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.

A216490 G.f. satisfies A(x) = 1 + x*A(x)^2 + x^3*A(x)^5.

Original entry on oeis.org

1, 1, 2, 6, 21, 78, 302, 1210, 4979, 20913, 89284, 386308, 1690221, 7465594, 33243970, 149080710, 672682035, 3051859515, 13913105076, 63704186436, 292825392338, 1350782681404, 6251139672412, 29014088205700, 135029529235367, 629978080093921, 2945889534054758
Offset: 0

Views

Author

Paul D. Hanna, Sep 07 2012

Keywords

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 21*x^4 + 78*x^5 + 302*x^6 + 1210*x^7 +...
Related expansions:
A(x)^2 = 1 + 2*x + 5*x^2 + 16*x^3 + 58*x^4 + 222*x^5 + 880*x^6 + 3588*x^7 +...
A(x)^5 = 1 + 5*x + 20*x^2 + 80*x^3 + 330*x^4 + 1391*x^5 + 5950*x^6 +...
Given (1) A(x) = 1 + x*A(x)^2 + x^3*A(x)^5,
suppose (2) A(x) = 1/A(-x*A(x)^3),
then substituting x in (1) with -x*A(x)^3 yields:
1/A(x) = 1 - x*A(x)^3/A(x)^2 - x^3*A(x)^9/A(x)^5,
which illustrates that (2) is consistent with (1).

Links

Crossrefs

Cf. A112806.

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^2+x^3*AGF^5-AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}]; Flatten[{1,aa}] (* Vaclav Kotesovec, Sep 10 2013 *)
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=1+x*A^2+x^3*A^5 +x*O(x^n));polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

Formula

G.f. satisfies: A(x) = 1/A(-x*A(x)^3); note that the Catalan function C(x) = 1 + x*C(x)^2 also satisfies this condition: C(x) = 1/C(-x*C(x)^3).
Recurrence: 8*n*(2*n+1)*(4*n+1)*(4*n+3)*(18857*n^5 - 123695*n^4 + 307075*n^3 - 357745*n^2 + 193668*n - 38880)*a(n) = (29662061*n^9 - 194572235*n^8 + 475490846*n^7 - 510273230*n^6 + 169643069*n^5 + 97131085*n^4 - 80249496*n^3 + 9135180*n^2 + 4205520*n - 777600)*a(n-1) - 3*(13463898*n^9 - 108514077*n^8 + 347609938*n^7 - 547442512*n^6 + 395255792*n^5 - 25515983*n^4 - 129824028*n^3 + 55362252*n^2 + 3442320*n - 3888000)*a(n-2) + 5*(5*n - 12)*(5*n - 11)*(5*n - 9)*(5*n - 8)*(18857*n^5 - 29410*n^4 + 865*n^3 + 9880*n^2 - 1092*n - 720)*a(n-3). - Vaclav Kotesovec, Sep 10 2013
a(n) ~ c*d^n/n^(3/2), where d = 1/768*((2775409885 + 28964352 * sqrt(8502))^(2/3) + 1573*(2775409885 + 28964352 * sqrt(8502))^(1/3) + 829273)/(2775409885 + 28964352 * sqrt(8502))^(1/3) = 4.952774592083496... is the root of the equation -3125 + 2142*d - 1573*d^2 + 256*d^3 = 0 and c = 0.3472109760934755295223550512446412412267... - Vaclav Kotesovec, Sep 10 2013
a(n) = Sum_{k=0..floor(n/3)} binomial(n-2*k,k) * binomial(2*n-k+1,n-2*k)/(2*n-k+1). - Seiichi Manyama, Aug 28 2023

A364833 G.f. satisfies A(x) = 1 + x*A(x)^2/(1 - x^3*A(x)^3).

Original entry on oeis.org

1, 1, 2, 5, 15, 49, 168, 595, 2160, 7997, 30083, 114660, 441840, 1718531, 6737820, 26600784, 105659970, 421949492, 1693120779, 6823018035, 27602090087, 112053680381, 456343848121, 1863893501065, 7633232165286, 31337360839387, 128944120202510
Offset: 0

Views

Author

Seiichi Manyama, Aug 28 2023

Keywords

Crossrefs

Cf. A112806, A219537.
Cf. A218251, A364161, A365247.
Cf. A001003, A049124.

