+1)*a(n) + 10*(1-2n)*a(n-1) + 64*(n-2)*a(n-2) = 0. - R. J. Mathar, Dec 08 2011 Recurrence follows from the D.E. (x-20*x^2+64*x^3)*y' + (1-10*x)*y - 1 - 8*x = 0 satisfied by the g.f.. - Robert Israel, Mar 16 2018
a(n) ~ sqrt(3)*2^(4*n+1)/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 14 2012
G.f.: 1/(1 - 8*x - x/(1 - 8*x - x/(1 - 8*x - x/(1 - 8*x - x/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Apr 04 2018

A121576 Riordan array (2-2x-sqrt(1-8x+4x^2), (1-2x-sqrt(1-8x+4x^2))/2).

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 24, 24, 8, 1, 114, 123, 51, 11, 1, 600, 672, 312, 87, 14, 1, 3372, 3858, 1914, 618, 132, 17, 1, 19824, 22992, 11904, 4218, 1068, 186, 20, 1, 120426, 140991, 75183, 28383, 8043, 1689, 249, 23, 1, 749976, 884112, 481704, 190347, 58398, 13929, 2508, 321, 26, 1

Offset: 0

Paul Barry, Aug 08 2006

Inverse of Riordan array (1/(1+2*x), x*(1-x)/(1+2*x)).
Row sums are A047891; first column is A054872. Signed version given by A121575.
Triangle T(n,k), 0 <= k <= n, read by rows, given by [2, 1, 3, 1, 3, 1, 3, 1, 3, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 09 2006

			Triangle begins
     1;
     2,    1;
     6,    5,    1;
    24,   24,    8,   1;
   114,  123,   51,  11,   1;
   600,  672,  312,  87,  14,  1;
  3372, 3858, 1914, 618, 132, 17, 1;
From _Paul Barry_, Apr 27 2009: (Start)
Production matrix is
  2, 1,
  2, 3, 1,
  2, 3, 3, 1,
  2, 3, 3, 3, 1,
  2, 3, 3, 3, 3, 1,
  2, 3, 3, 3, 3, 3, 1,
  2, 3, 3, 3, 3, 3, 3, 1
In general, the production matrix of the inverse of (1/(1-rx),x(1-x)/(1-rx)) is
  -r, 1,
  -r, 1 - r, 1,
  -r, 1 - r, 1 - r, 1,
  -r, 1 - r, 1 - r, 1 - r, 1,
  -r, 1 - r, 1 - r, 1 - r, 1 - r, 1,
  -r, 1 - r, 1 - r, 1 - r, 1 - r, 1 - r, 1,
  -r, 1 - r, 1 - r, 1 - r, 1 - r, 1 - r, 1 - r, 1 (End)

G. C. Greubel, Rows n=0..100 of triangle, flattened

[[(&+[ 2^j*Binomial(n,j)*Binomial(2*n-k-j, n)*(4-9*j+3*j^2-6*(j-1)*n + 2*n^2)/((n-j+2)*(n-j+1))/2: j in [0..(n-k)]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Nov 02 2018

Flatten[Table[Sum[Binomial[n,i]Binomial[2n-k-i,n](4-9i+3i^2-6(i-1)n+2n^2)/((n-i+2)(n-i+1))2^i,{i,0,n-k}]/2,{n,0,8},{k,0,n}]]
(* Emanuele Munarini, May 18 2011 *)

create_list(sum(binomial(n,i)*binomial(2*n-k-i,n)*(4-9*i+3*i^2-6*(i-1)*n+2*n^2)/((n-i+2)*(n-i+1))*2^i,i,0,n-k)/2,n,0,9,k,0,n); /* Emanuele Munarini, May 18 2011 */

for(n=0,10, for(k=0,n, print1(sum(j=0, n-k, 2^j*binomial(n,j) *binomial(2*n-k-j, n)*(4-9*j+3*j^2-6*(j-1)*n + 2*n^2)/((n-j+2)*(n-j+1)))/2, ", "))) \\ G. C. Greubel, Nov 02 2018

T(n,k) = [x^(n-k)](1-2*x-2*x^2)*(1+2*x)^n/(1-x)^(n+1) = (1/2)*Sum_{i=0..n-k} binomial(n,i) * binomial(2*n-k-i,n) * (4 - 9*i + 3*i^2 - 6*(i-1)*n + 2*n^2)/((n-i+2)*(n-i+1))*2^i. - Emanuele Munarini, May 18 2011

A219535 G.f. satisfies A(x) = 1 + x(2A(x)^2 + A(x)^3).

