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 1-3 of 3 results.

A200754 G.f. satisfies A(x) = 1 + x*A(x)^4 - x^2*A(x)^5.

Original entry on oeis.org

1, 1, 3, 13, 67, 380, 2288, 14351, 92737, 613063, 4126289, 28179766, 194780822, 1360053081, 9578997279, 67971291791, 485464864401, 3487203531460, 25176899072984, 182598098616625, 1329716528758651, 9718954060263384, 71273846758123552, 524279847227139350
Offset: 0

Views

Author

Paul D. Hanna, Nov 21 2011

Keywords

Comments

Compare to the g.f. G(x) for the ternary tree numbers (A001764): G(x) = 1 + x*G(x)^4 - x^2*G(x)^6 = 1 + x*G(x)^3.

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 67*x^4 + 380*x^5 + 2288*x^6 +...
Related expansions:
A(x)^4 = 1 + 4*x + 18*x^2 + 92*x^3 + 515*x^4 + 3068*x^5 + 19092*x^6 +...
A(x)^5 = 1 + 5*x + 25*x^2 + 135*x^3 + 780*x^4 + 4741*x^5 + 29915*x^6 +...
where a(2) = 4 - 1; a(3) = 18 - 5; a(4) = 92 - 25; a(5) = 515 - 135; ...

Links

Crossrefs

Cf. A200753, A200755, A365180.

Programs

  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=1+x*A^4-x^2*A^5+x*O(x^n));polcoeff(A,n)}
  • PARI
    a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n+3*k+1, k)*binomial(k, n-k)/(n+3*k+1)); \\ Seiichi Manyama, Nov 01 2023

Formula

Recurrence: 3*n*(n+1)*(3*n - 1)*(3*n + 1)*(1280*n^7 - 17280*n^6 + 94466*n^5 - 266799*n^4 + 407516*n^3 - 311946*n^2 + 81648*n + 12420)*a(n) = n*(281600*n^10 - 3942400*n^9 + 22465080*n^8 - 65757900*n^7 + 98435187*n^6 - 46293273*n^5 - 73831865*n^4 + 131625925*n^3 - 85490202*n^2 + 25470288*n - 2915460)*a(n-1) + (263680*n^11 - 4087040*n^10 + 26682396*n^9 - 96110406*n^8 + 210443037*n^7 - 293037231*n^6 + 270493529*n^5 - 181862299*n^4 + 103208358*n^3 - 47593224*n^2 + 12881700*n - 1360800)*a(n-2) - 5*(640000*n^11 - 10880000*n^10 + 78609000*n^9 - 311686500*n^8 + 721945299*n^7 - 918913929*n^6 + 367548335*n^5 + 642045653*n^4 - 1081692258*n^3 + 697174344*n^2 - 208955268*n + 24222240)*a(n-3) + 5*(5*n - 18)*(5*n - 16)*(5*n - 14)*(5*n - 12)*(1280*n^7 - 8320*n^6 + 17666*n^5 - 8869*n^4 - 15820*n^3 + 22148*n^2 - 9282*n + 1305)*a(n-4). - Vaclav Kotesovec, Nov 18 2017
a(n) ~ s*sqrt((2*r*s-1) / (2*Pi*(5*r*s-3))) / (2*n^(3/2)*r^n), where r = 0.1272568969777848138753091632571986265610307654216... and s = 1.358291097397172238669759690645074441686961930838... are roots of the system of equations s + r^2*s^5 = 1 + r*s^4, 1 + 5*r^2*s^4 = 4*r*s^3. - Vaclav Kotesovec, Nov 18 2017
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n+3*k+1,k) * binomial(k,n-k)/(n+3*k+1). - Seiichi Manyama, Nov 01 2023

A137265 G.f. y(x) is solution of x y^3 - (1 + x^2) y + 1 = 0 with y(0) = 1.

