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.

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A200075 G.f. satisfies A(x) = (1 + x*A(x)^2)*(1 + x^2*A(x)^3).

Original entry on oeis.org

1, 1, 3, 11, 45, 198, 914, 4367, 21414, 107155, 544987, 2808978, 14640073, 77025373, 408544815, 2182206259, 11727989593, 63373962690, 344109933186, 1876562458845, 10273572074493, 56443282489240, 311097732946200, 1719707775782826, 9531914043637385, 52963938340248863, 294966593345731623
Offset: 0

Views

Author

Paul D. Hanna, Nov 13 2011

Keywords

Comments

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

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 11*x^3 + 45*x^4 + 198*x^5 + 914*x^6 +...
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 28*x^3 + 121*x^4 + 552*x^5 + 2615*x^6 +...
A(x)^3 = 1 + 3*x + 12*x^2 + 52*x^3 + 237*x^4 + 1122*x^5 + 5463*x^6 +...
A(x)^5 = 1 + 5*x + 25*x^2 + 125*x^3 + 630*x^4 + 3211*x^5 + 16545*x^6 +...
where A(x) = 1 + x*A(x)^2 + x^2*A(x)^3 + x^3*A(x)^5.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x*A)*x*A + (1 + 2^2*x*A + x^2*A^2)*x^2*A^2/2 +
(1 + 3^2*x*A + 3^2*x^2*A^2 + x^3*A^3)*x^3*A^3/3 +
(1 + 4^2*x*A + 6^2*x^2*A^2 + 4^2*x^3*A^3 + x^4*A^4)*x^4*A^4/4 +
(1 + 5^2*x*A + 10^2*x^2*A^2 + 10^2*x^3*A^3 + 5^2*x^4*A^4 + x^5*A^5)*x^5*A^5/5 +
(1 + 6^2*x*A + 15^2*x^2*A^2 + 20^2*x^3*A^3 + 15^2*x^4*A^4 + 6^2*x^5*A^5 + x^6*A^6)*x^6*A^6/6 +...
more explicitly,
log(A(x)) = x + 5*x^2/2 + 25*x^3/3 + 129*x^4/4 + 686*x^5/5 + 3713*x^6/6 + 20350*x^7/7 +...
Given G(x) where 1+x*G(x) is the g.f. of A004148, then the coefficients in the powers of G(x) begin:
1: [(1), 1, 2, 4, 8, 17, 37, 82, 185, 423, 978, ...];
2: [1,(2), 5, 12, 28, 66, 156, 370, 882, 2112, ...];
3: [1, 3,(9), 25, 66, 171, 437, 1107, 2790, 7009, ...];
4: [1, 4, 14,(44), 129, 364, 1000, 2696, 7172, 18892, ...];
5: [1, 5, 20, 70,(225), 686, 2015, 5760, 16135, 44500, ...];
6: [1, 6, 27, 104, 363,(1188), 3713, 11214, 32994, 95106, ...];
7: [1, 7, 35, 147, 553, 1932,(6398), 20350, 62734, 188650, ...];
8: [1, 8, 44, 200, 806, 2992, 10460,(34936), 112585, 352560, ...];
9: [1, 9, 54, 264, 1134, 4455, 16389, 57330,(192726), 627406, ...]; ...;
the coefficients in parenthesis form the initial terms of this sequence:
[1/1, 2/2, 9/3, 44/4, 225/5, 1188/6, 6398/7, 34936/8, 192726/9, ...].
The coefficients in the logarithm of the g.f. is also a diagonal in the above table.
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[1/x*InverseSeries[Series[x*(1-x-x^2 + Sqrt[(1+x+x^2)*(1-3*x+x^2)])/2,{x,0,20}],x],x] (* Vaclav Kotesovec, Sep 19 2013 *)
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=(1+x*A^2)*(1+x^2*A^3)+x*O(x^n));polcoeff(A,n)}
    
  • PARI
    {a(n)=polcoeff(1/x*serreverse(x*(1-x-x^2 + sqrt((1+x+x^2)*(1-3*x+x^2)+x*O(x^n)))/2),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^j*A^j)*(x*A+x*O(x^n))^m/m))); polcoeff(A, n, x)}
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (1-x*A)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j*A^j)*x^m*A^m/m))); polcoeff(A, n, x)}

