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.

A192415 G.f. satisfies: A(x) = (1 + x*A(x))*(1 + x^3*A(x)^2).

Original entry on oeis.org

1, 1, 1, 2, 5, 11, 23, 51, 120, 286, 681, 1636, 3985, 9803, 24257, 60338, 150931, 379501, 958360, 2429294, 6179380, 15769380, 40361087, 103579221, 266471500, 687098810, 1775440421, 4596689688, 11922774513, 30977768907, 80615085087, 210103228155, 548352756656, 1433053608502
Offset: 0

Views

Author

Paul D. Hanna, Nov 02 2011

Keywords

Examples

			G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 5*x^4 + 11*x^5 + 23*x^6 + 51*x^7 +...
Related expansions:
A(x)^2 = 1 + 2*x + 3*x^2 + 6*x^3 + 15*x^4 + 36*x^5 + 82*x^6 + 190*x^7 +...
A(x)^3 = 1 + 3*x + 6*x^2 + 13*x^3 + 33*x^4 + 84*x^5 + 205*x^6 + 498*x^7 +...
where A(x) = 1 + x*A(x) + x^3*A(x)^2 + x^4*A(x)^3.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x^2*A(x))*x + (1 + 2^2*x^2*A(x) + x^4*A(x)^2)*x^2/2 +
(1 + 3^2*x^2*A(x) + 3^2*x^4*A(x)^2 + x^6*A(x)^3)*x^3/3 +
(1 + 4^2*x^2*A(x) + 6^2*x^4*A(x)^2 + 4^2*x^6*A(x)^3 + x^8*A(x)^4)*x^4/4 +
(1 + 5^2*x^2*A(x) + 10^2*x^4*A(x)^2 + 10^2*x^6*A(x)^3 + 5^2*x^8*A(x)^4 + x^10*A(x)^5)*x^5/5 +...
Explicitly,
log(A(x)) = x + x^2/2 + 4*x^3/3 + 13*x^4/4 + 31*x^5/5 + 70*x^6/6 + 176*x^7/7 + 469*x^8/8 + 1228*x^9/9 + 3161*x^10/10 +...

Links

Crossrefs

Cf. A198951, A198953, A198957, A007863, A036765, A104545.

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)*(1 + x^3*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)=local(A=1+x); for(i=1, n, A=(1 + x*A)*(1 + x^3*(A+x*O(x^n))^2)); 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^(2*j)*(A+x*O(x^n))^j)*x^m/m))); polcoeff(A, n, x)}
  • PARI
    x='x; y='y; Fxy = (1+x*y) * (1 + x^3*y^2) - y;
seq(N) = {
  my(y0 = 1 + O('x^N), y1=0);
  for (k = 1, N,
    y1 = y0 - subst(Fxy, y, y0)/subst(deriv(Fxy,y), y, y0);
    if (y1 == y0, break()); y0 = y1);
  Vec(y0);
};
seq(34)  \\ Gheorghe Coserea, Nov 30 2016

Formula

G.f. satisfies: A(x) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * x^(2*k) * A(x)^k ).
D-finite with recurrence: 4*(n+1)*(n+2)*(217*n^3 - 1239*n^2 + 1838*n - 336)*a(n) = 6*(n+1)*(434*n^4 - 2261*n^3 + 2339*n^2 + 1792*n - 1344)*a(n-1) - (n-1)*(2821*n^4 - 13286*n^3 + 7829*n^2 + 18464*n - 4032)*a(n-2) + 6*(868*n^5 - 6258*n^4 + 13981*n^3 - 7438*n^2 - 7769*n + 6136)*a(n-3) + 2*(n-3)*(2*n - 5)*(434*n^3 - 1393*n^2 + 211*n + 1048)*a(n-4) + 2*(n-4)*(2*n - 7)*(217*n^3 - 588*n^2 + 11*n + 480)*a(n-5). - Vaclav Kotesovec, Sep 19 2013
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 2.730683387097269698... is the root of the equation -4 - 8*d - 24*d^2 + 13*d^3 - 12*d^4 + 4*d^5 = 0 and c = 2.078548317061344694159945441842754... is the root of the equation -1 - 67*c^2 - 19811*c^4 + 36463*c^6 - 41664*c^8 + 7936*c^10 = 0. - Vaclav Kotesovec, Sep 19 2013, updated Nov 28 2016
a(n) = Sum_{k=0..n/2+1} C(n-k+2,k-1)*C(n-k+2,2*k-1)/(n-k+2). - Vladimir Kruchinin, Feb 12 2019

A198957 G.f. satisfies: A(x) = (1 + x*A(x))*(1 + x^2*A(x)^4).

