cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A036765 Number of ordered rooted trees with n non-root nodes and all outdegrees <= three.

Original entry on oeis.org

1, 1, 2, 5, 13, 36, 104, 309, 939, 2905, 9118, 28964, 92940, 300808, 980864, 3219205, 10626023, 35252867, 117485454, 393133485, 1320357501, 4449298136, 15038769672, 50973266380, 173214422068, 589998043276, 2014026871496, 6889055189032, 23608722350440
Offset: 0

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Author

Keywords

Comments

Number of Dyck n-paths that avoid UUUU. For example, a(4)=13 counts all 14 Dyck 4-paths except UUUUDDDD. - David Callan, Dec 09 2004
Number of restricted growth strings for Dyck paths with at most 2 consecutive rises (this is equivalent to the comment above, see example). - Joerg Arndt, Oct 31 2012
Let A(x) be the g.f. for the sequence of numbers of Dyck words with at most k consecutive ones (paths with at most k consecutive up-steps 'U', Restricted Growth Strings with at most k-1 consecutive rises), then B(x) := x*A(x) is the series reversion of x/(1+x+x^2+...+x^k). - Joerg Arndt, Oct 31 2012
a(n) is the number of ordered unlabeled rooted trees on n+1 nodes where each node has no more than 3 children. - Geoffrey Critzer, Jan 05 2013
a(n) = number of noncrossing partitions of [n] in which all blocks are of size <= 3. - David Callan, Aug 27 2014

Examples

			a(4) = 13 since the top row of M^4 = (13, 8, 4, 1, 1).
From _Joerg Arndt_, Oct 31 2012: (Start)
a(5)=36 because there are 36 Dyck words of length 5 that avoid "1111":
[ #]      RGS                rises         Dyck word
[ 1]    [ . . . . . ]     [ . . . . . ]    1.1.1.1.1.
[ 2]    [ . . . . 1 ]     [ . . . . 1 ]    1.1.1.11..
[ 3]    [ . . . 1 . ]     [ . . . 1 . ]    1.1.11..1.
[ 4]    [ . . . 1 1 ]     [ . . . 1 . ]    1.1.11.1..
[ 5]    [ . . . 1 2 ]     [ . . . 1 2 ]    1.1.111...
[ 6]    [ . . 1 . . ]     [ . . 1 . . ]    1.11..1.1.
[ 7]    [ . . 1 . 1 ]     [ . . 1 . 1 ]    1.11..11..
[ 8]    [ . . 1 1 . ]     [ . . 1 . . ]    1.11.1..1.
[ 9]    [ . . 1 1 1 ]     [ . . 1 . . ]    1.11.1.1..
[10]    [ . . 1 1 2 ]     [ . . 1 . 1 ]    1.11.11...
[11]    [ . . 1 2 . ]     [ . . 1 2 . ]    1.111...1.
[12]    [ . . 1 2 1 ]     [ . . 1 2 . ]    1.111..1..
[13]    [ . . 1 2 2 ]     [ . . 1 2 . ]    1.111.1...
[--]    [ . . 1 2 3 ]     [ . . 1 2 3 ]    1.1111....
[14]    [ . 1 . . . ]     [ . 1 . . . ]    11..1.1.1.
[15]    [ . 1 . . 1 ]     [ . 1 . . 1 ]    11..1.11..
[16]    [ . 1 . 1 . ]     [ . 1 . 1 . ]    11..11..1.
[17]    [ . 1 . 1 1 ]     [ . 1 . 1 . ]    11..11.1..
[18]    [ . 1 . 1 2 ]     [ . 1 . 1 2 ]    11..111...
[19]    [ . 1 1 . . ]     [ . 1 . . . ]    11.1..1.1.
[20]    [ . 1 1 . 1 ]     [ . 1 . . 1 ]    11.1..11..
[21]    [ . 1 1 1 . ]     [ . 1 . . . ]    11.1.1..1.
[22]    [ . 1 1 1 1 ]     [ . 1 . . . ]    11.1.1.1..
[23]    [ . 1 1 1 2 ]     [ . 1 . . 1 ]    11.1.11...
[24]    [ . 1 1 2 . ]     [ . 1 . 1 . ]    11.11...1.
[25]    [ . 1 1 2 1 ]     [ . 1 . 1 . ]    11.11..1..
[26]    [ . 1 1 2 2 ]     [ . 1 . 1 . ]    11.11.1...
[27]    [ . 1 1 2 3 ]     [ . 1 . 1 2 ]    11.111....
[28]    [ . 1 2 . . ]     [ . 1 2 . . ]    111...1.1.
[29]    [ . 1 2 . 1 ]     [ . 1 2 . 1 ]    111...11..
[30]    [ . 1 2 1 . ]     [ . 1 2 . . ]    111..1..1.
[31]    [ . 1 2 1 1 ]     [ . 1 2 . . ]    111..1.1..
[32]    [ . 1 2 1 2 ]     [ . 1 2 . 1 ]    111..11...
[33]    [ . 1 2 2 . ]     [ . 1 2 . . ]    111.1...1.
[34]    [ . 1 2 2 1 ]     [ . 1 2 . . ]    111.1..1..
[35]    [ . 1 2 2 2 ]     [ . 1 2 . . ]    111.1.1...
[36]    [ . 1 2 2 3 ]     [ . 1 2 . 1 ]    111.11....
[--]    [ . 1 2 3 . ]     [ . 1 2 3 . ]    1111....1.
[--]    [ . 1 2 3 1 ]     [ . 1 2 3 . ]    1111...1..
[--]    [ . 1 2 3 2 ]     [ . 1 2 3 . ]    1111..1...
[--]    [ . 1 2 3 3 ]     [ . 1 2 3 . ]    1111.1....
[--]    [ . 1 2 3 4 ]     [ . 1 2 3 4 ]    11111.....
(Dots are used for zeros for readability.)
(End)
		