Formula

G.f.: (1/x)*Series_Reversion( x*(1-x-x^2 + sqrt((1+x+x^2)*(1-3*x+x^2)))/2 ).
a(n) = [x^n] G(x)^(n+1)/(n+1), where 1+x*G(x) is the g.f. of A004148.
G.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion( x/G(x) ) where 1+x*G(x) is the g.f. of A004148.
(2) A(x) = G(x*A(x)) where G(x) = A(x/G(x)) 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)^k) * x^n*A(x)^n/n ).
(4) 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)^n/n ).
Recurrence: 8*n*(2*n+1)*(4*n+1)*(4*n+3)*(1557671*n^7 - 18939961*n^6 + 94817789*n^5 - 252067387*n^4 + 381880748*n^3 - 327052012*n^2 + 145198992*n - 25583040)*a(n) = (2026529971*n^11 - 24640889261*n^10 + 122927623620*n^9 - 322351865586*n^8 + 467303512311*n^7 - 343677276405*n^6 + 61590777290*n^5 + 76066203476*n^4 - 45605627832*n^3 + 4625651136*n^2 + 1916801280*n - 338688000)*a(n-1) + 2*(800642894*n^11 - 10936104295*n^10 + 62803409541*n^9 - 196202081616*n^8 + 357730085364*n^7 - 370711524567*n^6 + 174415015309*n^5 + 25877389846*n^4 - 63266190708*n^3 + 19055552472*n^2 + 1313789760*n - 861840000)*a(n-2) + 6*(308418858*n^11 - 4675368852*n^10 + 30103912361*n^9 - 106665982366*n^8 + 223860428776*n^7 - 274000455628*n^6 + 166116940489*n^5 - 2432493994*n^4 - 54297743044*n^3 + 22033617000*n^2 + 936446400*n - 1315440000)*a(n-3) + 6*(n-2)*(2*n-7)*(3*n-10)*(3*n-8)*(1557671*n^7 - 8036264*n^6 + 13889114*n^5 - 7559372*n^4 - 2491645*n^3 + 2975476*n^2 - 179460*n - 187200)*a(n-4). - Vaclav Kotesovec, Sep 19 2013
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 1301/1024 + 1/(1024*sqrt(3/(7183147 - (2002819072*2^(2/3))/(3725055779 + 42057117*sqrt(16305))^(1/3) + 1024*(7450111558 + 84114234*sqrt(16305))^(1/3)))) + (1/2)*sqrt(7183147/393216 - (3725055779 + 42057117*sqrt(16305))^(1/3)/(384*2^(2/3)) + 977939/(192*(7450111558 + 84114234*sqrt(16305))^(1/3)) + (1/131072)*(4194454317*sqrt(3/(7183147 - (2002819072*2^(2/3))/(3725055779 + 42057117*sqrt(16305))^(1/3) + 1024*(7450111558 + 84114234*sqrt(16305))^(1/3))))) = 5.89828930084513611... is the root of the equation -108 - 1188*d - 1028*d^2 - 1301*d^3 + 256*d^4 = 0 and c = 0.656947859044624009263362998790812821830934... - Vaclav Kotesovec, Sep 19 2013
a(n) = Sum_{k=0..floor(n/2)} binomial(2*n-k+1,k) * binomial(2*n-k+1,n-2*k) / (2*n-k+1). - Seiichi Manyama, Jul 18 2023

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

A215576 G.f. satisfies A(x) = (1 + x^2)*(1 + x*A(x)^2).

Original entry on oeis.org

1, 1, 3, 8, 24, 80, 278, 997, 3670, 13782, 52588, 203314, 794726, 3135540, 12470444, 49942305, 201233170, 815205699, 3318291966, 13565162636, 55669063762, 229257178198, 947142023262, 3924380904498, 16303716754884, 67899954924360, 283425070356740, 1185551594834910
Offset: 0

Views

Author

Paul D. Hanna, Aug 16 2012

Keywords

Comments

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

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 8*x^3 + 24*x^4 + 80*x^5 + 278*x^6 + 997*x^7 +...
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 22*x^3 + 73*x^4 + 256*x^5 + 924*x^6 + 3414*x^7 +...
where A(x) = 1+x^2 + x*(1+x^2)*A(x)^2.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x/A^2)*A*x + (1 + 2^2*x/A^2 + x^2/A^4)*A^2*x^2/2 +
 (1 + 3^2*x/A^2 + 3^2*x^2/A^4 + x^3/A^6)*A^3*x^3/3 +
 (1 + 4^2*x/A^2 + 6^2*x^2/A^4 + 4^2*x^3/A^6 + x^4/A^8)*A^4*x^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)*A^5*x^5/5 +
 (1 + 6^2*x/A^2 + 15^2*x^2/A^4 + 20^2*x^3/A^6 + 15^2*x^4/A^8 + 6^2*x^5/A^10 + x^6/A^12)*A^6*x^6/6 +...
more explicitly,
log(A(x)) = x + 5*x^2/2 + 16*x^3/3 + 57*x^4/4 + 231*x^5/5 + 938*x^6/6 + 3830*x^7/7 + 15833*x^8/8 +...
		

Crossrefs

Programs

  • Mathematica
    nmax=20;aa=ConstantArray[0,nmax]; aa[[1]]=1;Do[AGF=1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j; sol=Solve[Coefficient[(1+x^2)*(1+x*AGF^2)-AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}];Flatten[{1,aa}] (* Vaclav Kotesovec, Aug 19 2013 *)
    CoefficientList[Series[(1 - Sqrt[1 - 4*x*(1 + x^2)^2]) / (2*x*(1 + x^2)), {x, 0, 30}], x] (* Vaclav Kotesovec, Oct 11 2018 *)
    Table[Sum[Binomial[2*n - 4*i + 1, i] * Binomial[2*n - 4*i + 1, n - 2*i]/(2*n - 4*i + 1), {i, 0, Floor[n/2]}], {n, 0, 30}] (* Vaclav Kotesovec, Oct 11 2018, after Vladimir Kruchinin *)
  • PARI
    {a(n)=polcoeff((1 - sqrt(1 - 4*x*(1+x^2 +x*O(x^n))^2)) / (2*x*(1+x^2 +x*O(x^n))),n)}
    for(n=0,31,print1(a(n),", "))
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=(1+x*A^2)*(1+x^2)+x*O(x^n)); 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^j/A^(2*j))*(x*A+x*O(x^n))^m/m))); polcoeff(A, n, x)}
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, (1-x/A^2)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j/A^(2*j))*x^m*A^m/m))); polcoeff(A, n, x)}