Programs

  • Maple
    A364833 := proc(n)
    add( binomial(n-2*k-1,k)*binomial(2*n-3*k+1,n-3*k)/ (2*n-3*k+1),k=0..floor(n/3)) ;
end proc:
seq(A364833(n),n=0..80); # R. J. Mathar, Aug 29 2023
  • PARI
    a(n) = sum(k=0, n\3, binomial(n-2*k-1, k)*binomial(2*n-3*k+1, n-3*k)/(2*n-3*k+1));

Formula

a(n) = Sum_{k=0..floor(n/3)} binomial(n-2*k-1,k) * binomial(2*n-3*k+1,n-3*k)/(2*n-3*k+1).
D-finite with recurrence 31*n*(626109182191*n-1858292669035) *(n-1)*(n+1) *a(n) -n*(n-1) *(244150473843619*n^2 -1454194662255591*n +2175006457069082) *a(n-1) +3*(n-1) *(292927551362415*n^3 -2593205532882651*n^2 +7084566217454162*n -5823331737745632)*a(n-2) +(-843955616916167*n^4 +9932491073296715*n^3 -42016891739306929*n^2 +76184884157722453*n -50166914106142776) *a(n-3) +18*(1509721335071*n^4 -40413442328880*n^3 +330301781039401*n^2 -1078322794857576*n +1231650372542192) *a(n-4) +18*(39673125909769*n^4 -598320530478001*n^3 +3228489073613917*n^2 -7321259523567459*n +5788776339353646) *a(n-5) +27*(n-5) *(3102413205331*n^3 -35996479327373*n^2 +114122791959960*n -64735736097804) *a(n-6) -243*(n-6) *(n-7)*(475638134099*n^2 -2399948859181*n +2877042451214) *a(n-7) -243*(45857481910*n -35520400961) *(n-5) *(n-7) *(n-8)*a(n-8)=0. - R. J. Mathar, Aug 29 2023
G.f.: (1/x) * Series_Reversion( x*(1 - x / (1 - x^3)) ). - Seiichi Manyama, Sep 28 2024

A228987 G.f. satisfies A(x) = 1 + x*A(x)^3 + x^3*A(x)^8.

Original entry on oeis.org

1, 1, 3, 13, 66, 364, 2116, 12768, 79222, 502297, 3240120, 21196593, 140295584, 937787728, 6321624862, 42926227470, 293350136170, 2015999854478, 13923926272607, 96598395025615, 672852440805930, 4703751150849738, 32991286060134402, 232091541493091566
Offset: 0

Views

Author

Paul D. Hanna, Sep 10 2013

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 66*x^4 + 364*x^5 + 2116*x^6 +...
Related expansions:
A(x)^3 = 1 + 3*x + 12*x^2 + 58*x^3 + 312*x^4 + 1788*x^5 + 10686*x^6 +...
A(x)^8 = 1 + 8*x + 52*x^2 + 328*x^3 + 2082*x^4 + 13384*x^5 + 87124*x^6 +...
Given (1) A(x) = 1 + x*A(x)^3 + x^3*A(x)^8,
suppose (2) A(x) = 1/A(-x*A(x)^5),
then substituting x in (1) with -x*A(x)^5 yields:
1/A(x) = 1 - x*A(x)^5/A(x)^3 - x^3*A(x)^15/A(x)^8,
which illustrates that (2) is consistent with (1).

Links

Crossrefs

Cf. A216490, A112806.

Programs

  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=1+x*A^3+x^3*A^8 +x*O(x^n)); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. satisfies: A(x) = 1/A(-x*A(x)^5); note that the function G(x) = 1 + x*G(x)^3 (g.f. of A001764) also satisfies this condition: G(x) = 1/G(-x*G(x)^5).
a(n) ~ sqrt((3 - r*s^2)/(2*Pi*(3 + 28*r^2*s^5))) / (4*n^(3/2)*r^(n + 1/2)), where r = 0.1331154541373089587498695338172936885734070972340... and s = 1.408602671059676188189711196409966797670750551605... are real roots of the system of equations 1 + r*s^3 + r^3*s^8 = s, 3*r*s^2 + 8*r^3*s^7 = 1. - Vaclav Kotesovec, Nov 22 2017
a(n) = Sum_{k=0..floor(n/3)} binomial(n-2*k,k) * binomial(3*n-k+1,n-2*k)/(3*n-k+1). - Seiichi Manyama, Aug 28 2023
Showing 1-3 of 3 results.

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