Crossrefs

Right hand column of triangle A064580. The sequence of sequences A000007 (0^n), A000012 (constant 1), A001006 (Motzkin), A036765, A036766, ... tends to A000108 (Catalan).
Column k=3 of A288942.

Programs

  • Magma
    [&+[Binomial(n+1, n-2*k)*Binomial(n+1, k)/(n+1): k in [0..n]]: n in [0..30]]; // Vincenzo Librandi, Oct 16 2018
  • Maple
    r := 3; [ seq((1/n)*add( (-1)^j*binomial(n,j)*binomial(2*n-2-j*(r+1), n-1),j=0..floor((n-1)/(r+1))), n=1..30) ];
    # second Maple program:
    b:= proc(u, o) option remember; `if`(u+o=0, 1,
          add(b(u-j, o+j-1), j=1..min(1, u))+
          add(b(u+j-1, o-j), j=1..min(3, o)))
        end:
    a:= n-> b(0, n):
    seq(a(n), n=0..30);  # Alois P. Heinz, Aug 28 2017
  • Mathematica
    InverseSeries[Series[y/(1+y+y^2+y^3), {y, 0, 24}], x] (* then A(x)=y(x)/x *) (* Len Smiley, Apr 11 2000 *)
    b[u_, o_, k_] := b[u, o, k] = If[u + o == 0, 1, Sum[b[u - j, o + j - 1, k], {j, 1, Min[1, u]}] + Sum[b[u + j - 1, o - j, k], {j, 1, Min[k, o]}]];
    a[n_] := b[0, n, 3];
    Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Nov 05 2017, after Alois P. Heinz *)
    Table[HypergeometricPFQ[{-n-1, (1-n)/2, -n/2}, {1, 3/2}, -1], {n, 0, 28}] (* Vladimir Reshetnikov, Oct 15 2018 *)
  • PARI
    {a(n)=sum(j=0,n\2,binomial(n+1, n-2*j)*binomial(n+1,j))/(n+1)}
    
  • PARI
    {a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=1+x*A+(x*A)^2+(x*A)^3);polcoeff(A,n)}
    
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*(x*A+x*O(x^n))^j)*x^m/m)));polcoeff(A, n, x)}
    
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=exp(sum(m=1, n, sum(j=0, n, binomial(m+j, j)^2*(x*A+x*O(x^n))^j)*(1-x*A)^(2*m+1)*x^m/m)));polcoeff(A, n, x)}
    
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=1/(1-x+x*O(x^n))*exp(sum(m=1,n,A^m*sum(k=0,m-1,binomial(m-1,k)*binomial(m,k)*x^k)/(1-x)^(2*m)*x^(2*m)/m) +x*O(x^n)));polcoeff(A,n)} /* Paul D. Hanna */
    
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=1/(1-x+x*O(x^n))*exp(sum(m=1,n,A^m*sum(k=0,n,binomial(m+k-1,k)*binomial(m+k,k)*x^k)*x^(2*m)/m) +x*O(x^n)));polcoeff(A,n)} /* Paul D. Hanna */
    
  • PARI
    Vec(serreverse(x/(1+x+x^2+x^3)+O(x^66))/x) /* Joerg Arndt, Jun 10 2011 */
    