Formula

G.f. satisfies:
(1) A(x) = (1 - sqrt(1 - 4*x*(1+x^2)^2)) / (2*x*(1+x^2)).
(2) A(x) = exp( Sum_{n>=1} (x^n/n) * A(x)^n * (Sum_{k=0..n} C(n,k)^2 * x^k / A(x)^(2*k)) ).
(3) 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)^n/n ).
(4) A(x) = x / Series_Reversion( x*G(x) ) where G(x) is the g.f. of A200717.
(5) A(x) = G(x/A(x)) where G(x) = A(x*G(x)) is the g.f. of A200717.
Recurrence: (n+1)*a(n) = 2*(2*n-1)*a(n-1) - (n+1)*a(n-2) + 6*(2*n-5)*a(n-3) + 6*(2*n-9)*a(n-5) + 2*(2*n-13)*a(n-7). - Vaclav Kotesovec, Aug 19 2013
a(n) ~ c*d^n/n^(3/2), where d = 4.41997678... is the root of the equation -4-8*d^2-4*d^4+d^5=0 and c = sqrt(d*(8 + 16*d^2 + 8*d^4 + 3*d^5 + d^7) / (Pi*(1 + d^2)^3))/4 = 0.648259186485429075561822659694489853... - Vaclav Kotesovec, Aug 19 2013, updated Oct 11 2018
a(n) = Sum_{i=0..floor(n/2)} C(2*n-4*i+1,i)*C(2*n-4*i+1,n-2*i)/(2*n-4*i+1). - Vladimir Kruchinin, Oct 11 2018

A200717 G.f. satisfies: A(x) = (1 + x*A(x)^3) * (1 + x^2*A(x)^2).

Original entry on oeis.org

1, 1, 4, 18, 93, 521, 3073, 18806, 118297, 760162, 4968480, 32928392, 220766739, 1494635330, 10203884795, 70167751762, 485574854049, 3379064343829, 23631314301088, 165998001901786, 1170706810318259, 8286253163771045, 58842370488310336, 419102145275264242, 2993221125640617827
Offset: 0

Views

Author

Paul D. Hanna, Nov 21 2011

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)).

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 18*x^3 + 93*x^4 + 521*x^5 + 3073*x^6 +...
Related expansions:
A(x)^2 = 1 + 2*x + 9*x^2 + 44*x^3 + 238*x^4 + 1372*x^5 + 8256*x^6 +...
A(x)^3 = 1 + 3*x + 15*x^2 + 79*x^3 + 447*x^4 + 2655*x^5 + 16324*x^6 +...
A(x)^5 = 1 + 5*x + 30*x^2 + 180*x^3 + 1110*x^4 + 7006*x^5 + 45075*x^6 +...
where A(x) = 1 + x*A(x)^3 + x^2*A(x)^2 + x^3*A(x)^5.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x/A)*x*A^2 + (1 + 2^2*x/A + x^2/A^2)*x^2*A^4/2 +
(1 + 3^2*x/A + 3^2*x^2/A^2 + x^3/A^3)*x^3*A^6/3 +
(1 + 4^2*x/A + 6^2*x^2/A^2 + 4^2*x^3/A^3 + x^4/A^4)*x^4*A^8/4 +
(1 + 5^2*x/A + 10^2*x^2/A^2 + 10^2*x^3/A^3 + 5^2*x^4/A^4 + x^5/A^5)*x^5*A^10/5 + ...
		

Crossrefs

Programs

  • Mathematica
    nmax=20; aa=ConstantArray[0,nmax]; aa[[1]]=1; Do[AGF=1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j; sol=Solve[Coefficient[(1 + x*AGF^3) * (1 + x^2*AGF^2) - AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}]; Flatten[{1,aa}] (* Vaclav Kotesovec, Sep 19 2013 *)
  • PARI
    {a(n)=polcoeff((1/x)*serreverse( x*(1 + sqrt(1 - 4*x*(1+x^2)^2 +x*O(x^n))) / (2*(1+x^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) = (1/x)*Series_Reversion( x*(1 + sqrt(1 - 4*x*(1+x^2)^2)) / (2*(1+x^2)) ).
(2) A(x) = exp( Sum_{n>=1} x^n * A(x)^(2*n)/n * [Sum_{k=0..n} C(n,k)^2 * x^k / A(x)^k] ).
(3) A(x) = exp( Sum_{n>=1} x^n * A(x)^(2*n)/n * [(1-x/A(x)^2)^(2*n+1) * Sum_{k>=0} C(n+k,k)^2*x^k / A(x)^k] ).
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 7.60435909657327146... is the root of the equation -108 + 27*d^2 + 1620*d^3 - 216*d^4 - 1456*d^5 - 2556*d^6 - 716*d^7 + 20*d^8 + 16*d^9 = 0 and c = 0.45780648099092640511434469483084555191269495951... - Vaclav Kotesovec, Sep 19 2013

A215623 G.f. satisfies A(x) = (1 + x*A(x)) * (1 + x*A(x)^4).