Original entry on oeis.org

1, 1, 2, 7, 26, 102, 424, 1827, 8078, 36466, 167376, 778718, 3664164, 17407068, 83375616, 402198915, 1952296598, 9528757098, 46735576816, 230227356906, 1138609205372, 5651170500612, 28138939936704, 140527262919342, 703704207921932, 3532664478314484, 17775185122527776
Offset: 0

Views

Author

Paul D. Hanna, Nov 01 2011

Keywords

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 26*x^4 + 102*x^5 + 424*x^6 + 1827*x^7 +...
Related expansions:
A(x)^4 = 1 + 4*x + 14*x^2 + 56*x^3 + 237*x^4 + 1028*x^5 + 4570*x^6 +...
A(x)^5 = 1 + 5*x + 20*x^2 + 85*x^3 + 375*x^4 + 1681*x^5 + 7660*x^6 +...
where A(x) = 1 + x*A(x) + x^2*A(x)^4 + x^3*A(x)^5.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + x*A(x)^3)*x + (1 + 2^2*x*A(x)^3 + x^2*A(x)^6)*x^2/2 +
(1 + 3^2*x*A(x)^3 + 3^2*x^2*A(x)^6 + x^3*A(x)^9)*x^3/3 +
(1 + 4^2*x*A(x)^3 + 6^2*x^2*A(x)^6 + 4^2*x^3*A(x)^9 + x^4*A(x)^12)*x^4/4 +
(1 + 5^2*x*A(x)^3 + 10^2*x^2*A(x)^6 + 10^2*x^3*A(x)^9 + 5^2*x^4*A(x)^12 + x^5*A(x)^15)*x^5/5 +...
more explicitly,
log(A(x)) = x + 3*x^2/2 + 16*x^3/3 + 75*x^4/4 + 356*x^5/5 + 1746*x^6/6 + 8660*x^7/7 + 43299*x^8/8 +...
Also, g.f. A(x) = G(x*A(x)) where G(x) = A(x/G(x)) (g.f. of A104545) begins:
G(x) = 1 + x + x^2 + 3*x^3 + 5*x^4 + 11*x^5 + 25*x^6 + 55*x^7 + 129*x^8 +...

Crossrefs

Cf. A198953, A198951, A007863, A036765, A104545.

Programs

  • Mathematica
    CoefficientList[1/x*InverseSeries[Series[2*x^3*(1+x)/(1 - Sqrt[1-4*x^2*(1+x)^2]), {x, 0, 20}], x],x] (* Vaclav Kotesovec, May 28 2014 *)
  • Maxima
    a(n):=sum(binomial(2*j+n,j)*binomial(2*j+n+1,4*j+1)/(n+j+1),j,0,(n)/2); /* Vladimir Kruchinin, May 28 2014 */
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=(1 + x*A)*(1 + x^2*(A+x*O(x^n))^4)); polcoeff(A, n)}
  • PARI
    {a(n)=polcoeff((1/x)*serreverse( 2*x^3*(1+x)/(1 - sqrt(1-4*x^2*(1+x +x^3*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^3+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, (1-x*A^3)^(2*m+1)*sum(j=0, n, binomial(m+j, j)^2*x^j*(A^3+x*O(x^n))^j)*x^m/m))); polcoeff(A, n, x)}
  • PARI
    x='x; y='y; Fxy = (1 + x*y)*(1 + x^2*y^4) - y;
seq(N) = {
  my(y0 = 1 + O('x^N), y1=0);
  for (k = 1, N,
    y1 = y0 - subst(Fxy, y, y0)/subst(deriv(Fxy, y), y, y0);
    if (y1 == y0, break()); y0 = y1);
  Vec(y0);
};
seq(27) \\ Gheorghe Coserea, Nov 30 2016