Original entry on oeis.org

1, 3, 21, 192, 2001, 22539, 267276, 3287496, 41556585, 536565225, 7046232285, 93820316412, 1263673602300, 17186898452772, 235709926636296, 3256050894487824, 45263067114496665, 632721425905230213, 8888476706476318047, 125418490224196533096, 1776734673565844413929

Offset: 0

Paul D. Hanna, Nov 21 2012

nonn

			G.f.: A(x) = 1 + 3*x + 21*x^2 + 192*x^3 + 2001*x^4 + 22539*x^5 +...
Related expansions:
A(x)^2 = 1 + 6*x + 51*x^2 + 510*x^3 + 5595*x^4 + 65148*x^5 +...
A(x)^3 = 1 + 9*x + 90*x^2 + 981*x^3 + 11349*x^4 + 136980*x^5 +...
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + 3*x + 12*x^2 + 57*x^3 + 300*x^4 + 1686*x^5 +...+ A047891(n+1)*x^n +...

Michael De Vlieger, Table of n, a(n) for n = 0..799
Elżbieta Liszewska, Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.

Column k=2 of A336575.

Cf. A047891, A219534, A219536.

CoefficientList[1/x*InverseSeries[Series[2*x^2/(1-2*x-Sqrt[1-8*x+4*x^2]), {x, 0, 21}], x],x] (* Vaclav Kotesovec, Dec 28 2013 *)

/* Formula A(x) = 1 + x*(2*A(x)^2 + A(x)^3): */
{a(n)=my(A=1);for(i=1,n,A=1+x*(2*A^2+A^3) +x*O(x^n));polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

/* Formula using Series Reversion: */
{a(n)=my(A=1,G=(1-2*x-sqrt(1-8*x+4*x^2+x^3*O(x^n)))/(2*x));A=(1/x)*serreverse(x/G);polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

a(n) = sum(k=0, n, 2^(n-k)*binomial(n, k)*binomial(2*n+k+1, n)/(2*n+k+1)); \\ Seiichi Manyama, Jul 28 2020

a(n) = sum(k=0, n, 2^k*binomial(2*n+1, k)*binomial(3*n-k, n-k))/(2*n+1); \\ Seiichi Manyama, Jul 28 2020

Let G(x) = (1-2*x - sqrt(1 - 8*x + 4*x^2)) / (2*x), then g.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion(x/G(x)),
(2) A(x) = G(x*A(x)) and G(x) = A(x/G(x)),
where x*G(x) is the g.f. of A047891.
Recurrence: 2*n*(2*n+1)*(11*n - 16)*a(n) = (649*n^3 - 1593*n^2 + 1130*n - 240)*a(n-1) + 16*(n-2)*(2*n-3)*(11*n-5)*a(n-2). - Vaclav Kotesovec, Dec 28 2013
a(n) ~ sqrt((33+17*sqrt(33))/11) * ((59+11*sqrt(33))/8)^n / (4 * sqrt(2*Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 28 2013
From Seiichi Manyama, Jul 28 2020: (Start)
a(n) = Sum_{k=0..n} 2^(n-k) * binomial(n,k) * binomial(2*n+k+1,n)/(2*n+k+1).
a(n) = (1/(2*n+1)) * Sum_{k=0..n} 2^k * binomial(2*n+1,k) * binomial(3*n-k,n-k). (End)
From Seiichi Manyama, Aug 10 2023: (Start)
a(n) = (1/n) * Sum_{k=0..n-1} (-2)^k * 3^(n-k) * binomial(n,k) * binomial(3*n-k,n-1-k) for n > 0.
a(n) = (1/n) * Sum_{k=1..n} 3^k * binomial(n,k) * binomial(2*n,k-1) for n > 0. (End)

A371391 Expansion of (1/x) * Series_Reversion( x * (1-x) / (1+2*x)^2 ).

