A030266 -id:A030266 - cp's OEIS frontend

A110447 Number of permutations containing 3241 patterns only as part of 35241 patterns.

1, 1, 2, 6, 23, 104, 531, 2982, 18109, 117545, 808764, 5862253, 44553224, 353713232, 2924697019, 25124481690, 223768976093, 2062614190733, 19646231085928, 193102738376890, 1956191484175505, 20401540100814142, 218825717967033373, 2411606083999341827

Offset: 0

David Callan, Jul 20 2005

nonn

a(n) = # permutations on [n] in which the (scattered) pattern 3241 only occurs as part of a 35241 pattern. For example, a(5) counts all 24 permutations on [4] except 3241 and the permutation p = 42531 is not counted by a(6) because the entries 4251 form a 3241 pattern but p fails to contain an entry larger than 5 between its entries 4 and 2.

a(n) = # (31-4-2)-avoiding perms on [n]. (31-4-2)-avoiding means the "3" and "1" must be consecutive in the permutation while the "4" and "2' may be scattered. For example, 35142 contains the (scattered) pattern 3-1-4-2 but avoids 31-4-2. - David Callan, Oct 11 2005

David Callan, A combinatorial interpretation of the eigensequence for composition, arXiv:math/0507169 [math.CO], 2005.
David Callan, A Combinatorial Interpretation of the Eigensequence for Composition, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.4.
David Callan, A Wilf equivalence related to two stack sortable permutations, arXiv:math/0510211 [math.CO], 2005.
David Callan, Lagrange Inversion Counts 3bar-5241-Avoiding Permutations, J. Int. Seq. 14 (2011) # 11.9.4
Lara Pudwell, Enumeration schemes for permutations avoiding barred patterns, El. J. Combinat. 17 (1) (2010) R29.

This sequence is A030266 shifted left.

A:= proc(n) option remember; unapply(`if`(n=0, x,
      A(n-1)(x)+coeff(A(n-1)(A(n-1)(x)), x, n) *x^(n+1)), x)
    end:
a:= n-> coeff(A(1+n)(x), x, 1+n):
seq(a(n), n=0..23);  # Alois P. Heinz, Jul 10 2023

(* The following recurrence for a(n) is derived in the first linked paper *)
a[0]=c[1]=1
a[n_]/;n>=1 := a[n] = Sum[a[i]c[n-i], {i, 0, n-1}]
c[n_]/;n>=2 := c[n] = Sum[i a[n-1, i], {i, n-1}]
a[n_, k_]/;1<=k<=n-1 := a[n, k] = Sum[a[i]a[n-1-i, j], {i, 0, k-1}, {j, k-i, n-1-i}]
a[ n_, n_ ]/;n>=1 := a[n, n] = a[n-1]
(* The following Mathematica code generates all the permutations counted by a(n).
Run the code; then Aset[n] returns the permutations counted by a(n). *)
Aset[0] = { { } }
Aset[1] = { {1} }
Cset[1] = { {1} }
Aset[n_, n_ ]/;n>=1 := Aset[n, n ] = Map[Join[{n}, # ]&, Aset[n-1 ] ]
processBn[n_, single_, i_] := Module[{base=Drop[Range[n], {i}]}, Join[{i}, base[[single]] ] ]
Cset[n_]/;n>=2 := Cset[n] = Flatten[Table[Map[processBn[n, #, i]&, Aset[n-1, j-1]], {j, 2, n}, {i, j-1}], 2]
processAn[pair_, j_]:=Module[{p1=pair[[1]], p2=pair[[2]]}, Flatten[Insert[j+p2, p1, 2] ] ]
Aset[ n_ ]/;n>=2 := Aset[ n ] = Flatten[ Table[ Map[ processAn[ #, j ]&, CartesianProduct[ Aset[ j ], Cset[ n-j ] ] ], {j, 0, n-1} ], 1 ]
processAnk[n_, k_, pair_, j_]:=Module[{p1=pair[[1]], p2=pair[[2]], base}, base=Complement[Range[j+1, n], {k}]; Join[{k}, p1, base[[p2]]] ]
Aset[ n_, k_ ]/;1<=k<=n-1 := Aset[ n, k ] = Flatten[ Table[ Map[ processAnk[ n, k, #, j ]&, CartesianProduct[ Aset[ j ], Aset[ n-1-j, r ] ] ], {j, 0, k-1}, {r, k-j, n-1-j} ], 2 ]