Original entry on oeis.org

1, 2, 11, 89, 836, 8551, 92445, 1039030, 12019135, 142151324, 1711116646, 20894534324, 258195565959, 3222677162409, 40569811695707, 514520507077695, 6567611974106756, 84310605465652750, 1087798325715407703, 14098475168420865396, 183465816241394787196
Offset: 0

Views

Author

Paul D. Hanna, Aug 17 2012

Keywords

Comments

The radius of convergence of g.f. A(x) is r = 0.0712256396327314729661274986100... with A(r) = 1.4248895273944523042559975726479124492235978714420... where y=A(r) satisfies 3*y^7 - 4*y^6 + 16*y^5 - 28*y^4 + 8*y^3 - 4 = 0.

Examples

			G.f.: A(x) = 1 + 2*x + 11*x^2 + 89*x^3 + 836*x^4 + 8551*x^5 + 92445*x^6 + ...
Related expansions.
A(x)^4 = 1 + 8*x + 68*x^2 + 652*x^3 + 6750*x^4 + 73544*x^5 + 831078*x^6 + ...
A(x)^5 = 1 + 10*x + 95*x^2 + 965*x^3 + 10350*x^4 + 115507*x^5 + ...
where A(x) = 1 + x*(A(x) + A(x)^4) + x^2*A(x)^5.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + A(x)^3)*x + (1 + 2^2*A(x)^3 + A(x)^6)*x^2/2 +
  (1 + 3^2*A(x)^3 + 3^2*A(x)^6 + A(x)^9)*x^3/3 +
  (1 + 4^2*A(x)^3 + 6^2*A(x)^6 + 4^2*A(x)^9 + A(x)^12)*x^4/4 +
  (1 + 5^2*A(x)^3 + 10^2*A(x)^6 + 10^2*A(x)^9 + 5^2*A(x)^12 + A(x)^15)*x^5/5 + ...
more explicitly,
log(A(x)) = 2*x + 18*x^2/2 + 209*x^3/3 + 2550*x^4/4 + 32082*x^5/5 + 411705*x^6/6 + 5356416*x^7/7 + ....
		

Crossrefs

Programs

  • 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*(A+x*O(x^n))^(3*j))*x^m/m))); polcoeff(A, n)}
    
  • PARI
    {a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1+x*A)*(1+x*A^4)+x*O(x^n)); polcoeff(A, n)}
    for(n=0,21,print1(a(n),", "))

Formula

G.f. satisfies A(x) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * A(x)^(3*k)).
The formal inverse of g.f. A(x) is (sqrt((1-x^3)^2 + 4*x^4) - (1+x^3))/(2*x^4).
a(n) = Sum_{k=0..n} binomial(n+3*k+1,k) * binomial(n+3*k+1,n-k) / (n+3*k+1). - Seiichi Manyama, Jul 19 2023
From Peter Bala, Sep 10 2024: (Start)
x/series_reversion(x*A(x)) = 1 + 2*x + 7*x^2 + 39*x^3 + 242*x^4 + 1634*x^5 + ..., the g.f. of A364336.
(1/x) * series_reversion(x/A(x)) = 1 + 2*x + 15*x^2 + 163*x^3 + 2070*x^4 + 28698*x^5 + ..., the g.f. of A364331. (End)

A215624 G.f. satisfies A(x) = (1 + x*A(x)) * (1 + x*A(x)^5).

Original entry on oeis.org

1, 2, 13, 130, 1518, 19358, 261323, 3670828, 53100530, 785657529, 11834135909, 180863294507, 2797643204500, 43715591710804, 689030031494554, 10941710269299893, 174889301792724294, 2811464199460768704, 45426696813655278251
Offset: 0

Views

Author

Paul D. Hanna, Aug 17 2012

Keywords

Comments

The radius of convergence of g.f. A(x) is r = 0.05685644444171304880925020950930... with A(r) = 1.3208055627586104770123863310077013110788003146438630... where y=A(r) satisfies 4*y^9 - 5*y^8 + 25*y^6 - 40*y^5 + 10*y^4 - 5 = 0.

Examples

			G.f.: A(x) = 1 + 2*x + 13*x^2 + 130*x^3 + 1518*x^4 + 19358*x^5 +...
Related expansions.
A(x)^5 = 1 + 10*x + 105*x^2 + 1250*x^3 + 16120*x^4 + 219162*x^5 +...
A(x)^6 = 1 + 12*x + 138*x^2 + 1720*x^3 + 22803*x^4 + 315840*x^5 +...
where A(x) = 1 + x*(A(x) + A(x)^5) + x^2*A(x)^6.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + A(x)^4)*x + (1 + 2^2*A(x)^4 + A(x)^8)*x^2/2 +
  (1 + 3^2*A(x)^4 + 3^2*A(x)^8 + A(x)^12)*x^3/3 +
  (1 + 4^2*A(x)^4 + 6^2*A(x)^8 + 4^2*A(x)^12 + A(x)^16)*x^4/4 +
  (1 + 5^2*A(x)^4 + 10^2*A(x)^8 + 10^2*A(x)^12 + 5^2*A(x)^16 + A(x)^20)*x^5/5 +...
more explicitly,
log(A(x)) = 2*x + 22*x^2/2 + 320*x^3/3 + 4886*x^4/4 + 76962*x^5/5 + 1236784*x^6/6 + 20152260*x^7/7 +...
		