Original entry on oeis.org

1, 5, 34, 269, 2326, 21314, 203428, 2000957, 20142862, 206524790, 2149261852, 22644243218, 241061343004, 2589022298084, 28019201644744, 305254481274269, 3345077342003134, 36846738570089774, 407754101877613804, 4531049315843043974, 50538820796852529364

Offset: 0

Seiichi Manyama, Mar 21 2024

nonn

Index entries for reversions of series

Cf. A047891, A371392.

Table[Sum[2^k*Binomial[2*(n+1), k]*Binomial[2*n-k, n-k]/(n+1), {k,0,n}], {n,0,30}] (* Vaclav Kotesovec, Jul 31 2025 *)

my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x)/(1+2*x)^2)/x)

a(n) = sum(k=0, n, 2^k*binomial(2*(n+1), k)*binomial(2*n-k, n-k))/(n+1);

a(n) = (1/(n+1)) * Sum_{k=0..n} 2^k * binomial(2*(n+1),k) * binomial(2*n-k,n-k).
a(n) = (1/(n+1)) * [x^n] ( (1+2*x)^2 / (1-x) )^(n+1). - Seiichi Manyama, Jul 31 2025
a(n) ~ 2^(2*n-2) * 3^(n+2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Jul 31 2025

A348793 G.f. A(x) satisfies A(x) = (1 + x * A(x)^3) / (1 - 2 * x).

Original entry on oeis.org

1, 3, 15, 102, 807, 6951, 63240, 597864, 5815167, 57815553, 584919951, 6002197914, 62321630100, 653553174756, 6912106219176, 73642451396160, 789642274208271, 8515008918555573, 92281921130853213, 1004600177464845450, 10980406558088695599, 120454756647900759543

Offset: 0

Ilya Gutkovskiy, Nov 03 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..937

Cf. A001764, A047891, A346626.

nmax = 21; A[] = 0; Do[A[x] = (1 + x A[x]^3)/(1 - 2 x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[0] = 1; a[n_] := a[n] = 2 a[n - 1] + Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 21}]

a(n) = sum(k=0, n, 2^(n-k)*binomial(n, k)*binomial(n+2*k+1, n)/(n+2*k+1)); \\ Seiichi Manyama, Jul 24 2023

a(0) = 1; a(n) = 2 * a(n-1) + Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1).
a(n) ~ sqrt((2 + s^3)/(3*Pi*s*(1 - 2*r))) / (2*n^(3/2)*r^n), where r = (2 + (3*(-2 + sqrt(6))^(1/3))/2^(2/3) - 3/(2*(-2 + sqrt(6)))^(1/3)) / 4 = 0.084819663336750180604484695162155813902734598764355... and s = 1/2 + (-sqrt(2) + sqrt(3))/(2^(5/6)*(-2 + sqrt(6))^(1/3)) + 1/(2*(-2 + sqrt(6)))^(2/3) = 1.8064439323587723772036249693148814564378856424032... - Vaclav Kotesovec, Nov 04 2021
a(n) = Sum_{k=0..n} 2^(n-k) * binomial(n,k) * binomial(n+2*k+1,n) / (n+2*k+1). - Seiichi Manyama, Jul 24 2023

A172455 The case S(6,-4,-1) of the family of self-convolutive recurrences studied by Martin and Kearney.