Formula

a(n) = (1/(n+1))*Sum_{j=0..floor(n/2)} binomial(n+1, n-2*j)*binomial(n+1, j). G.f. A(z) satisfies A=1+z*A+(z*A)^2+(z*A)^3. - Emeric Deutsch, Nov 29 2003
G.f.: F(x)/x where F(x) is the reversion of x/(1+x+x^2+x^3). - Joerg Arndt, Jun 10 2011
From Paul D. Hanna, Feb 13 2011: (Start)
O.g.f.: A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^2*x^k*A(x)^k) * x^n/n ).
O.g.f.: A(x) = exp( Sum_{n>=1} (Sum_{k>=0} C(n+k,k)^2*x^k*A(x)^k)*(1-x*A(x))^(2*n+1)* x^n/n ). (End)
From Paul D. Hanna, Feb 24 2011: (Start)
O.g.f.: A(x) = (1/(1-x))*exp( Sum_{n>=1} A(x)^n*(Sum_{k=0..n-1} C(n-1,k)*C(n,k)*x^k)/(1-x)^(2*n) * x^(2*n)/n ).
O.g.f.: A(x) = (1/(1-x))*exp( Sum_{n>=1} A(x)^n*(Sum_{k>=0} C(n+k-1,k)*C(n+k,k)*x^k) * x^(2*n)/n ). (End)
Let M = an infinite quadradiagonal matrix with all 1's in every diagonal: (sub, main, super, and super-super), and the rest zeros. V = vector [1,0,0,0,...]. The sequence = left column terms of M*V iterates. - Gary W. Adamson, Jun 06 2011
An infinite square production matrix M for the sequence is:
1, 1, 0, 0, 0, 0, ...
1, 0, 1, 0, 0, 0, ...
2, 1, 0, 1, 0, 0, ...
3, 2, 1, 0, 1, 0, ...
4, 3, 2, 1, 0, 1, ...
5, 4, 3, 2, 1, 0, ...
..., such that a(n) is the top left term of M^n. - Gary W. Adamson, Feb 21 2012
D-finite with recurrence: 2*(n+1)*(2*n+3)*(13*n-1)*a(n) = (143*n^3 + 132*n^2 - 17*n - 18)*a(n-1) + 4*(n-1)*(26*n^2 + 11*n - 6)*a(n-2) + 16*(n-2)*(n-1)*(13*n + 12)*a(n-3). - Vaclav Kotesovec, Sep 09 2013
a(n) ~ c*d^n/n^(3/2), where d = 1/12*((6371+624*sqrt(78))^(2/3)+11*(6371+624*sqrt(78))^(1/3)+217)/(6371+624*sqrt(78))^(1/3) = 3.610718613276... is the root of the equation -16-8*d-11*d^2+4*d^3=0 and c = sqrt(f/Pi) = 0.9102276936417..., where f = 1/9984*(9295 + (13*(45085576939 - 795629568*sqrt(78)))^(1/3) + (13*(45085576939 + 795629568*sqrt(78)))^(1/3)) is the root of the equation -128+1696*f-9295*f^2+3328*f^3=0. - Vaclav Kotesovec, Sep 10 2013
From Peter Bala, Jun 21 2015: (Start)
The coefficient of x^n in A(x)^r equals r/(n + r)*Sum_{k = 0..floor(n/2)} binomial(n + r,k)*binomial(n + r,n - 2*k) by the Lagrange-Bürmann formula.
O.g.f. A(x) = exp(Sum_{n >= 1} A005725(n)*x^n/n), where A005725(n) = Sum_{k = 0..floor(n/2)} binomial(n,k)*binomial(n,n - 2*k). Cf. A186241, A198951, A200731. (End)
a(n) = hypergeom([-n-1, (1-n)/2, -n/2], [1, 3/2], -1). - Vladimir Reshetnikov, Oct 15 2018

Extensions

Name clarified by Andrew Howroyd, Dec 04 2017

A198951 G.f. satisfies: A(x) = (1 + x*A(x))*(1 + x^3*A(x)^3).

Original entry on oeis.org

1, 1, 1, 2, 6, 16, 39, 99, 271, 763, 2146, 6062, 17359, 50337, 147057, 431874, 1275273, 3786649, 11298031, 33846202, 101762937, 306997821, 929038518, 2819426688, 8578433304, 26163061776, 79970186791, 244938841096, 751646959402, 2310683396056, 7115199919151
Offset: 0

Views

Author

Paul D. Hanna, Oct 31 2011

Keywords

Comments

a(n) is also the number of rooted labeled trees on n nodes such that each node has 0, 1, 3, or 4 children. - Patrick Devlin, Mar 04 2012

Examples

			G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 6*x^4 + 16*x^5 + 39*x^6 + 99*x^7 + ...
Related expansions:
A(x)^3 = 1 + 3*x + 6*x^2 + 13*x^3 + 36*x^4 + 105*x^5 + 292*x^6 + ...
A(x)^4 = 1 + 4*x + 10*x^2 + 24*x^3 + 67*x^4 + 200*x^5 + 582*x^6 + ...
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x^2*A(x)^2)*x + (1 + 2^2*x^2*A(x)^2 + x^4*A(x)^4)*x^2/2 +
(1 + 3^2*x^2*A(x)^2 + 3^2*x^4*A(x)^4 + x^6*A(x)^6)*x^3/3 +
(1 + 4^2*x^2*A(x)^2 + 6^2*x^4*A(x)^4 + 4^2*x^6*A(x)^6 + x^8*A(x)^8)*x^4/4 +
(1 + 5^2*x^2*A(x)^2 + 10^2*x^4*A(x)^4 + 10^2*x^6*A(x)^6 + 5^2*x^8*A(x)^8 + x^10*A(x)^10)*x^5/5 + ...
more explicitly,
log(A(x)) = x + x^2/2 + 4*x^3/3 + 17*x^4/4 + 51*x^5/5 + 136*x^6/6 + 393*x^7/7 + 1233*x^8/8 + ...
		

Crossrefs

Programs

  • Maple
    a:= n-> coeff(series(RootOf(A=(1+x*A)*(1+x^3*A^3), A), x, n+1), x, n):
    seq(a(n), n=0..30);  # Alois P. Heinz, May 16 2012
  • Mathematica
    InverseSeries[ Series[ x/((1 + x)*(1 + x^3)), {x, 0, 31}], x] // CoefficientList[#, x]& // Rest (* Jean-François Alcover, Sep 10 2013 *)
  • PARI
    {a(n)=local(A=1/x*serreverse(x/(1+x+x^3+x^4+x*O(x^n)))); polcoeff(A, n)}
    