Formula

G.f. A(x) satisfies:
(1) A(x) = exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * x^k * A(x)^(3*k) ).
(2) A(x) = (1/x)*Series_Reversion( 2*x^3*(1+x)/(1 - sqrt(1-4*x^2*(1+x)^2)) ).
(3) A(x) = G(x*A(x)) where G(x) = A(x/G(x)) is the g.f. of A104545 (Motzkin paths of length n having no consecutive (1,0) steps).
(4) A(x) = exp( Sum_{n>=1} x^n/n * (1-x*A(x)^3)^(2*n+1)*Sum_{k>=0} C(n+k,k)^2 * x^k * A(x)^(3*k) ).
a(n) = sum(j=0..n/2, binomial(2*j+n,j)*binomial(2*j+n+1,4*j+1)/(n+j+1)). - Vladimir Kruchinin, May 28 2014
a(n) ~ sqrt((1 + 2*r*s^3 + 3*r^2*s^4)/(2*Pi*s*(3 + 5*r*s))) / (2*n^(3/2)*r^(n+1/2)), where r = 0.187614989725738719..., s = 1.61178302212918247... are roots of the system of equations r + 4*r^2*s^3 + 5*r^3*s^4 = 1, (1+r*s)*(1+r^2*s^4) = s. - Vaclav Kotesovec, May 28 2014

A200718 G.f. satisfies A(x) = (1 + x*A(x)^2) * (1 + x^2*A(x)^6).

Original entry on oeis.org

1, 1, 3, 14, 75, 433, 2636, 16668, 108399, 720431, 4871555, 33409042, 231817448, 1624503716, 11480658056, 81731416480, 585579734959, 4219179476875, 30552067317233, 222225174139730, 1622894404239115, 11894991079960721, 87472260252499560, 645183802300787356, 4771926560361458884
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 + 3*x^2 + 14*x^3 + 75*x^4 + 433*x^5 + 2636*x^6 +...
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 34*x^3 + 187*x^4 + 1100*x^5 + 6784*x^6 +...
A(x)^6 = 1 + 6*x + 33*x^2 + 194*x^3 + 1200*x^4 + 7674*x^5 + 50317*x^6 +...
A(x)^8 = 1 + 8*x + 52*x^2 + 336*x^3 + 2210*x^4 + 14776*x^5 + 100216*x^6 +...
where A(x) = 1 + x*A(x)^2 + x^2*A(x)^6 + x^3*A(x)^8.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x*A^4)*x*A + (1 + 2^2*x*A^4 + x^2*A^8)*x^2*A^2/2 +
(1 + 3^2*x*A^4 + 3^2*x^2*A^8 + x^3*A^12)*x^3*A^3/3 +
(1 + 4^2*x*A^4 + 6^2*x^2*A^8 + 4^2*x^3*A^12 + x^4*A^16)*x^4*A^4/4 +
(1 + 5^2*x*A^4 + 10^2*x^2*A^8 + 10^2*x^3*A^12 + 5^2*x^4*A^16 + x^5*A^20)*x^5*A^5/5 + ...
The g.f. of A104545, G(x) = A(x/G(x)^2) where A(x) = G(x*A(x)^2), begins:
G(x) = 1 + x + x^2 + 3*x^3 + 5*x^4 + 11*x^5 + 25*x^6 + 55*x^7 + 129*x^8 +...

Links

Crossrefs

Cf. A104545, A200716, A200717, A199876, A199877, A198951, A198953, A198957, A192415, A198888, A036765.
Cf. A186241, A199874, A200074, A200075, A200719, A215576.