A379598 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where column k is the expansion of B(x)^k, where B(x) is the g.f. of A110447.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 5, 6, 0, 1, 4, 9, 16, 23, 0, 1, 5, 14, 31, 62, 104, 0, 1, 6, 20, 52, 123, 278, 531, 0, 1, 7, 27, 80, 213, 552, 1398, 2982, 0, 1, 8, 35, 116, 340, 964, 2750, 7718, 18109, 0, 1, 9, 44, 161, 513, 1561, 4784, 14976, 46083, 117545, 0

Offset: 0

Seiichi Manyama, Feb 27 2025

			Square array begins:
  1,   1,    1,    1,    1,    1,     1, ...
  0,   1,    2,    3,    4,    5,     6, ...
  0,   2,    5,    9,   14,   20,    27, ...
  0,   6,   16,   31,   52,   80,   116, ...
  0,  23,   62,  123,  213,  340,   513, ...
  0, 104,  278,  552,  964, 1561,  2400, ...
  0, 531, 1398, 2750, 4784, 7755, 11987, ...

Columns k=0..1 give A000007, A110447 (A030266(n+1)).

Cf. A379599, A380178.

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

See A030266.

A001028 E.g.f. satisfies A'(x) = 1 + A(A(x)), A(0)=0.

Original entry on oeis.org

1, 1, 2, 7, 37, 269, 2535, 29738, 421790, 7076459, 138061343, 3089950076, 78454715107, 2238947459974, 71253947372202, 2511742808382105, 97495087989736907, 4145502184671892500, 192200099033324115855, 9676409879981926733908, 527029533717566423156698

Offset: 1

Peter J. Cameron

The e.g.f. is diverging (see the Math Overflow link). - Pietro Majer, Jan 29 2017

This functional equation (for f(x)=1+A(x-1)) was the subject of problem B5 of the 2010 Putnam exam.

Vaclav Kotesovec, Table of n, a(n) for n = 1..320 (first 100 terms from Alois P. Heinz)
P. J. Cameron, Sequence operators from groups, Linear Alg. Applic., 226-228 (1995), 109-113.
Math Overflow, f' = exp(f^(-1)), again, January 2017.

Cf. A030266, A035049.

A:= proc(n) option remember; local T; if n=0 then 0 else T:= A(n-1); unapply(convert(series(Int(1+T(T(x)), x), x, n+1), polynom), x) fi end: a:= n-> coeff(A(n)(x), x, n)*n!: seq(a(n), n=1..22); # Alois P. Heinz, Aug 23 2008

terms = 21; A[] = 0; Do[A[x] = x + Integrate[A[A[x]], x] + O[x]^(n+1) // Normal, {n, terms}];
Rest[CoefficientList[A[x], x]]*Range[terms]! (* Jean-François Alcover, Dec 07 2011, updated Jan 10 2018 *)

Co(n,k,a):= if k=1 then a(n) else sum(a(i+1)*Co(n-i-1,k-1,a), i,0,n-k); a(n):= if n=1 then 1 else (1/n)*sum(Co(n-1,k,a)*a(k),k,1,n-1); makelist(n!*a(n),n,1,7); /* Vladimir Kruchinin, Jun 30 2011 */

{a(n) = my(A=x); for(i=1,n, A = serreverse(intformal(1/(1+A) +x*O(x^n)))); n!*polcoeff(A,n)}
for(n=1,25,print1(a(n),", ")) \\ Paul D. Hanna, Jun 27 2015

E.g.f. satisfies: A(x) = Series_Reversion( Integral 1/(1 + A(x)) dx ). - Paul D. Hanna, Jun 27 2015

More terms from Christian G. Bower, Oct 15 1998
Corrected by Alois P. Heinz, Aug 23 2008
Two more terms from Sean A. Irvine, Feb 22 2012

A120970 G.f. A(x) satisfies A(x/A(x)^2) = 1 + x ; thus A(x) = 1 + Series_Reversion(x/A(x)^2).