  • PARI
    {a(n)=polcoeff((1+x+x^3+x^4+x*O(x^n))^(n+1)/(n+1), n)}
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n,sum(j=0, m, binomial(m, j)^2*x^(2*j)*(A+x*O(x^n))^(2*j))*x^m/m))); polcoeff(A, n)}
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (1-x^2*A^2)^(2*m+1)*sum(j=0, n\2, binomial(m+j, j)^2*x^(2*j)*(A^2+x*O(x^n))^j)*x^m/m))); polcoeff(A, n, x)}

Formula

G.f. satisfies:
(1) A(x) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * x^(2*k) * A(x)^(2*k) ).
(2) A(x) = (1/x)*Series_Reversion(x/((1+x)*(1+x^3))).
(3) a(n) = [x^n] (1 + x + x^3 + x^4)^(n+1) / (n+1).
(4) A(x) = exp( Sum_{n>=1} x^n/n * (1-x^2*A(x)^2)^(2*n+1)*Sum_{k>=0} C(n+k,k)^2 * x^(2*k) * A(x)^(2*k) ).
D-finite with recurrence: 3*(n+1)*(3*n+2)*(3*n+4)*(119*n^3 - 210*n^2 + 73*n - 6)*a(n) = 2*(6664*n^6 - 1764*n^5 - 11585*n^4 + 426*n^3 + 4129*n^2 - 102*n - 288)*a(n-1) - 18*(n-1)*(1190*n^5 - 910*n^4 - 1937*n^3 + 895*n^2 + 606*n - 216)*a(n-2) + 162*(n-2)*(n-1)*(2*n-3)*(119*n^3 + 147*n^2 + 10*n - 24)*a(n-3). - Vaclav Kotesovec, Sep 09 2013
a(n) ~ c*d^n/n^(3/2), where d = 1/81*((2144134+520506*sqrt(17))^(2/3)+112*(2144134+520506*sqrt(17))^(1/3)-2036)/(2144134+520506*sqrt(17))^(1/3) = 3.23407602060970245... is the root of the equation -324 + 180*d - 112*d^2 + 27*d^3 = 0 and c = 0.6286981954423757284622435... - Vaclav Kotesovec, Sep 09 2013
A(1/d) = 370/243 + (3*sqrt(17)/509 - 3070/123687)*(2144134+520506*sqrt(17))^(1/3) + (141*sqrt(17)/2072648 - 129529/503653464)*(2144134+520506*sqrt(17))^(2/3) = 2.053716618436594614948796... - Vaclav Kotesovec, Sep 10 2013
From Peter Bala, Jun 21 2015: (Start)
a(n) = 1/(n + 1)*Sum_{k = 0..floor(n/3)} binomial(n + 1,k)* binomial(n + 1,n - 3*k). Applying Maple's sumrecursion command to this formula gives the above recurrence of Kotesovec.
More generally, the coefficient of x^n in A(x)^r equals r/(n + r)*Sum_{k = 0..floor(n/3)} binomial(n + r,k)*binomial(n + r,n - 3*k) by the Lagrange-Bürmann formula.
O.g.f. A(x) = exp(Sum_{n >= 1} A228960(n)*x^n/n), where A228960(n) = Sum_{k = 0..floor(n/3)} binomial(n,k)*binomial(n,3*k). Cf. A036765, A186241 and A200731. (End)

A186241 G.f. satisfies A(x) = 1 + x*A(x)^2 + x^2*A(x)^4 + x^3*A(x)^6.

Original entry on oeis.org

1, 1, 3, 12, 54, 262, 1337, 7072, 38426, 213197, 1202795, 6879160, 39794416, 232429030, 1368806610, 8118934656, 48458809586, 290832756606, 1754059333738, 10625545472716, 64620970743082, 394409682103262, 2415084675723048, 14832185219521152, 91339478577683664
Offset: 0

Views

Author

Vladimir Kruchinin, Feb 15 2011

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 54*x^4 + 262*x^5 + 1337*x^6 +...
where A(x) = (1 + x*A(x)^2)*(1 + x^2*A(x)^4).
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 30*x^3 + 141*x^4 + 704*x^5 + 3666*x^6 +...
A(x)^4 = 1 + 4*x + 18*x^2 + 88*x^3 + 451*x^4 + 2392*x^5 + 13022*x^6 +...
A(x)^6 = 1 + 6*x + 33*x^2 + 182*x^3 + 1014*x^4 + 5718*x^5 + 32623*x^6 +...
where A(x) = 1 + x*A(x)^2 + x^2*A(x)^4 + x^3*A(x)^6.
From _Paul D. Hanna_, Nov 11 2011: (Start)
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x*A^2)*x*A + (1 + 2^2*x*A^2 + x^2*A^4)*x^2*A^2/2 +
(1 + 3^2*x*A^2 + 3^2*x^2*A^4 + x^3*A^6)*x^3*A^3/3 +
(1 + 4^2*x*A^2 + 6^2*x^2*A^4 + 4^2*x^3*A^6 + x^4*A^8)*x^4*A^4/4 +
(1 + 5^2*x*A^2 + 10^2*x^2*A^4 + 10^2*x^3*A^6 + 5^2*x^4*A^8 + x^5*A^10)*x^5*A^5/5 + ...
which involves squares of binomial coefficients. (End)
		