Crossrefs

Programs

  • 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*(A+x*O(x^n))^(4*j))*x^m/m))); polcoeff(A, n)}
    
  • PARI
    {a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1+x*A)*(1+x*A^5)+x*O(x^n)); polcoeff(A, n)}
    for(n=0,21,print1(a(n),", "))

Formula

G.f. satisfies A(x) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * A(x)^(4*k)).
The formal inverse of g.f. A(x) is (sqrt((1-x^4)^2 + 4*x^5) - (1+x^4))/(2*x^5).
a(n) = Sum_{k=0..n} binomial(n+4*k+1,k) * binomial(n+4*k+1,n-k) / (n+4*k+1). - Seiichi Manyama, Jul 19 2023

A200716 G.f. satisfies: A(x) = (1 + x*A(x)^3) * (1 + x^2*A(x)).

Original entry on oeis.org

1, 1, 4, 17, 84, 453, 2574, 15185, 92119, 571022, 3600981, 23029021, 149000790, 973581692, 6415198045, 42580369370, 284427460919, 1910594331920, 12898153658337, 87461992473577, 595455441375978, 4068652368270955, 27891991988552554, 191783482751813061, 1322319472577803761
Offset: 0

Views

Author

Paul D. Hanna, Nov 21 2011

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)).

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 17*x^3 + 84*x^4 + 453*x^5 + 2574*x^6 +...
Related expansions:
A(x)^3 = 1 + 3*x + 15*x^2 + 76*x^3 + 414*x^4 + 2370*x^5 + 14047*x^6 +...
A(x)^4 = 1 + 4*x + 22*x^2 + 120*x^3 + 685*x^4 + 4048*x^5 + 24558*x^6 +...
where A(x) = 1 + x*A(x)^3 + x^2*A(x) + x^3*A(x)^4.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x/A^2)*x*A^2 + (1 + 2^2*x/A^2 + x^2/A^4)*x^2*A^4/2 +
(1 + 3^2*x/A^2 + 3^2*x^2/A^4 + x^3/A^6)*x^3*A^6/3 +
(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^8/4 +
(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^10/5 + ...
		

Crossrefs

Programs

  • Mathematica
    nmax=20; aa=ConstantArray[0,nmax]; aa[[1]]=1; Do[AGF=1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j; sol=Solve[Coefficient[(1 + x*AGF^3) * (1 + x^2*AGF) - AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}]; Flatten[{1,aa}] (* Vaclav Kotesovec, Sep 19 2013 *)
  • 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. A(x) satisfies:
(1) A(x) = exp( Sum_{n>=1} x^n * A(x)^(2*n)/n * [Sum_{k=0..n} C(n,k)^2 * x^k / A(x)^(2*k)] ).
(2) A(x) = exp( Sum_{n>=1} x^n * A(x)^(2*n)/n * [(1-x/A(x)^2)^(2*n+1) * Sum_{k>=0} C(n+k,k)^2 * x^k/A(x)^(2*k)] ).
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 7.342019160707096169... is the root of the equation -27 + 108*d^2 - 162*d^4 + 54*d^5 + 108*d^6 + 216*d^7 - 27*d^8 - 18*d^9 - 27*d^10 + 4*d^11 = 0 and c = 0.468554406193087607276981923311829947714908080994... - Vaclav Kotesovec, Sep 19 2013

A200731 G.f. satisfies: A(x) = (1 + x*A(x)^3) * (1 + x^2*A(x)^6).

Original entry on oeis.org

1, 1, 4, 22, 139, 953, 6894, 51796, 400269, 3161262, 25403536, 207043048, 1707345547, 14219399626, 119431172630, 1010495472960, 8604568715969, 73683710894255, 634142349130800, 5482062214763436, 47582484748270453, 414503778412715065, 3622792181209018168, 31758958747482608912
Offset: 0

Views

Author

Paul D. Hanna, Nov 21 2011

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, q=3.

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 22*x^3 + 139*x^4 + 953*x^5 + 6894*x^6 +...
where A(x) = (1 + x*A(x)^3)*(1 + x^2*A(x)^6).
Related expansions:
A(x)^3 = 1 + 3*x + 15*x^2 + 91*x^3 + 609*x^4 + 4335*x^5 + 32197*x^6 +...
A(x)^6 = 1 + 6*x + 39*x^2 + 272*x^3 + 1989*x^4 + 15054*x^5 + 116955*x^6 +...
A(x)^9 = 1 + 9*x + 72*x^2 + 570*x^3 + 4545*x^4 + 36639*x^5 + 298662*x^6 +...
where A(x) = 1 + x*A(x)^3 + x^2*A(x)^6 + x^3*A(x)^9.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x*A^3)*x*A^2 + (1 + 2^2*x*A^3 + x^2*A^6)*x^2*A^4/2 +
(1 + 3^2*x*A^3 + 3^2*x^2*A^6 + x^3*A^9)*x^3*A^6/3 +
(1 + 4^2*x*A^3 + 6^2*x^2*A^6 + 4^2*x^3*A^9 + x^4*A^12)*x^4*A^8/4 +
(1 + 5^2*x*A^3 + 10^2*x^2*A^6 + 10^2*x^3*A^9 + 5^2*x^4*A^12 + x^5*A^15)*x^5*A^10/5 + ...
which involves squares of binomial coefficients.
		