Programs

  • Mathematica
    a[n_] := Sum[Binomial[2*n + 2*k + 1, k]*Binomial[2*n + 2*k + 1, n - 2*k]/ (2*n + 2*k + 1), {k, 0, n/2}];
Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Jan 09 2018, after Vladimir Kruchinin *)
  • Maxima
    a(n):=sum((binomial(2*n+2*k+1,k)*binomial(2*n+2*k+1,n-2*k))/(2*n+2*k+1),k,0,(n)/2); /* Vladimir Kruchinin, Mar 11 2016 */
  • PARI
    {a(n)=polcoeff(sqrt( (1/x)*serreverse( 2*x^5*(1+x)^2/(1 - 2*x^2*(1+x)^2 - sqrt(1 - 4*x^2*(1+x)^2+O(x^(n+6)))) ) ),n)}
  • PARI
    {a(n)=local(p=1,q=4,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=1,q=4,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=1,q=4,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( 2*x^5*(1+x)^2/(1 - 2*x^2*(1+x)^2 - sqrt(1 - 4*x^2*(1+x)^2)) ) ).
(2) A(x) = G(x*A(x)^2) where G(x) = A(x/G(x)^2) is the g.f. of A104545 (Motzkin paths of length n having no consecutive (1,0) steps).
(3) A(x) = exp( Sum_{n>=1} x^n * A(x)^n/n * [Sum_{k=0..n} C(n,k)^2 * x^k * A(x)^(4*k)] ).
(4) A(x) = exp( Sum_{n>=1} x^n * A(x)^n/n * [(1-x/A(x)^4)^(2*n+1) * Sum_{k>=0} C(n+k,k)^2*x^k * A(x)^(4*k)] ).
a(n) = Sum_{k=0..floor(n/2)}((binomial(2*n+2*k+1,k)*binomial(2*n+2*k+1,n-2*k))/(2*n+2*k+1)). - Vladimir Kruchinin, Mar 11 2016

A200719 G.f. satisfies A(x) = (1 + x*A(x)^2) * (1 + x^2*A(x)^5).

Original entry on oeis.org

1, 1, 3, 13, 64, 340, 1903, 11053, 65993, 402527, 2497439, 15712220, 100001459, 642719263, 4165537744, 27193644061, 178654643151, 1180282875483, 7836312619243, 52259258911091, 349902441457427, 2351240866736891, 15851508780927739, 107187240225220684, 726784821098903319
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 + 3*x^2 + 13*x^3 + 64*x^4 + 340*x^5 + 1903*x^6 +...
Related expansions:
A(x)^2 = 1 + 2*x + 7*x^2 + 32*x^3 + 163*x^4 + 886*x^5 + 5039*x^6 +...
A(x)^5 = 1 + 5*x + 25*x^2 + 135*x^3 + 765*x^4 + 4481*x^5 + 26920*x^6 +...
A(x)^7 = 1 + 7*x + 42*x^2 + 252*x^3 + 1533*x^4 + 9457*x^5 + 59101*x^6 +...
where A(x) = 1 + x*A(x)^2 + x^2*A(x)^5 + x^3*A(x)^7.
The logarithm of the g.f. A = A(x) equals the series:
log(A(x)) = (1 + x*A^3)*x*A + (1 + 2^2*x*A^3 + x^2*A^6)*x^2*A^2/2 +
(1 + 3^2*x*A^3 + 3^2*x^2*A^6 + x^3*A^9)*x^3*A^3/3 +
(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^4/4 +
(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^5/5 + ...

Links

Crossrefs

Cf. A200716, A200717, A200718, A200074, A200075, A199876, A199874, A199877, A198951, A198953, A198957, A192415, A198888, A036765, A104545.
Cf. A186241, A199874, A200074, A200075, A200718, A215576.