Original entry on oeis.org

1, 1, 2, 9, 60, 504, 4946, 54430, 655362, 8496454, 117311198, 1711459903, 26228829200, 420370445830, 7021029571856, 121859518887327, 2192820745899978, 40831103986939664, 785429260324068156, 15585831041632684997, 318649154587152781210, 6704504768568697046504

Offset: 0

Paul D. Hanna, Jul 20 2006

nonn

From Paul D. Hanna, Nov 16 2008: (Start)
More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
A(x) = 1 + x*G(x)^(m+k) where G(x) = A(x*G(x)^k) and G(x/A(x)^k) = A(x);
thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1. (End)

			G.f.: A(x) = 1 + x + 2*x^2 + 9*x^3 + 60*x^4 + 504*x^5 + 4946*x^6 + ...
Related expansions.
A(x)^2 = 1 + 2*x + 5*x^2 + 22*x^3 + 142*x^4 + 1164*x^5 + 11221*x^6 + ...
A(A(x)-1) = 1 + x + 4*x^2 + 26*x^3 + 218*x^4 + 2151*x^5 + 23854*x^6 + ...
A(A(x)-1)^2 = 1 + 2*x + 9*x^2 + 60*x^3 + 504*x^4 + 4946*x^5 + ...
x/A(x)^2 = x - 2*x^2 - x^3 - 10*x^4 - 73*x^5 - 662*x^6 - 6842*x^7 - ...
Series_Reversion(x/A(x)^2) = x + 2*x^2 + 9*x^3 + 60*x^4 + 504*x^5 + 4946*x^6 + ...
To illustrate the formula a(n) = [x^(n-1)] 2*A(x)^(2*n)/(2*n),
form a table of coefficients in A(x)^(2*n) as follows:
  A^2:  [(1), 2,   5,   22,   142,   1164,   11221,   121848, ...];
  A^4:  [ 1, (4), 14,   64,   397,   3116,   29002,   306468, ...];
  A^6:  [ 1,  6, (27), 134,   825,   6270,   56492,   580902, ...];
  A^8:  [ 1,  8,  44, (240), 1502,  11200,   98144,   983016, ...];
  A^10: [ 1, 10,  65,  390, (2520), 18672,  160115,  1565260, ...];
  A^12: [ 1, 12,  90,  592,  3987, (29676), 250730,  2399388, ...];
  A^14: [ 1, 14, 119,  854,  6027,  45458, (381010), 3582266, ...]; ...
in which the main diagonal forms the initial terms of this sequence:
[2/2*(1), 2/4*(4), 2/6*(27), 2/8*(240), 2/10*(2520), 2/12*(29676), ...].

Vaclav Kotesovec, Table of n, a(n) for n = 0..350

Cf. A120971; variants: A120972, A120974, A120976, A030266, A067145, A107096.
Cf. related variants: A145347, A145348, A147664, A145349, A145350. - Paul D. Hanna, Nov 16 2008
Cf. A381602.

terms = 21; A[] = 1; Do[A[x] = 1 + x*A[A[x] - 1]^2 + O[x]^j // Normal, {j, terms}]; CoefficientList[A[x], x] (* Jean-François Alcover, Jan 15 2018 *)

{a(n)=local(A=[1,1]);for(i=2,n,A=concat(A,0); A[ #A]=-Vec(subst(Ser(A),x,x/Ser(A)^2))[ #A]);A[n+1]}
for(n=0,30,print1(a(n),", "))

/* This sequence is generated when k=2, m=0: A(x/A(x)^k) = 1 + x*A(x)^m */ {a(n,k=2,m=0)=local(A=sum(i=0,n-1,a(i,k,m)*x^i));if(n==0,1,polcoeff((m+k)/(m+k*n)*A^(m+k*n),n-1))} \\ Paul D. Hanna, Nov 16 2008
for(n=0,30,print1(a(n),", "))

b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(2*n+k, j)/(2*n+k)*b(n-j, 2*j)));
a(n) = if(n==0, 1, b(n-1, 2)); \\ Seiichi Manyama, Jun 04 2025