Crossrefs

Programs

  • Maple
    F:= proc(n) if n::even then
      simplify((1/2)*hypergeom([-(1/2)*n, -2*n-1, -(1/2)*n+1/2], [(1/2)*n+1, 3/2+(1/2)*n], -1)*(2*n+2)!/((2*n+1)*((n+1)!)^2))
      else
      simplify((1/2)*hypergeom([-(1/2)*n, -2*n-1, -(1/2)*n+1/2], [(1/2)*n+1, 3/2+(1/2)*n], -1)*(2*n+2)!/((2*n+1)*((n+1)!)^2))
      fi
    end proc:
    map(F, [$0..30]); # Robert Israel, Jun 22 2015
  • Mathematica
    a[n_] := 1/(2n + 1) Sum[Binomial[2n + 1, k] Binomial[2n + 1, n - 2k], {k, 0, n/2}];
    (* or: *)
    a[n_] := (Binomial[2n + 1, n] HypergeometricPFQ[{-2n - 1, 1/2 - n/2, -n/2}, {n/2 + 1, n/2 + 3/2}, -1])/(2n + 1);
    Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Nov 17 2017 *)
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=(1+x*A^2)*(1+x^2*A^4)+x*O(x^n));polcoeff(A,n)} /* Paul D. Hanna */
    
  • PARI
    {a(n)=polcoeff(sqrt((1/x)*serreverse(x/(1 + x + x^2 + x^3 +x*O(x^n))^2)), n)} /* Paul D. Hanna */
    
  • PARI
    {a(n)=polcoeff( (1 + x + x^2 + x^3+x*O(x^n))^(2*n+1)/(2*n+1), n)} /* Paul D. Hanna */
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (x*A+x*O(x^n))^m/m*sum(j=0, m, binomial(m, j)^2*x^j*A^(2*j))))); polcoeff(A, n, x)} /* Paul D. Hanna */
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n,x^m*A^m/m*(1-x*A^2)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j*A^(2*j))))); polcoeff(A, n, x)} /* Paul D. Hanna */

Formula

a(n) = (1/(2*n-1))*Sum_{j=0..2*n-1} binomial(2*n-1,j)*Sum_{i=j..n+j-1} binomial(j,i-j)*binomial(2*n-j-1,3*j-3*n-i+1), n>0.
From Paul D. Hanna, Nov 11 2011: (Start)
G.f. A(x) satisfies:
(1) A(x) = sqrt( (1/x)*Series_Reversion( x/(1 + x + x^2 + x^3)^2 ) ).
(2) A( x/(1 + x + x^2 + x^3)^2 ) = 1 + x + x^2 + x^3.
(3) A(x) = G(x*A(x)) where G(x) = A(x/G(x)) = g.f. of A036765 (number of rooted trees with a degree constraint).
(4) a(n) = [x^n] (1 + x + x^2 + x^3)^(2*n+1) / (2*n+1).
(5) A(x) = exp( Sum_{n>=1} x^n*A(x)^n/n * [Sum_{k=0..n} C(n,k)^2 * x^k*A(x)^(2*k)] ).
(6) A(x) = exp( Sum_{n>=1} x^n*A(x)^n/n * [(1-x*A(x)^2)^(2*n+1)*Sum_{k>=0} C(n+k,k)^2*x^k*A(x)^(2*k)] ).
(End)
From Peter Bala, Jun 21 2015: (Start)
a(n) = 1/(2*n + 1)*Sum_{k = 0..floor(n/2)} binomial(2*n + 1,k)*binomial(2*n + 1,n - 2*k).
More generally, the coefficient of x^n in A(x)^r equals r/(2*n + r)*Sum_{k = 0..floor(n/2)} binomial(2*n + r,k)*binomial(2*n + r,n - 2*k) by the Lagrange-Bürmann formula.
O.g.f. A(x) = exp(Sum_{n >= 1} 1/2*b(n)*x^n/n), where b(n) = Sum_{k = 0..floor(n/2)} binomial(2*n,k)*binomial(2*n,n - 2*k). Cf. A036765, A198951, A200731. (End)
Recurrence: 5*n*(5*n - 1)*(5*n + 1)*(5*n + 2)*(5*n + 3)*(13144*n^4 - 57784*n^3 + 90149*n^2 - 59354*n + 13980)*a(n) = 8*(2*n - 1)*(16259128*n^8 - 71478808*n^7 + 108653137*n^6 - 60530902*n^5 - 2811173*n^4 + 12694433*n^3 - 2398482*n^2 - 352503*n + 78570)*a(n-1) + 128*(n-1)*(2*n - 3)*(2*n - 1)*(52576*n^6 - 178560*n^5 + 136156*n^4 + 22938*n^3 - 16067*n^2 - 3138*n - 405)*a(n-2) + 2048*(n-2)*(n-1)*(2*n - 5)*(2*n - 3)*(2*n - 1)*(13144*n^4 - 5208*n^3 - 4339*n^2 + 168*n + 135)*a(n-3). - Vaclav Kotesovec, Nov 17 2017
A(x^2) = (1/x) * series reversion of x/(1 + x^2 + x^4 + x^6). - Peter Bala, Jul 27 2023

A211248 G.f. satisfies: A(x) = (1 + x*A(x)^3) * (1 + x^2*A(x)^4).