Crossrefs

Programs

  • Mathematica
    nmax = 23; sol = {a[0] -> 1};
    Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x] - (1 + x A[x]^3)*(1 + x^2 A[x]^6) + 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 *)
  • PARI
    {a(n)=polcoeff( ((1/x)*serreverse(x/(1 + x + x^2 + x^3 +x*O(x^n))^3))^(1/3), n)}
    
  • PARI
    {a(n)=polcoeff( (1 + x + x^2 + x^3 +x*O(x^n))^(3*n+1)/(3*n+1), n)}
    
  • PARI
    {a(n)=local(p=2,q=3,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=3,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=3,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(n) = [x^n] (1 + x + x^2 + x^3)^(3*n+1) / (3*n+1).
(2) A(x) = ( (1/x)*Series_Reversion( x/(1 + x + x^2 + x^3)^3 ) )^(1/3).
(3) A( x/(1 + x + x^2 + x^3)^3 ) = 1 + x + x^2 + x^3.
(4) A(x) = G(x*A(x)^2) where G(x) = A(x/G(x)^2) = g.f. of A036765 (number of rooted trees with a degree constraint).
(5) A(x) = exp( Sum_{n>=1} x^n*A(x)^(2*n)/n * [Sum_{k=0..n} C(n,k)^2 * x^k*A(x)^(3*k)] ).
(6) A(x) = exp( Sum_{n>=1} x^n*A(x)^(2*n)/n * [(1-x*A(x)^2)^(2*n+1)*Sum_{k>=0} C(n+k,k)^2*x^k*A(x)^(3*k) )] ).
From Peter Bala, Jun 21 2015: (Start)
a(n) = 1/(3*n + 1)*Sum_{k = 0..floor(n/2)} binomial(3*n + 1,k)*binomial(3*n + 1,n - 2*k).
More generally, the coefficient of x^n in A(x)^r equals r/(3*n + r)*Sum_{k = 0..floor(n/2)} binomial(3*n + r,k)*binomial(3*n + r,n - 2*k) by the Lagrange-Bürmann formula.
O.g.f. A(x) = exp(Sum_{n >= 1} 1/3*b(n)x^n/n), where b(n) = Sum_{k = 0..floor(n/2)} binomial(3*n,k)*binomial(3*n,n - 2*k). Cf. A036765, A186241, A198951. (End)
Recurrence: 128*n*(2*n - 1)*(4*n - 1)*(4*n + 1)*(8*n - 3)*(8*n - 1)*(8*n + 1)*(8*n + 3)*(511073753*n^7 - 4871850365*n^6 + 19478089219*n^5 - 42349790393*n^4 + 54094962928*n^3 - 40605677522*n^2 + 16589611340*n - 2846611200)*a(n) = 3*(3*n - 2)*(3*n - 1)*(3047149994898003*n^13 - 32094344705469618*n^12 + 145743661212727337*n^11 - 373710048777443810*n^10 + 593788894662012231*n^9 - 600683242386376410*n^8 + 377600776651518819*n^7 - 130595257353511374*n^6 + 11334217618972546*n^5 + 8004135084547148*n^4 - 2618300200112616*n^3 + 152383960257264*n^2 + 33025238671680*n - 3264156403200)*a(n-1) - 576*(n-1)*(3*n - 5)*(3*n - 4)*(3*n - 2)*(3*n - 1)*(495741540410*n^10 - 3982082543435*n^9 + 12891395244590*n^8 - 21360691645174*n^7 + 18695904340190*n^6 - 7495052530111*n^5 + 212344193250*n^4 + 656210670544*n^3 - 106487698440*n^2 - 7969373424*n + 1477828800)*a(n-2) + 110592*(n-2)*(n-1)*(3*n - 8)*(3*n - 7)*(3*n - 5)*(3*n - 4)*(3*n - 2)*(3*n - 1)*(511073753*n^7 - 1294334094*n^6 + 979535842*n^5 - 149518418*n^4 - 72732399*n^3 + 16154432*n^2 + 843684*n - 192240)*a(n-3). - Vaclav Kotesovec, Nov 17 2017
a(n) ~ s/(2*sqrt(3*Pi*(4 - 9*r*s^2*(1 + r*s^3)))*n^(3/2)*r^n), where r = 0.1068159753611743655799981945670627355827110854720... and s = 1.345561337338583233012136458010090420775336284226... are real roots of the system of equations (1 + r*s^3)*(1 + r^2*s^6) = s, 3*r*s^2*(1 + 2*r*s^3 + 3*r^2*s^6) = 1. - Vaclav Kotesovec, Nov 22 2017

A228960 a(n) = [x^n] (1 + x + x^3 + x^4)^n.

Original entry on oeis.org

1, 1, 4, 17, 51, 136, 393, 1233, 3865, 11851, 36301, 112520, 351352, 1098189, 3433704, 10758609, 33794505, 106344793, 335061790, 1056924667, 3338026857, 10554163533, 33402840615, 105809430024, 335444908176, 1064268538776, 3379009937161, 10735253448349, 34127137228747
Offset: 1

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Author

Paul D. Hanna, Sep 10 2013

Keywords

Comments

Equals the Logarithmic derivative of A198951, where the g.f. of A198951 satisfies: G(x) = (1 + x*G(x))*(1 + x^3*G(x)^3).