G.f. satisfies: A(x) = 1 + x*A(A(x) - 1)^2.
Let B(x) be the g.f. of A120971, then B(x) and g.f. A(x) are related by:
(a) B(x) = A(A(x)-1),
(b) B(x) = A(x*B(x)^2),
(c) A(x) = B(x/A(x)^2),
(d) A(x) = 1 + x*B(x)^2,
(e) B(x) = 1 + x*B(x)^2*B(A(x)-1)^2,
(f) A(B(x)-1) = B(A(x)-1) = B(x*B(x)^2).
a(n) = [x^(n-1)] (1/n)*A(x)^(2n) for n>=1 with a(0)=1; i.e., a(n) equals 1/n times the coefficient of x^(n-1) in A(x)^(2n) for n>=1. [Paul D. Hanna, Nov 16 2008]
From Seiichi Manyama, Jun 04 2025: (Start)
Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120971.
b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(2*n+k,j)/(2*n+k) * b(n-j,2*j).
a(n) = b(n-1,2) for n > 0. (End)

A120972 G.f. A(x) satisfies A(x/A(x)^3) = 1 + x ; thus A(x) = 1 + series_reversion(x/A(x)^3).

Original entry on oeis.org

1, 1, 3, 21, 217, 2814, 42510, 718647, 13270944, 263532276, 5567092665, 124143735663, 2905528740060, 71058906460091, 1809695198254281, 47861102278428198, 1311488806252697283, 37164457324943708739, 1087356593493807164289, 32801308084353988297404

Offset: 0

Paul D. Hanna, Jul 20 2006

nonn

More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then
A(x) = 1 + x*G(x)^(m+k) where G(x) = A(x*G(x)^k) and G(x/A(x)^k) = A(x);
thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1.

			G.f.: A(x) = 1 + x + 3*x^2 + 21*x^3 + 217*x^4 + 2814*x^5 + 42510*x^6 +...
Related expansions.
A(x)^3 = 1 + 3*x + 12*x^2 + 82*x^3 + 813*x^4 + 10212*x^5 + 150699*x^6 +...
A(A(x)-1) = 1 + x + 6*x^2 + 60*x^3 + 776*x^4 + 11802*x^5 + 201465*x^6 +...
A(A(x)-1)^3 = 1 + 3*x + 21*x^2 + 217*x^3 + 2814*x^4 + 42510*x^5 +...
x/A(x)^3 = x - 3*x^2 - 3*x^3 - 37*x^4 - 420*x^5 - 5823*x^6 -...
Series_Reversion(x/A(x)^3) = x + 3*x^2 + 21*x^3 + 217*x^4 + 2814*x^5 + 42510*x^6 +...
To illustrate the formula a(n) = [x^(n-1)] 3*A(x)^(3*n)/(3*n),
form a table of coefficients in A(x)^(3*n) as follows:
  A^3:  [(1), 3,  12,   82,    813,   10212,   150699,   2503233, ...];
  A^6:  [ 1, (6), 33,  236,   2262,   27270,   388906,   6289080, ...];
  A^9:  [ 1,  9, (63), 489,   4671,   54684,   756012,  11904813, ...];
  A^12: [ 1, 12, 102, (868),  8445,   97260,  1310040,  20112516, ...];
  A^15: [ 1, 15, 150, 1400, (14070), 161343,  2130505,  31961175, ...];
  A^18: [ 1, 18, 207, 2112,  22113, (255060), 3324003,  48876264, ...];
  A^21: [ 1, 21, 273, 3031,  33222,  388563, (5030529), 72769014, ...]; ...
in which the main diagonal forms the initial terms of this sequence:
[3/3*(1), 3/6*(6), 3/9*(63), 3/12*(868), 3/15*(14070), 3/18*(255060), ...].

Cf. A120973; variants: A120970, A120974, A120976, A030266, A067145, A107096.

Cf. A381603.

terms = 18; A[] = 1; Do[A[x] = 1 + x*A[A[x] - 1]^3 + O[x]^j // Normal, {j, terms}]; CoefficientList[A[x], x] (* Jean-François Alcover, Jan 15 2018 *)

{a(n)=local(A=[1,1]);for(i=2,n,A=concat(A,0); A[ #A]=-Vec(subst(Ser(A),x,x/Ser(A)^3))[ #A]);A[n+1]}

{a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=1+x*subst(A^3,x,A-1+x*O(x^n)));polcoeff(A,n)}

/* This sequence is generated when k=3, m=0: A(x/A(x)^k) = 1 + x*A(x)^m */
{a(n, k=3, m=0)=local(A=sum(i=0, n-1, a(i, k, m)*x^i)); if(n==0, 1, polcoeff((m+k)/(m+k*n)*A^(m+k*n), n-1))}
for(n=0,25,print1(a(n),", "))

b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(3*n+k, j)/(3*n+k)*b(n-j, 3*j)));
a(n) = if(n==0, 1, b(n-1, 3)); \\ Seiichi Manyama, Jun 04 2025

