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 1-2 of 2 results.

A199475 G.f. satisfies A(x) = Sum_{n>=0} x^n * (1 - A(x)^(2*n+2))/(1 - A(x)^2).

Original entry on oeis.org

1, 2, 7, 34, 195, 1225, 8146, 56336, 401005, 2918308, 21614216, 162385693, 1234515922, 9479336440, 73410868630, 572719097908, 4496923141241, 35509806367132, 281814387290431, 2246608404455588, 17982234787607464, 144458551104066553, 1164342291135424494
Offset: 0

Views

Author

Paul D. Hanna, Nov 08 2011

Keywords

Comments

Compare to g.f. B(x) of A007317 (binomial transform of Catalan numbers):
B(x) = Sum_{n>=0} x^n * (1 - B(x)^(n+1))/(1 - B(x)).

Examples

			G.f.: A(x) = 1 + 2*x + 7*x^2 + 34*x^3 + 195*x^4 + 1225*x^5 +...
where g.f. A = A(x) satisfies the equivalent expressions:
A = 1 + x*(1-A^4)/(1-A^2) + x^2*(1-A^6)/(1-A^2) + x^3*(1-A^8)/(1-A^2) +...
A = 1 + x*(1 + A^2) + x^2*(1 + A^2 + A^4) + x^3*(1 + A^2 + A^4 + A^6) +...

Links

Crossrefs

Cf. A007317, A171199, A199548, A349289, A349290, A349291, A349292, A349293.

Programs

  • Mathematica
    Rest[CoefficientList[InverseSeries[Series[(2*x^2)/(1 + x^2 - Sqrt[1 - 4*x - 2*x^2 + x^4]), {x, 0, 30}], x], x]] (* Vaclav Kotesovec, Jul 30 2021 *)
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, x^m*sum(k=0, m, A^(2*k))+x*O(x^n))); polcoeff(A, n)}
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=1/((1-x)*(1 - x*A^2+x*O(x^n))));polcoeff(A,n)}
  • PARI
    {a(n)=polcoeff(1/x*serreverse(2*x^2/(1+x^2-sqrt(1-4*x-2*x^2+x^4+x^3*O(x^n)))),n)}

Formula

G.f. satisfies: A(x) = 1/((1-x)*(1 - x*A(x)^2)).
G.f.: A(x) = (1/x)*Series_Reversion( 2*x^2/(1+x^2 - sqrt(1-4*x-2*x^2+x^4)) ).
G.f. satisfies: A(x) = G(x*A(x)) and G(x) = A(x/G(x)) = g.f. of A171199, where G(x) = exp( Sum_{n>=1} [G(x)^n + G(x)^-n]*x^n/n ).
a(n) = 1 + Sum_{i=0..n-1} Sum_{j=0..n-i-1} a(i) * a(j) * a(n-i-j-1). - Ilya Gutkovskiy, Jul 25 2021
a(n) ~ sqrt(387 + 35*sqrt(129)) * (35 + 3*sqrt(129))^n / (9 * sqrt(Pi) * n^(3/2) * 2^(3*n + 5/2)). - Vaclav Kotesovec, Jul 30 2021
a(n) = Sum_{k=0..n} binomial(n+k,n-k) * binomial(3*k,k)/(2*k+1). - Seiichi Manyama, Oct 03 2023
D-finite with recurrence 2*n*(2*n+1)*a(n) +3*(-13*n^2+11*n-2)*a(n-1) +(35*n^2-23*n-42)*a(n-2) +(35*n^2-257*n+426)*a(n-3) +3*(-13*n^2+93*n-166)*a(n-4) +2*(n-4)*(2*n-9)*a(n-5)=0. - R. J. Mathar, Feb 10 2024

A228907 G.f. satisfies: A(x) = 1 + Sum_{n>=0} x^n * (1 - A(x)^(2*n))/(1 - A(x)).