Examples

			L.g.f.: L(x) = x + x^2/2 + 4*x^3/3 + 17*x^4/4 + 51*x^5/5 + 136*x^6/6 +...
Given G(x) = exp(L(x)), which is the g.f. of A198951:
G(x) = 1 + x + x^2 + 2*x^3 + 6*x^4 + 16*x^5 + 39*x^6 + 99*x^7 + 271*x^8 + 763*x^9 + 2146*x^10 +...+ A198951(n)*x^n +...
then the l.g.f. L(x) satisfies the series:
L(x) = (1 + x^2*G(x)^2)*x
+ (1 + 2^2*x^2*G(x)^2 + x^4*G(x)^4)*x^2/2
+ (1 + 3^2*x^2*G(x)^2 + 3^2*x^4*G(x)^4 + x^6*G(x)^6)*x^3/3
+ (1 + 4^2*x^2*G(x)^2 + 6^2*x^4*G(x)^4 + 4^2*x^6*G(x)^6 + x^8*G(x)^8)*x^4/4
+ (1 + 5^2*x^2*G(x)^2 + 10^2*x^4*G(x)^4 + 10^2*x^6*G(x)^6 + 5^2*x^8*G(x)^8 + x^10*G(x)^10)*x^5/5 +...
The table of coefficients in (1 + x + x^3 + x^4)^n begins:
n=1: [1,(1), 0,  1,   1,   0,   0,    0,    0,    0,    0, ...];
n=2: [1, 2, (1), 2,   4,   2,   1,    2,    1,    0,    0, ...];
n=3: [1, 3,  3, (4),  9,   9,   6,    9,    9,    4,    3, ...];
n=4: [1, 4,  6,  8, (17), 24,  22,   28,   36,   28,   22, ...];
n=5: [1, 5, 10, 15,  30, (51), 60,   75,  105,  110,  100, ...];
n=6: [1, 6, 15, 26,  51,  96,(136), 180,  261,  326,  345, ...];
n=7: [1, 7, 21, 42,  84, 168, 273, (393), 588,  819,  987, ...];
n=8: [1, 8, 28, 64, 134, 280, 504,  792,(1233),1848, 2472, ...];
n=9: [1, 9, 36, 93, 207, 450, 876, 1494, 2439,(3865),5616, ...]; ...
the terms in parenthesis forms the initial terms of this sequence.
		

Crossrefs

Programs

  • Maple
    A228960 := proc(n)
        (1+x+x^3+x^4)^n ;
        coeftayl(%,x=0,n) ;
    end proc: # R. J. Mathar, Sep 15 2013
  • Mathematica
    Table[Coefficient[(1 + x + x^3 + x^4)^n,x,n],{n,1,30}] (* Vaclav Kotesovec, Dec 27 2013 *)
    Table[HypergeometricPFQ[{1/3 - n/3, 2/3 - n/3, -n, -n/3}, {1/3, 2/3, 1}, 1], {n, 20}] (* Vladimir Reshetnikov, Oct 04 2016 *)
  • PARI
    {a(n)=polcoeff((1+x+x^3+x^4+x*O(x^n))^n, n)}
    for(n=1,30,print1(a(n),", "))
    
  • PARI
    {a(n)=local(A=1/x*serreverse(x/(1+x+x^3+x^4+x*O(x^n)))); n*polcoeff(log(A), n)}
    for(n=1,30,print1(a(n),", "))
    
  • PARI
    {a(n)=local(A=x); for(i=1, n, A=sum(m=1, n, sum(j=0, m, binomial(m, j)^2*x^(2*j)*exp(2*j*A+x*O(x^n)))*x^m/m)); n*polcoeff(A, n)}
    for(n=1,30,print1(a(n),", "))
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=sum(m=1, n, (1-x^2*exp(2*A))^(2*m+1)*sum(j=0, n\2, binomial(m+j, j)^2*x^(2*j)*exp(2*j*A+x*O(x^n)))*x^m/m)); n*polcoeff(A, n, x)}
    for(n=1,30,print1(a(n),", "))

Formula

L.g.f. L(x) satisfies:
(1) L(x) = log( (1/x)*Series_Reversion(x/((1+x)*(1+x^3))) ).
(2) L(x) = Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * x^(2*k) * exp(2*k*L(x)).
(3) L(x) = Sum_{n>=1} x^n/n * (1 - x^2*exp(2*L(x)))^(2*n+1) * Sum_{k>=0} C(n+k,k)^2 * x^(2*k) * exp(2*k*L(x)).
Conjecture: 3*n*(1661*n-1820) *(3*n-1) *(3*n-2) *a(n) +(1927258*n^4 -17091925*n^3 +50171975*n^2 -59448794*n +24442440) *a(n-1) +12*(-705605*n^4 +6363374*n^3 -20228575*n^2 +27219817*n -13185456) *a(n-2) +54*(n-2) *(250934*n^3 -1892927*n^2 +4367836*n -3055963) *a(n-3) -486*(n-2)*(n-3) *(26090*n-34343) *(2*n-7) *a(n-4)=0. - R. J. Mathar, Sep 15 2013.
Recurrence (of order 3): 3*n*(3*n-2)*(3*n-1)*(238*n^3 - 1302*n^2 + 2285*n - 1293)*a(n) = 2*(13328*n^6 - 92904*n^5 + 249452*n^4 - 329211*n^3 + 224408*n^2 - 74649*n + 9360)*a(n-1) - 18*(n-1)*(2380*n^5 - 15400*n^4 + 36710*n^3 - 39398*n^2 + 18345*n - 2880)*a(n-2) + 162*(n-2)*(n-1)*(2*n-5)*(238*n^3 - 588*n^2 + 395*n - 72)*a(n-3). - Vaclav Kotesovec, Dec 27 2013
a(n) ~ c*d^n/sqrt(n), where d = 1/81*((2144134 + 520506*sqrt(17))^(2/3) - 2036 + 112*(2144134 + 520506*sqrt(17))^(1/3))*(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 = 1/102*sqrt(17)*sqrt((34102 + 8262*sqrt(17))^(1/3)*((34102+8262*sqrt(17))^(2/3) + 136 + 136*(34102 + 8262*sqrt(17))^(1/3)))/((34102 + 8262*sqrt(17))^(1/3)*sqrt(Pi)) = 0.3061270429417747... - Vaclav Kotesovec, Dec 27 2013
From Peter Bala, Jun 15 2015: (Start)
a(n) = [x^(3*n)](1 + x + x^3 + x^4)^n.
a(n) = Sum_{k = 0..floor(n/3)} binomial(n,k)*binomial(n,3*k). Applying Maple's sumrecursion command to this formula gives the above recurrence of Kotesovec. (End)
a(n) = hypergeom([1/3-n/3, 2/3-n/3, -n, -n/3], [1/3, 2/3, 1], 1). - Vladimir Reshetnikov, Oct 04 2016
From Peter Bala, Apr 15 2023: (Start)
Conjecture 1: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for positive integers n and r and all primes p >= 5. Cf. A350383.
Conjecture 2: let k be a positive integer, m an integer and let f(x) = g(x)/h(x), where both g(x) and h(x) are finite products of cyclotomic polynomials. Then the same supercongruences hold, except for a finite number of primes p depending on f(x), for the sequence {a_(k,m,f)(n): n >= 0} defined by a_(k,m,f)(n) = [x^(k*n)] f(x)^(m*n). (End)