G.f. satisfies: A(x) = 1 + x*A(A(x) - 1)^3.
a(n) = [x^(n-1)] A(x)^(3*n)/n for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in A(x)^(3*n)/n for n>=1 (see comment).
Let B(x) be the g.f. of A120973, then B(x) and g.f. A(x) are related by:
(a) B(x) = A(A(x)-1),
(b) B(x) = A(x*B(x)^3),
(c) A(x) = B(x/A(x)^3),
(d) A(x) = 1 + x*B(x)^3,
(e) B(x) = 1 + x*B(x)^3*B(A(x)-1)^3,
(f) A(B(x)-1) = B(A(x)-1) = B(x*B(x)^3).
From Seiichi Manyama, Jun 04 2025: (Start)
Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120973.
b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(3*n+k,j)/(3*n+k) * b(n-j,3*j).
a(n) = b(n-1,3) for n > 0. (End)

A121687 G.f. satisfies A(x) = 1 + xA(x) A( x*A(x) )^2.

Original entry on oeis.org

1, 1, 3, 14, 83, 574, 4432, 37244, 335153, 3194510, 32001596, 335019839, 3649450270, 41227610316, 481724831132, 5809341783543, 72177761136925, 922539273876404, 12115001489115910, 163284755614174305

Offset: 0

Paul D. Hanna, Aug 15 2006, Aug 20 2008

nonn

			G.f. A(x) = 1 + x + 3*x^2 + 14*x^3 + 83*x^4 + 574*x^5 + 4432*x^6 +...
A(x)^2 = 1 + 2*x + 7*x^2 + 34*x^3 + 203*x^4 + 1398*x^5 + 10706*x^6 +...
A(x*A(x)) = 1 + x + 4*x^2 + 23*x^3 + 160*x^4 + 1259*x^5 + 10833*x^6 +...
A(x*A(x))^2 = 1 + 2*x + 9*x^2 + 54*x^3 + 382*x^4 + 3022*x^5 + 25993*x^6 +...
A(x)*A(x*A(x))^2 = 1 + 3*x + 14*x^2 + 83*x^3 + 574*x^4 + 4432*x^5 +...
The logarithm of the g.f. is given by:
log(A(x)) = A(x)^2*x + {d/dx x*A(x)^4}*x^2/2! + {d^2/dx^2 x^2*A(x)^6}*x^3/3! + {d^3/dx^3 x^3*A(x)^8}*x^4/4! +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..150

Cf. A383563, variants: A030266, A182953, A182954.

{a(n)=local(A=1+x);for(i=0,n,A=serreverse(x/(1+x*(A +x*O(x^n))^2))/x); polcoeff(A,n)}

{a(n)=local(A=1+x+x*O(x^n));for(i=0,n,A=1/(1-x*subst(A^2,x,x*A)));polcoeff(A,n)}

{a(n)=local(A=1+sum(i=1,n-1,a(i)*x^i+x*O(x^n)));
for(i=1,n,A=exp(sum(m=1,n,sum(k=0,n-m,binomial(m+k-1,k)*polcoeff(A^(2*m),k)*x^k)*x^m/m)+x*O(x^n)));polcoeff(A,n)} \\ Paul D. Hanna, Dec 15 2010

{a(n, m=1)=if(n==0, 1, if(m==0, 0^n, sum(k=0, n, m*binomial(n+m, k)/(n+m)*a(n-k, 3*k))))} \\ Paul D. Hanna, Dec 15 2010

G.f. satisfies G(x) = x/(1 + x*A(x)^2) where G(x*A(x)) = x.
G.f. satisfies A(x) = 1/(1 - x*A(x*A(x))^2).
G.f. satisfies the following two equations (which are equivalent)
A(x) = exp( Sum_{m>=0} {d^m/dx^m x^m*A(x)^(2m+2)} * x^(m+1)/(m+1)! )
A(x) = exp( Sum_{m>=1} [Sum_{k>=0} C(m+k-1,k)*{[y^k] A(y)^(2m)}*x^k]*x^m/m). - Paul D. Hanna, Dec 15 2010
Recurrence:
Let A(x)^m = Sum_{n>=0} a(n,m)*x^n with a(0,m)=1, then
a(n,m) = Sum_{k=0..n} m*C(n+m,k)/(n+m) * a(n-k,2k). - Paul D. Hanna, Dec 15 2010

A120974 G.f. A(x) satisfies A(x/A(x)^4) = 1 + x; thus A(x) = 1 + series_reversion(x/A(x)^4).