Original entry on oeis.org

1, 1, 4, 20, 114, 703, 4565, 30752, 212921, 1505916, 10833164, 79018804, 583062388, 4344431508, 32641910199, 247033970128, 1881402836376, 14408753414558, 110897147057354, 857307054338476, 6653979156676983, 51831065993122915, 405060413133136902, 3175019470333290488
Offset: 0

Views

Author

Paul D. Hanna, Apr 05 2012

Keywords

Comments

More generally, for fixed parameters p and q, if F(x) satisfies:
F(x) = exp( Sum_{n>=1} x^n * F(x)^(n*p)/n * [Sum_{k=0..n} C(n,k)^2 * x^k * F(x)^(k*q)] ),
then F(x) = (1 + x*F(x)^(p+1))*(1 + x^2*F(x)^(p+q+1)); here p=2 and q=1.

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 20*x^3 + 114*x^4 + 703*x^5 + 4565*x^6 +...
where A( x*(1-x-x^3)^2/(1+x^2)^2 ) = (1+x^2)/(1-x-x^3).
Related expansions:
A(x)^3 = 1 + 3*x + 15*x^2 + 85*x^3 + 522*x^4 + 3381*x^5 + 22735*x^6 +...
A(x)^4 = 1 + 4*x + 22*x^2 + 132*x^3 + 841*x^4 + 5588*x^5 + 38288*x^6 +...
A(x)^7 = 1 + 7*x + 49*x^2 + 343*x^3 + 2429*x^4 + 17430*x^5 +...
where A(x) = 1 + x*A(x)^3 + x^2*A(x)^4 + x^3*A(x)^7.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x*A(x))*x*A(x)^2 + (1 + 2^2*x*A(x) + x^2*A(x)^2)*x^2*A(x)^4/2 +
(1 + 3^2*x*A(x) + 3^2*x^2*A(x)^2 + x^3*A(x)^3)*x^3*A(x)^6/3 +
(1 + 4^2*x*A(x) + 6^2*x^2*A(x)^2 + 4^2*x^3*A(x)^3 + x^4*A(x)^4)*x^4*A(x)^8/4 +
(1 + 5^2*x*A(x) + 10^2*x^2*A(x)^2 + 10^2*x^3*A(x)^3 + 5^2*x^4*A(x)^4 + x^5*A(x)^5)*x^5*A(x)^10/5 +
(1 + 6^2*x*A(x) + 15^2*x^2*A(x)^2 + 20^2*x^3*A(x)^3 + 15^2*x^4*A(x)^4 + 6^2*x^5*A(x)^5 + x^6*A(x)^6)*x^6*A(x)^12/6 +...
more explicitly,
log(A(x)) = x + 7*x^2/2 + 49*x^3/3 + 359*x^4/4 + 2706*x^5/5 + 20767*x^6/6 +...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Sqrt[1/x * InverseSeries[Series[x*(1 - x - x^3)^2/(1 + x^2)^2, {x, 0, 20}], x]], x] (* Vaclav Kotesovec, Nov 22 2017 *)
  • PARI
    {a(n)=polcoeff(sqrt( (1/x)*serreverse( x*(1-x-x^3)^2/(1+x^2+x*O(x^n))^2 ) ), n)}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    {a(n)=local(p=2, q=1, A=1+x); for(i=1, n, A=(1+x*A^(p+1))*(1+x^2*A^(p+q+1))+x*O(x^n)); polcoeff(A, n)}
    
  • PARI
    {a(n)=local(p=2, q=1, A=1+x); for(i=1, n, A=exp(sum(m=1, n, x^m*(A+x*O(x^n))^(p*m)/m*sum(j=0, m, binomial(m, j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}
    
  • PARI
    {a(n)=local(p=2, q=1, A=1+x); for(i=1, n, A=exp(sum(m=1, n, x^m*(A+x*O(x^n))^(p*m)/m*(1-x*A^q)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}

Formula

G.f. A(x) satisfies:
(1) A(x) = sqrt( (1/x)*Series_Reversion( x*(1-x-x^3)^2/(1+x^2)^2 ) ).
(2) A( x*(1-x-x^3)^2/(1+x^2)^2 ) = (1+x^2)/(1-x-x^3).
(3) a(n) = [x^n] ((1+x^2)/(1-x-x^3))^(2*n+2) / (n+1).
(4) A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^2 * x^k*A(x)^k) * x^n*A(x)^(2*n)/n ).
(5) A(x) = exp( Sum_{n>=1} (1-x*A(x))^(2*n+1) * (Sum_{k>=0} C(n+k,k)^2*x^k*A(x)^k) * x^n*A(x)^(2*n)/n ).
(6) A(x) = (1/x)*Series_Reversion(x/G(x)) where A(x) = G(x*A(x)) and A(x/G(x)) = G(x) = (1 + x*G(x)^2)*(1 + x^2*G(x)^2) is the g.f. of A199874.
a(n) ~ s * sqrt((1 + 2*r*s + 3*r^2*s^4) / (3*Pi*(1 + 2*r*s + 7*r^2*s^4))) / (2*n^(3/2)*r^n), where r = 0.1194948955213353102456218138370139612914667337222... and s = 1.428770161302757679335810379290625953730830139744... are real roots of the system of equations (1 + r*s^3)*(1 + r^2*s^4) = s, r*s^2*(3 + 4*r*s + 7*r^2*s^4) = 1. - Vaclav Kotesovec, Nov 22 2017

A211249 G.f. satisfies: A(x) = (1 + x*A(x)^3) * (1 + x^2*A(x)^5).