Original entry on oeis.org

1, 1, 2, 8, 35, 163, 796, 4024, 20885, 110654, 596064, 3254752, 17974893, 100227022, 563482140, 3190633232, 18179765509, 104158703503, 599698459613, 3467978715612, 20134256546896, 117313279477959, 685756774642494, 4020515276730588, 23636036336651811
Offset: 0

Views

Author

Robert Israel, Mar 12 2008

Keywords

Examples

			a(3) = 8 because g(x) = 1 + x + 2 x^2 + 8 x^3 + O(x^4) satisfies x*g(x)^3 - (1 + x^2)*g(x) + 1 = O(x^4).

Links

Crossrefs

Cf. A000108, A200753, A200755.
Cf. A364474.

Programs

  • Maple
    f:= (x,y) -> x*y^3 - (1 + x^2)*y + 1; N:= (y,n) -> convert(normal(taylor(y-f(x,y)/D[2](f)(x,y),x=0,n)),polynom); Y:= 1; for j from 1 to 6 do Y:= N(Y,2^j) end do; seq(coeftayl(Y,x=0,j),j=0..2^6-1);
  • Mathematica
    max = 22; g[x_] := Sum[a[k]*x^k, {k, 0, max}]; coes = CoefficientList[ Series[ x*g[x]^3 - (1+x^2)*g[x] + 1, {x, 0, max}], x]; sol = First[ Solve[ Thread[ coes == 0 ] ] ]; Table[a[n] /. sol, {n, 0, max}](* Jean-François Alcover, Nov 28 2011 *)
terms = 25; y[] = 1; Do[y[x] = (1 + x*y[x]^3)/(1 + x^2) + O[x]^terms, terms]; CoefficientList[y[x], x] (* Jean-François Alcover, Jan 11 2018 *)
  • PARI
    a(n) = sum(k=0, n\2, (-1)^k*binomial(3*n-5*k, k)*binomial(3*n-6*k, n-2*k)/(2*n-4*k+1)); \\ Seiichi Manyama, Nov 02 2023

Formula

a(0) = 1, a(1) = 1, a(n) = -a(n-2) + sum_{i=0}^{n-1} sum_{j=0}^{n-1-i} a(i) a(j) a(n-1-i-j).
a(n) ~ sqrt(1 - (2*r)^(5/3)) / (2^(4/3) * sqrt(3*Pi) * n^(3/2) * r^(n + 1/3)), where r = 0.15978798947663136723274504893788499231133813071845... is the real root of the equation (1+r^2)^3 = 27*r/4. - Vaclav Kotesovec, May 03 2016
a(n) = Sum_{k=0..floor(n/2)} (-1)^k * binomial(3*n-5*k,k) * binomial(3*n-6*k,n-2*k) / (2*n-4*k+1). - Seiichi Manyama, Nov 02 2023

A367027 G.f. A(x) satisfies A(x) = 1 + x*A(x)^3 - x^2*A(x)^5.

Original entry on oeis.org

1, 1, 2, 4, 5, -13, -147, -816, -3534, -12650, -35420, -53040, 199056, 2391340, 14555740, 68264112, 261045693, 769660569, 1167906402, -5145668100, -61758940705, -385813067255, -1857144860445, -7266981925560, -21793022441775, -32643056947527, 161919845140752
Offset: 0

Views

Author

Seiichi Manyama, Nov 02 2023

Keywords

Crossrefs

Cf. A000108, A137265, A200753, A200755.
Cf. A361245, A368969.

Programs

  • PARI
    a(n) = sum(k=0, n\2, (-1)^k*binomial(3*n-k, k)*binomial(3*n-2*k, n-2*k))/(2*n+1);

Formula

a(n) = (1/(2*n+1)) * Sum_{k=0..floor(n/2)} (-1)^k * binomial(3*n-k,k) * binomial(3*n-2*k,n-2*k).
G.f.: ( (1/x) * Series_Reversion( x * (1-x+x^2)^2 ) )^(1/2). - Seiichi Manyama, Mar 08 2025
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.