Original entry on oeis.org

1, 1, 4, 38, 532, 9329, 190312, 4340296, 108043128, 2890318936, 82209697588, 2467155342740, 77676395612884, 2554497746708964, 87449858261161216, 3107829518797739032, 114399270654847628768, 4353537522757357068296, 171010040645759712226048

Offset: 0

Paul D. Hanna, Jul 20 2006

nonn

Robert Israel, Table of n, a(n) for n = 0..250

Cf. A120975; variants: A120970, A120972, A120976, A030266, A067145, A107096.

A:= x -> 1:
for m from 1 to 30 do
  Ap:= unapply(A(x)+c*x^m,x);
  S:= series(Ap(x/Ap(x)^4)-1-x, x, m+1);
  cs:= solve(convert(S,polynom),c);
  A:= subs(c=cs, eval(Ap));
od:
seq(coeff(A(x),x,m),m=0..30);# Robert Israel, Oct 25 2019

nmax = 17; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[ A[x/A[x]^4] - 1 - x + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 02 2019 *)

{a(n)=local(A=[1,1]);for(i=2,n,A=concat(A,0); A[ #A]=-Vec(subst(Ser(A),x,x/Ser(A)^4))[ #A]);A[n+1]}

b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(4*n+k, j)/(4*n+k)*b(n-j, 4*j)));
a(n) = if(n==0, 1, b(n-1, 4)); \\ Seiichi Manyama, Jun 04 2025

G.f. satisfies: A(x) = 1 + x*B(x)^4 = 1 + (1 + x*C(x)^4 )^4 where B(x) and C(x) satisfy: C(x) = B(x)*B(A(x)-1), B(x) = A(A(x)-1), B(A(x)-1) = A(B(x)-1), B(x/A(x)^4) = A(x), B(x) = A(x*B(x)^4) and B(x) is g.f. of A120975.
From Seiichi Manyama, Jun 04 2025: (Start)
Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120975.
b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(4*n+k,j)/(4*n+k) * b(n-j,4*j).
a(n) = b(n-1,4) for n > 0. (End)

A128329 Main diagonal of table A128325.

Original entry on oeis.org

1, 1, 4, 30, 321, 4389, 72512, 1399755, 30865353, 764755508, 21024535960, 634924059276, 20890221475598, 743727414390456, 28484480606420928, 1167761832049224515, 51022550712426870397, 2366859765773183488674

Offset: 0

Paul D. Hanna, Mar 11 2007

nonn

Cf. A030266; A128325 (table), A128326 (row 1), A128327 (row 2), A128328 (row 3).

{a(n)=local(A=1+x,B);for(i=0,n,A=1+x*A*subst(A,x,x*A+x*O(x^n))); B=A;for(i=1,n,B=subst(B,x,x*A+x*O(x^n)));polcoeff(B,n)}

a(n) = [x^n] {1 + H(x)}, where H(x) is the (n+2)-th self-composition of G(x) and G(x) = x + x*G(G(x)) is the g.f. of A030266.

A182954 G.f. satisfies: A(x) = 1 + xA(x) A( x*A(x) )^4.

Original entry on oeis.org

1, 1, 5, 39, 381, 4284, 53163, 710810, 10085621, 150326044, 2336828792, 37687170215, 628069684439, 10782885724300, 190248852445782, 3442896376032300, 63804661588968521, 1209314277690837796

Offset: 0

Paul D. Hanna, Dec 15 2010

nonn

			G.f.: A(x) = 1 + x + 5*x^2 + 39*x^3 + 381*x^4 + 4284*x^5 + ...
Related expansions:
A(x*A(x)) = 1 + x + 6*x^2 + 54*x^3 + 592*x^4 + 7331*x^5 + 98870*x^6 + ...
A(x*A(x))^4 = 1 + 4*x + 30*x^2 + 292*x^3 + 3305*x^4 + 41420*x^5 + ...
The g.f. satisfies:
log(A(x)) = A(x)^4*x + {d/dx x*A(x)^8}*x^2/2! + {d^2/dx^2 x^2*A(x)^12}*x^3/3! + {d^3/dx^3 x^3*A(x)^16}*x^4/4! + ...