Original entry on oeis.org

1, 1, 4, 21, 126, 819, 5611, 39900, 291719, 2179181, 16560175, 127617168, 994951887, 7833555324, 62196300997, 497425570173, 4003607595960, 32404662671330, 263586896132154, 2153631763231319, 17666722629907960, 145449082369322208, 1201414340736684702
Offset: 0

Views

Author

Paul D. Hanna, Apr 05 2012

Keywords

Comments

More generally, for fixed parameters p and q, if F(x) satisfies:
F(x) = exp( Sum_{n>=1} x^n * F(x)^(n*p)/n * [Sum_{k=0..n} C(n,k)^2 * x^k * F(x)^(k*q)] ),
then F(x) = (1 + x*F(x)^(p+1))*(1 + x^2*F(x)^(p+q+1)); here p=2 and q=2.

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 21*x^3 + 126*x^4 + 819*x^5 + 5611*x^6 +...
where A( x*(1-x-x^3)^2/(1+x^2)^2 ) = (1+x^2)/(1-x-x^3).
Related expansions:
A(x)^3 = 1 + 3*x + 15*x^2 + 88*x^3 + 564*x^4 + 3828*x^5 + 27040*x^6 +...
A(x)^5 = 1 + 5*x + 30*x^2 + 195*x^3 + 1335*x^4 + 9486*x^5 + 69305*x^6 +...
A(x)^8 = 1 + 8*x + 60*x^2 + 448*x^3 + 3374*x^4 + 25704*x^5 +...
where A(x) = 1 + x*A(x)^3 + x^2*A(x)^5 + x^3*A(x)^8.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x*A(x)^2)*x*A(x)^2 +
(1 + 2^2*x*A(x)^2 + x^2*A(x)^4)*x^2*A(x)^4/2 +
(1 + 3^2*x*A(x)^2 + 3^2*x^2*A(x)^4 + x^3*A(x)^6)*x^3*A(x)^6/3 +
(1 + 4^2*x*A(x)^2 + 6^2*x^2*A(x)^4 + 4^2*x^3*A(x)^6 + x^4*A(x)^8)*x^4*A(x)^8/4 +
(1 + 5^2*x*A(x)^2 + 10^2*x^2*A(x)^4 + 10^2*x^3*A(x)^6 + 5^2*x^4*A(x)^8 + x^5*A(x)^10)*x^5*A(x)^10/5 +
(1 + 6^2*x*A(x)^2 + 15^2*x^2*A(x)^4 + 20^2*x^3*A(x)^6 + 15^2*x^4*A(x)^8 + 6^2*x^5*A(x)^10 + x^6*A(x)^12)*x^6*A(x)^12/6 +...
more explicitly,
log(A(x)) = x + 7*x^2/2 + 52*x^3/3 + 403*x^4/4 + 3211*x^5/5 + 26050*x^6/6 +...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Sqrt[1/x * InverseSeries[Series[x*(1-2*x-x^2+x^4 + (1-x-x^2) * Sqrt[(1+x+x^2)*(1-3*x+x^2)])/2, {x, 0, 20}], x]], x] (* Vaclav Kotesovec, Nov 22 2017 *)
  • PARI
    {a(n)=polcoeff(sqrt( (1/x)*serreverse( x*(1-2*x-x^2+x^4 + (1-x-x^2)*sqrt( (1+x+x^2)*(1-3*x+x^2) +x*O(x^n)))/2 ) ), n)}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    {a(n)=local(p=2, q=2, A=1+x); for(i=1, n, A=(1+x*A^(p+1))*(1+x^2*A^(p+q+1))+x*O(x^n)); polcoeff(A, n)}
    
  • PARI
    {a(n)=local(p=2, q=2, A=1+x); for(i=1, n, A=exp(sum(m=1, n, x^m*(A+x*O(x^n))^(p*m)/m*sum(j=0, m, binomial(m, j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}
    
  • PARI
    {a(n)=local(p=2, q=2, A=1+x); for(i=1, n, A=exp(sum(m=1, n, x^m*(A+x*O(x^n))^(p*m)/m*(1-x*A^q)^(2*m+1)*sum(j=0, n, binomial(m+j,j)^2*x^j*(A+x*O(x^n))^(q*j))))); polcoeff(A, n, x)}

Formula

G.f.: sqrt( (1/x)*Series_Reversion( x*(1-2*x-x^2+x^4 + (1-x-x^2)*sqrt( (1+x+x^2)*(1-3*x+x^2) ))/2 ) ).
G.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion(x/G(x)) where A(x) = G(x*A(x)) and A(x/G(x)) = G(x) is the g.f. of A200075.
(2) A(x) = sqrt( (1/x)*Series_Reversion( x/G(x)^2 ) ) where A(x) = G(x*A(x)^2) and G(x) = A(x/G(x)^2) and 1+x*G(x) is the g.f. of A004148.
(3) A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^2 * x^k*A(x)^(2*k)) * x^n*A(x)^(2*n)/n ).
(4) A(x) = exp( Sum_{n>=1} (1-x*A(x)^2)^(2*n+1) * (Sum_{k>=0} C(n+k,k)^2*x^k*A(x)^(2*k)) * x^n*A(x)^(2*n)/n ).
a(n) ~ s * sqrt((1 + 2*r*s^2 + 3*r^2*s^5) / (Pi*(3 + 10*r*s^2 + 28*r^2*s^5))) / (2*n^(3/2)*r^n), where r = 0.1130413665724951344267888513870607581680912144315... and s = 1.385648922830296011590145919380626723251960276539... are real roots of the system of equations (1 + r*s^3)*(1 + r^2*s^5) = s, r*s^2*(3 + 5*r*s^2 + 8*r^2*s^5) = 1. - Vaclav Kotesovec, Nov 22 2017