Original entry on oeis.org

1, 7, 84, 1463, 33936, 990542, 34938624, 1445713003, 68639375616, 3676366634402, 219208706540544, 14397191399702118, 1032543050697424896, 80280469685284582812, 6725557192852592984064, 603931579625379293509683

Offset: 1

N. J. A. Sloane, Nov 20 2010

nonn

			G.f. = x + 7*x^2 + 84*x^3 + 1463*x^4 + 33936*x^5 + 990542*x^6 + 34938624*x^7 + ...
a(2) = 7 since (6*2 - 4) * a(2-1) - (a(1) * a(2-1)) = 7.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
R. J. Martin and M. J. Kearney, An exactly solvable self-convolutive recurrence, Aequat. Math., 80 (2010), 291-318. see p. 307.
R. J. Martin and M. J. Kearney, An exactly solvable self-convolutive recurrence, arXiv:1103.4936 [math.CO], 2011.
NIST Digital Library of Mathematical Functions, Airy Functions.
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.
Eric Weisstein's World of Mathematics, Airy Functions, contains the definitions of Ai(x), Bi(x).

Cf. A000079 S(1,1,-1), A000108 S(0,0,1), A000142 S(1,-1,0), A000244 S(2,1,-2), A000351 S(4,1,-4), A000400 S(5,1,-5), A000420 S(6,1,-6), A000698 S(2,-3,1), A001710 S(1,1,0), A001715 S(1,2,0), A001720 S(1,3,0), A001725 S(1,4,0), A001730 S(1,5,0), A003319 S(1,-2,1), A005411 S(2,-4,1), A005412 S(2,-2,1), A006012 S(-1,2,2), A006318 S(0,1,1), A047891 S(0,2,1), A049388 S(1,6,0), A051604 S(3,1,0), A051605 S(3,2,0), A051606 S(3,3,0), A051607 S(3,4,0), A051608 S(3,5,0), A051609 S(3,6,0), A051617 S(4,1,0), A051618 S(4,2,0), A051619 S(4,3,0), A051620 S(4,4,0), A051621 S(4,5,0), A051622 S(4,6,0), A051687 S(5,1,0), A051688 S(5,2,0), A051689 S(5,3,0), A051690 S(5,4,0), A051691 S(5,5,0), A053100 S(6,1,0), A053101 S(6,2,0), A053102 S(6,3,0), A053103 S(6,4,0), A053104 S(7,1,0), A053105 S(7,2,0), A053106 S(7,3,0), A062980 S(6,-8,1), A082298 S(0,3,1), A082301 S(0,4,1), A082302 S(0,5,1), A082305 S(0,6,1), A082366 S(0,7,1), A082367 S(0,8,1), A105523 S(0,-2,1), A107716 S(3,-4,1), A111529 S(1,-3,2), A111530 S(1,-4,3), A111531 S(1,-5,4), A111532 S(1,-6,5), A111533 S(1,-7,6), A111546 S(1,0,1), A111556 S(1,1,1), A143749 S(0,10,1), A146559 S(1,1,-2), A167872 S(2,-3,2), A172450 S(2,0,-1), A172485 S(-1,-2,3), A177354 S(1,2,1), A292186 S(4,-6,1), A292187 S(3, -5, 1).

a[1] = 1; a[n_]:= a[n] = (6*n-4)*a[n-1] - Sum[a[k]*a[n-k], {k, 1, n-1}]; Table[a[n], {n, 1, 20}] (* Vaclav Kotesovec, Jan 19 2015 *)

{a(n) = local(A); if( n<1, 0, A = vector(n); A[1] = 1; for( k=2, n, A[k] = (6 * k - 4) * A[k-1] - sum( j=1, k-1, A[j] * A[k-j])); A[n])} /* Michael Somos, Jul 24 2011 */

S(v1, v2, v3, N=16) = {
  my(a = vector(N)); a[1] = 1;
  for (n = 2, N, a[n] = (v1*n+v2)*a[n-1] + v3*sum(j=1,n-1,a[j]*a[n-j])); a;
};
S(6,-4,-1)
\\ test: y = x*Ser(S(6,-4,-1,201)); 6*x^2*y' == y^2 - (2*x-1)*y - x
\\ Gheorghe Coserea, May 12 2017