Paul D. Hanna, Table of n, a(n) for n = 0..300

Cf. A384265, A030266, A121687, A182953, A182955.

{a(n) = my(A = 1 + sum(i=1,n-1,a(i)*x^i+x*O(x^n)));
for(i=1,n, A = exp( sum(m=1,n, sum(k=0,n-m, binomial(m+k-1,k)*polcoef(A^(4*m),k)*x^k) * x^m/m ) + x*O(x^n))); polcoef(A,n)}

{a(n, m=1) = if(n==0, 1, if(m==0, 0^n, sum(k=0, n, m*binomial(n+m, k)/(n+m)*a(n-k, 4*k))))}

G.f. A(x) satisfies:
* A(x) = exp( Sum_{m>=0} {d^m/dx^m x^m*A(x)^(4m+4)} * x^(m+1)/(m+1)! );
* A(x) = exp( Sum_{m>=1} [Sum_{k>=0} C(m+k-1,k)*{[y^k] A(y)^(4m)}*x^k]*x^m/m);
which are equivalent.
Recurrence:
Let A(x)^m = Sum_{n>=0} a(n,m)*x^n with a(0,m)=1, then
a(n,m) = Sum_{k=0..n} m*C(n+m,k)/(n+m) * a(n-k,4k).

A213010 G.f. satisfies: A(x) = x+x^2 + x*A(A(x)).

Original entry on oeis.org

1, 2, 4, 16, 80, 480, 3296, 25152, 209600, 1884160, 18110080, 184898304, 1994964736, 22654449664, 269855506944, 3362350046208, 43715434232832, 591812683833344, 8326660788725760, 121550217508892672, 1838089917983911936, 28753297176215257088, 464675647688625364992

Offset: 1

Paul D. Hanna, Jun 01 2012

nonn

The half-iteration of the g.f. equals an integer series (A213009).

			G.f.: A(x) = x + 2*x^2 + 4*x^3 + 16*x^4 + 80*x^5 + 480*x^6 + 3296*x^7 +...
where
A(A(x)) = x + 4*x^2 + 16*x^3 + 80*x^4 + 480*x^5 + 3296*x^6 +...
Related expansions.
Let B(B(x)) = A(x), then B(x) is an integer series:
B(x) = x + x^2 + x^3 + 5*x^4 + 21*x^5 + 125*x^6 + 825*x^7 + 6133*x^8 +...
where the coefficients of B(x) are congruent to 1 modulo 4.

Paul D. Hanna, Table of n, a(n) for n = 1..256

Cf. A213009, A030266, A215114, A215116, A215118.

{a(n)=local(A=x+2*x^2);for(i=1,n,A=x+x^2+x*subst(A,x,A+x*O(x^n)));polcoeff(A,n)}
for(n=1,31,print1(a(n),", "))

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

cp's OEIS Frontend

A110447 Number of permutations containing 3241 patterns only as part of 35241 patterns.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

A379598 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where column k is the expansion of B(x)^k, where B(x) is the g.f. of A110447.

Views

Author

Keywords

Examples

Crossrefs

Programs

PARI

Formula

A001028 E.g.f. satisfies A'(x) = 1 + A(A(x)), A(0)=0.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

Maxima

PARI

Formula

Extensions

A120970 G.f. A(x) satisfies A(x/A(x)^2) = 1 + x ; thus A(x) = 1 + Series_Reversion(x/A(x)^2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

Formula

A120972 G.f. A(x) satisfies A(x/A(x)^3) = 1 + x ; thus A(x) = 1 + series_reversion(x/A(x)^3).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

PARI

Formula

A121687 G.f. satisfies A(x) = 1 + x*A(x) * A( x*A(x) )^2.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

PARI

PARI

PARI

Formula

A120974 G.f. A(x) satisfies A(x/A(x)^4) = 1 + x; thus A(x) = 1 + series_reversion(x/A(x)^4).

Views

Author

Keywords

A121687 G.f. satisfies A(x) = 1 + xA(x) A( x*A(x) )^2.

A182954 G.f. satisfies: A(x) = 1 + xA(x) A( x*A(x) )^4.