A262082 Array of coefficients A(n,k) of the formal power series P(n,x) read by upwards antidiagonals, where P(n,x) = Sum_{k>=0} A(n,k)*x^k = 1 + x*P(n,x)^(1*n) + x^2*P(n,x)^(2*n) + x^3*P(n,x)^(3*n) for n >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 5, 0, 1, 1, 4, 12, 13, 0, 1, 1, 5, 22, 54, 36, 0, 1, 1, 6, 35, 139, 262, 104, 0, 1, 1, 7, 51, 284, 953, 1337, 309, 0, 1, 1, 8, 70, 505, 2509, 6894, 7072, 939, 0, 1, 1, 9, 92, 818, 5455, 23426, 51796, 38426, 2905, 0, 1, 1, 10
Offset: 0

Views

Author

Werner Schulte, Sep 10 2015

Keywords

Comments

The terms define the array A(n,k):
n\k: 0 1 2 3 4 5 6 7 8 9 10 ...
0: 1 1 1 1 0 0 0 0 0 0 0 ...
1: 1 1 2 5 13 36 104 309 939 2905 ...
2: 1 1 3 12 54 262 1337 7072 38426 ...
3: 1 1 4 22 139 953 6894 51796 400269 ...
4: 1 1 5 35 284 2509 23426 ...
5: 1 1 6 51 505 5455 62336 ...
6: 1 1 7 70 818 ...
7: 1 1 8 92 ...
8: 1 1 9 ...
9: 1 1 10 ...
10: 1 1 ...
11: 1 ...
etc.
For row 1 see A036765, for row 2 see A186241, and for row 3 see A200731.
Conjecture 1: The A(n,k), here n > 0, are the number of lattice paths, if
(a) length of path is k*n (for the k-th term of row n),
(b) allowed steps are (1,-1), (1,-1+n), (1,-1+2*n), and (1,-1+3*n),
(c) you start at (0,0), end at (k*n,0), and
(d) never cross the x-axis.
Conjecture 2: The coefficients B(m,n,k) of the P(n,x)^m (see the formula below), m > 0 and n > 0, are the number of lattice paths, if
(a) length of path is k*n+m-1 (k-th coefficient of P(n,x)^m),
(b) allowed steps are (1,-1), (1,-1+n), (1,-1+2*n), and (1,-1+3*n),
(c) you start at (0,m-1), end at (k*n+m-1,0), and
(d) never cross the x-axis.

Examples

			The terms of the array A(n,k) read by upwards antidiagonals define the triangle T(n,m) = A(n-m,m) for 0 <= m <= n, i.e.,
n\m 0  1   2   3    4     5     6     7    8  9 ...
0:  1
1:  1  1
2:  1  1   1
3:  1  1   2   1
4:  1  1   3   5    0
5:  1  1   4  12   13     0
6:  1  1   5  22   54    36     0
7:  1  1   6  35  139   262   104     0
8:  1  1   7  51  284   953  1337   309    0
9:  1  1   8  70  505  2509  6894  7072  939  0
etc. [reformatted by _Wolfdieter Lang_, Oct 15 2015]
		

Crossrefs

Formula

A(n,k) = 1/(n*k+1) * Sum_{j=0..k} (-2)^j*binomial(n*k+1,j)* binomial(3*n*k+3-2*j,k-j) for n >= 0, and k >= 0. (conjectured)
A(n,0) = A(n,1) = 1, n >= 0;
A(n,2) = n+1, n >= 0;
A(n,3) = (3*n^2+5*n+2)/2, n >= 0;
A(n,4) = (8*n^3+18*n^2+13*n)/3, n >= 0;
A(n,5) = (125*n^4+350*n^3+355*n^2+34*n)/24, n >= 0.
The g.f. P(n,x) of row n of the array A(n,k) satisfy:
P(n,x) = P(n-1,x*P(n,x)), n > 0;
P(n,x) = P(n-2,x*P(n,x)^2), n > 1;
etc.
P(n,x) = P(0,x*P(n,x)^n), n >= 0.
The coefficients B(m,n,k) of the P(n,x)^m are:
B(m,n,k) = m/(n*k+m) * Sum_{j=0..k} (-2)^j*binomial(n*k+m,j)* binomial(3*n*k+3*m-2*j,k-j) for m > 0, n > 0, and k >= 0. (conjectured)
P(n,x) = exp(Sum_{k>=1} 1/(n*k)*(Sum_{j=0..k} (-2)^j*binomial(n*k,j)* binomial(3*n*k-2*j,k-j))) for n > 0 (conjectured); (see for n=1: A036765, for n=2: A186241, and for n=3: A200731).
P(n,x/(1+x+x^2+x^3)^n) = 1+x+x^2+x^3 for n >= 0. - Werner Schulte, Nov 20 2015
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