Original entry on oeis.org

1, 2, 6, 24, 114, 598, 3336, 19402, 116302, 713368, 4455650, 28240942, 181180912, 1174280146, 7677229718, 50570040088, 335289825874, 2235856077798, 14985808827416, 100900119437082, 682145490613118, 4628755102582328, 31514118237222850, 215214456560655070
Offset: 0

Views

Author

Paul D. Hanna, Sep 08 2013

Keywords

Comments

Conjectured to be the number of permutations of length n+1 avoiding the partially ordered pattern (POP) {1>4, 1>5, 2>4, 2>5, 5>3} of length 5. That is, conjectured to be the number of length n+1 permutations having no subsequences of length 5 in which the first and second elements are larger than the elements in positions 4 and 5, and the fifth element is larger than the element in position 3.- Sergey Kitaev, Dec 13 2020
This conjecture has been proven. It can be restated as the number of size n permutations avoiding 45123, 45132, 45213, 54123, 54132, 54213. - Christian Bean, Jul 24 2024

Examples

			G.f.: A(x) = 1 + 2*x + 6*x^2 + 24*x^3 + 114*x^4 + 598*x^5 + 3336*x^6 +...
where g.f. A = A(x) satisfies the equivalent expressions:
A = 1 + x*(1-A^2)/(1-A) + x^2*(1-A^4)/(1-A) + x^3*(1-A^6)/(1-A) +...
A = 1 + x*(1 + A) + x^2*(1 + A + A^2 + A^3) + x^3*(1 + A + A^2 + A^3 + A^4 + A^5) +...

Links

Crossrefs

Cf. A199548.

Programs

  • Mathematica
    nmax=20;aa=ConstantArray[0,nmax];aa[[1]]=2;Do[AGF=1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j;sol=Solve[Coefficient[1+x*AGF*(2-AGF+AGF^2)+x^2*AGF^2*(1-AGF)-AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}];Flatten[{1,aa}] (* Vaclav Kotesovec, Sep 09 2013 *)
  • PARI
    {a(n) = my(A=1+x); for(i=1, n, A = 1 + sum(m=0, n, x^m*sum(k=0, 2*m-1, A^k) + x*O(x^n))); polcoeff(A, n)}
for(n=0,25,print1(a(n),", "))
  • PARI
    {a(n) = my(A=1+x); for(i=1, n, A = 1+x*A*(2-A+A^2)+x^2*A^2*(1-A)+x*O(x^n)); polcoeff(A, n)}
for(n=0,25,print1(a(n),", "))

Formula

G.f. satisfies: A(x) = 1 + x*A(x)*(2 - A(x) + A(x)^2) + x^2*A(x)^2*(1 - A(x)).
G.f.: (1/x)*Series_Reversion( x*(1+x+x^2 + sqrt(1-2*x-5*x^2-2*x^3+x^4)) / (2*(1+x)^2) ).
Recurrence: n*(2*n+1)*(7*n^2 - 35*n + 34)*a(n) = (126*n^4 - 742*n^3 + 1193*n^2 - 601*n + 78)*a(n-1) - (182*n^4 - 1253*n^3 + 2788*n^2 - 2293*n + 450)*a(n-2) + (56*n^4 - 434*n^3 + 1189*n^2 - 1273*n + 318)*a(n-3) + (n-4)*(2*n-5)*(7*n^2 - 21*n + 6)*a(n-4). - Vaclav Kotesovec, Sep 09 2013
a(n) ~ c*d^n/n^(3/2), where d = 8/3+14/3*cos(arctan(3*sqrt(3)/13)/3) = 7.29589694323977237... is the root of the equation 1 + 5*d - 8*d^2 + d^3 = 0 and c = sqrt(1/3 + sqrt(7)*cos((4*Pi + arccos(-1/(2*sqrt(7))))/3)/6) / sqrt(Pi) = 0.33910585091755684322274547... - Vaclav Kotesovec, Sep 09 2013, updated Mar 17 2024
Showing 1-2 of 2 results.

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