Programs

  • PARI
    {a(n)=local(p=1,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=1,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=1,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)^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)^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: 4232*(n-2)*(n-1)*n*(2*n - 3)*(2*n - 1)*(2*n + 1)*(108983978975*n^7 - 1828734495225*n^6 + 13017379495661*n^5 - 50928975062019*n^4 + 118201965098732*n^3 - 162617590602876*n^2 + 122676758610192*n - 39103265134080)*a(n) = 8*(n-2)*(n-1)*(2*n - 3)*(2*n - 1)*(850837923857825*n^9 - 15127768128079400*n^8 + 116088908648008427*n^7 - 502364025369222635*n^6 + 1342887860190877280*n^5 - 2280899268898038065*n^4 + 2433907848768834828*n^3 - 1548429898790214180*n^2 + 521138603722292640*n - 68863424146977600)*a(n-1) - 30*(n-2)*(2*n - 3)*(155302170039375*n^11 - 3227155335853125*n^10 + 29807524885054600*n^9 - 161278340404759950*n^8 + 566950865855228019*n^7 - 1356848300481904461*n^6 + 2250482361655315470*n^5 - 2579665279074165840*n^4 + 1996011605601581864*n^3 - 988803599084885136*n^2 + 280851990522009984*n - 34444332223983360)*a(n-2) + 5*(7288303593953125*n^13 - 187891351713750000*n^12 + 2204843674914291875*n^11 - 15579013461781304250*n^10 + 73867718896175411475*n^9 - 247858726321141236540*n^8 + 604530296941440837821*n^7 - 1082990060568950070282*n^6 + 1421457900098213642392*n^5 - 1345695224728829837040*n^4 + 889319601933492222864*n^3 - 386196670582228097568*n^2 + 97916706472751405568*n - 10797892365692920320)*a(n-3) + 10*(n-3)*(2*n - 7)*(5*n - 18)*(5*n - 14)*(5*n - 12)*(5*n - 11)*(108983978975*n^7 - 1065846642400*n^6 + 4333636082786*n^5 - 9458655747964*n^4 + 11899609166891*n^3 - 8554104592084*n^2 + 3208950812340*n - 473478110640)*a(n-4). - Vaclav Kotesovec, Nov 17 2017
a(n) ~ s * sqrt((r*s*(r*s^3 - 1) - 3) / (7*Pi*(5*r*s*(1 + r*s^3) - 3))) / (2*n^(3/2)*r^n), where r = 0.1385102270697349252376651829944449360743895474888... and s = 1.450646440303399446510765649245639306003224666768... are real roots of the system of equations (1 + r*s^2)*(1 + r^2*s^5) = s, r*s*(2 + 5*r*s^3 + 7*r^2*s^5) = 1. - Vaclav Kotesovec, Nov 22 2017
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

A329673 Number of meanders of length n with Motzkin-steps avoiding the consecutive steps HH.

Original entry on oeis.org

1, 2, 4, 10, 24, 60, 152, 388, 1000, 2592, 6752, 17664, 46368, 122080, 322240, 852464, 2259552, 5999552, 15954560, 42486592, 113282048, 302386304, 807999744, 2161077120, 5785032448, 15498450944, 41551965184, 111478804480, 299274439680, 803905119232, 2160632498176, 5810087371520
Offset: 0

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Author

Valerie Roitner, Nov 26 2019

Keywords

Comments

The Motzkin step set is U=(1,1), H=(1,0) and D=(1,-1). A meander is a path starting at (0,0) and never crossing the x-axis, i.e., staying at nonnegative altitude.

Examples

			a(2)=4 since we have 4 meanders of length 2 avoiding HH, namely UU, UH, UD and HU.

Crossrefs

Cf. A104545 which counts excursions avoiding consecutive HH steps. Cf. A329672 and A329674 which count meanders avoiding consecutive UU and DD respectively.

Formula

G.f.: -(1-2*t-2*t^2-sqrt(1-4*t^2-8*t^3-4*t^4))/(2*t*(1-2*t-2*t^2)).
D-finite with recurrence (n+1)*a(n) -2*a(n-1) -4*n*a(n-2) +8*(-n+2)*a(n-3) +4*(-n+3)*a(n-4)=0. - R. J. Mathar, Jan 25 2023

A104544 Triangle read by rows: T(n,k) is the number of Motzkin paths of length n having k HH's, where H=(1,0).

Original entry on oeis.org

1, 1, 1, 3, 0, 1, 5, 3, 0, 1, 11, 6, 3, 0, 1, 25, 13, 9, 3, 0, 1, 55, 40, 16, 12, 3, 0, 1, 129, 95, 60, 20, 15, 3, 0, 1, 303, 250, 155, 80, 25, 18, 3, 0, 1, 721, 661, 415, 235, 100, 31, 21, 3, 0, 1, 1743, 1708, 1206, 620, 335, 120, 38, 24, 3, 0, 1, 4241, 4515, 3262, 1946, 875, 455
Offset: 1

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Author

Emeric Deutsch, Mar 14 2005

Keywords

Examples

			Triangle starts:
1;
1,1;
3,0,1;
5,3,0,1;
11,6,3,0,1;
T(4,1)=3 because we have HHUD, UDHH and UHHD, where U=(1,1), D=(1,-1) and H=(1,0).

Crossrefs

Column 0 yields A104545. Row sums yield the Motzkin numbers (A001006).

Formula

G.f.=G=G(t, z) satisfies z^2(1+z-tz)G^2-(1-tz)G+1+z-tz=0.
Showing 1-6 of 6 results.

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