a(n) = (6*n - 4) * a(n-1) - Sum_{k=1..n-1} a(k) * a(n-k) if n>1. - Michael Somos, Jul 24 2011
G.f.: x / (1 - 7*x / (1 - 5*x / (1 - 13*x / (1 - 11*x / (1 - 19*x / (1 - 17*x / ... )))))). - Michael Somos, Jan 03 2013
a(n) = 3/(2*Pi^2)*int((4*x)^((3*n-1)/2)/(Ai'(x)^2+Bi'(x)^2), x=0..inf), where Ai'(x), Bi'(x) are the derivatives of the Airy functions. [Vladimir Reshetnikov, Sep 24 2013]
a(n) ~ 6^n * (n-1)! / (2*Pi) [Martin + Kearney, 2011, p.16]. - Vaclav Kotesovec, Jan 19 2015
6*x^2*y' = y^2 - (2*x-1)*y - x, where y(x) = Sum_{n>=1} a(n)*x^n. - Gheorghe Coserea, May 12 2017
G.f.: x/(1 - 2*x - 5*x/(1 - 7*x/(1 - 11*x/(1 - 13*x/(1 - ... - (6*n - 1)*x/(1 - (6*n + 1)*x/(1 - .... Cf. A062980. - Peter Bala, May 21 2017

A364432 G.f. satisfies A(x) = 1 + xA(x)(2 + A(x)^3).

Original entry on oeis.org

1, 3, 18, 162, 1728, 20169, 249318, 3207600, 42500700, 576012060, 7947785448, 111269613006, 1576658688480, 22568473199358, 325855352769588, 4740157737123696, 69405108247439676, 1022070746845708740, 15127922880893671704, 224931239520535651464

Offset: 0

Seiichi Manyama, Jul 24 2023

nonn

Cf. A047891, A348793.

Cf. A364430, A364431.

A364432 := proc(n)
    add(2^(n-k)* binomial(n,k) * binomial(n+3*k+1,n) / (n+3*k+1),k=0..n) ;
end proc:
seq(A364432(n),n=0..70); # R. J. Mathar, Jul 25 2023

a(n) = sum(k=0, n, 2^(n-k)*binomial(n, k)*binomial(n+3*k+1, n)/(n+3*k+1));

a(n) = Sum_{k=0..n} 2^(n-k) * binomial(n,k) * binomial(n+3*k+1,n) / (n+3*k+1).

D-finite with recurrence 3*n*(3*n-1)*(3*n+1)*a(n) +2*(-74*n^3 -375*n^2+ 665*n -252)*a(n-1) +12*(-337*n^3 +1941*n^2 -2984*n +1092)*a(n-2) +144*(-70*n^3 +861*n^2 -3347*n +4152)*a(n-3) +432*(n-4)*(31*n^2 -314*n +735)*a(n-4) -2592*(10*n-51) *(n-4)*(n-5)*a(n-5) +15552*(n-5)*(n-6) *(n-4)*a(n-6)=0. - R. J. Mathar, Jul 25 2023

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A054872 Number of (12345, 13245, 21345, 23145, 31245, 32145)-avoiding permutations.

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A082302 Expansion of g.f.: (1 - 5*x - sqrt(25*x^2 - 14*x + 1))/(2*x).

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A082366 G.f.: (1 - 7*x - sqrt(49*x^2 - 18*x + 1))/(2*x).

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A082367 G.f.: (1-8*x-sqrt(64*x^2-20*x+1))/(2*x).

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A121576 Riordan array (2-2*x-sqrt(1-8*x+4*x^2), (1-2*x-sqrt(1-8*x+4*x^2))/2).

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A219535 G.f. satisfies A(x) = 1 + x*(2*A(x)^2 + A(x)^3).

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A082302 Expansion of g.f.: (1 - 5x - sqrt(25x^2 - 14x + 1))/(2x).

A082366 G.f.: (1 - 7x - sqrt(49x^2 - 18x + 1))/(2x).

A082367 G.f.: (1-8x-sqrt(64x^2-20x+1))/(2x).

A121576 Riordan array (2-2x-sqrt(1-8x+4x^2), (1-2x-sqrt(1-8x+4x^2))/2).

A219535 G.f. satisfies A(x) = 1 + x(2A(x)^2 + A(x)^3).

A364432 G.f. satisfies A(x) = 1 + xA(x)(2 + A(x)^3).