A200725 G.f. A(x) satisfies A(x) = (1+x^2)*(1 + x*A(x)^3).

Original entry on oeis.org

1, 1, 4, 16, 76, 399, 2206, 12664, 74790, 451420, 2772313, 17267652, 108821293, 692609446, 4445642625, 28744599748, 187047449289, 1224027357216, 8050074481917, 53179900898596, 352726704965748, 2348036826102013, 15682048658695168, 105052549830928908, 705678173069959645
Offset: 0

Views

Author

Paul D. Hanna, Nov 21 2011

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, q=-3.

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 16*x^3 + 76*x^4 + 399*x^5 + 2206*x^6 +...
Related expansion:
A(x)^3 = 1 + 3*x + 15*x^2 + 73*x^3 + 384*x^4 + 2133*x^5 + 12280*x^6 +...
where a(3) = 1 + 15; a(4) = 3 + 73; a(5) = 15 + 384; a(6) = 73 + 2133; ...
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x/A^3)*x*A^2 + (1 + 2^2*x/A^3 + x^2/A^6)*x^2*A^4/2 +
(1 + 3^2*x/A^3 + 3^2*x^2/A^6 + x^3/A^9)*x^3*A^6/3 +
(1 + 4^2*x/A^3 + 6^2*x^2/A^6 + 4^2*x^3/A^9 + x^4/A^12)*x^4*A^8/4 +
(1 + 5^2*x/A^3 + 10^2*x^2/A^6 + 10^2*x^3/A^9 + 5^2*x^4/A^12 + x^5/A^15)*x^5*A^10/5 + ...
which involves the squares of the binomial coefficients C(n,k).
		

Crossrefs

Programs

  • Mathematica
    nmax=20;aa=ConstantArray[0,nmax]; aa[[1]]=1;Do[AGF=1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j; sol=Solve[Coefficient[(1+x^2)*(1+x*AGF^3)-AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}];Flatten[{1,aa}] (* Vaclav Kotesovec, Aug 19 2013 *)
  • PARI
    {a(n)=local(p=2,q=-3,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=-3,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=-3,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) = exp( Sum_{n>=1} x^n * A(x)^(2*n)/n * (Sum_{k=0..n} C(n,k)^2 * x^k / A(x)^(3*k)) ).
(2) A(x) = exp( Sum_{n>=1} x^n * A(x)^(2*n)/n * (1 - x/A(x)^3)^(2*n+1) * Sum_{k>=0} C(n+k,k)^2*x^k / A(x)^(3*k) ).
Recurrence: 2*(n-4)*(n-2)*n*(2*n+1)*a(n) = 3*(n-4)*(n-2)*(3*n-2)*(3*n-1)*a(n-1) - 2*(n-4)*(n-2)*n*(2*n-1)*a(n-2) + 6*(n-4)*(3*n-8)*(6*n^2 - 17*n + 2)*a(n-3) + 6*(3*n-14)*(9*n^3 - 66*n^2 + 114*n - 4)*a(n-5) + 6*n*(3*n-20)*(6*n^2 - 47*n + 78)*a(n-7) + 3*(n-2)*n*(3*n-26)*(3*n-19)*a(n-9). - Vaclav Kotesovec, Aug 19 2013
a(n) ~ c*d^n/n^(3/2), where d = 7.1535029565... is the root of the equation -27 - 81*d^2 - 81*d^4 - 27*d^6 + 4*d^7 = 0 and c = 0.26300783791885411389369671... - Vaclav Kotesovec, Aug 19 2013
a(n) = Sum_{k=0..floor(n/2)} binomial(3*n-6*k+1,k) * binomial(3*n-6*k+1,n-2*k)/(3*n-6*k+1). - Seiichi Manyama, Dec 17 2024
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