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 1-5 of 5 results.

A364338 G.f. satisfies A(x) = (1 + x) * (1 + x*A(x)^5).

Original entry on oeis.org

1, 2, 11, 105, 1140, 13555, 170637, 2235472, 30161255, 416248640, 5848462880, 83378361111, 1203100853951, 17537182300140, 257858115407535, 3819894878557990, 56958234329850060, 854192593184162160, 12875579347191388830, 194963091634569681550, 2964229359714424159370, 45234864131654311730160
Offset: 0

Views

Author

Seiichi Manyama, Jul 19 2023

Keywords

Crossrefs

Programs

  • Mathematica
    terms = 22; A[] = 0; Do[A[x] = (1+x)(1+x*A[x]^5) + O[x]^terms // Normal, terms]; CoefficientList[A[x], x] (* Stefano Spezia, Mar 24 2025 *)
  • PARI
    a(n) = sum(k=0, n, binomial(5*k+1, k)*binomial(5*k+1, n-k)/(5*k+1));

Formula

a(n) = Sum_{k=0..n} binomial(5*k+1,k) * binomial(5*k+1,n-k) / (5*k+1).

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)

A364333 G.f. satisfies A(x) = (1 + x*A(x)^2) * (1 + x*A(x)^6).

Original entry on oeis.org

1, 2, 17, 216, 3224, 52640, 910452, 16392140, 303996224, 5767278431, 111401778266, 2183535060362, 43319505976084, 868220464851417, 17552981176788200, 357544690982030744, 7330803752675100908, 151172599088871911072, 3133367418601958989295, 65242183918761533467216
Offset: 0

Views

Author

Seiichi Manyama, Jul 18 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(2*n+4*k+1, k)*binomial(2*n+4*k+1, n-k)/(2*n+4*k+1));

Formula

a(n) = Sum_{k=0..n} binomial(2*n+4*k+1,k) * binomial(2*n+4*k+1,n-k) / (2*n+4*k+1).

A216359 G.f. satisfies: A(x) = (1 + x*A(x)^2) * (1 + x/A(x)).

Original entry on oeis.org

1, 2, 3, 13, 32, 147, 445, 2067, 7019, 32590, 119209, 551551, 2125429, 9795863, 39221165, 180177403, 742575760, 3403131833, 14342166121, 65626369612, 281459129188, 1286834885967, 5596229192396, 25580269950635, 112492633046446, 514323765191879, 2282371511598955
Offset: 0

Views

Author

Paul D. Hanna, Sep 04 2012

Keywords

Comments

The radius of convergence of g.f. A(x) is r = 0.209619875959405379599013693... with A(r) = 2.36951367232829409921688546894691317519410... where y=A(r) satisfies y^7 - 2*y^6 - 4*y^4 + 4*y^3 + 4*y - 2 = 0.

Examples

			G.f.: A(x) = 1 + 2*x + 3*x^2 + 13*x^3 + 32*x^4 + 147*x^5 + 445*x^6 +...
Related expansions.
A(x)^2 = 1 + 4*x + 10*x^2 + 38*x^3 + 125*x^4 + 500*x^5 + 1839*x^6 +...
A(x)^3 = 1 + 6*x + 21*x^2 + 83*x^3 + 315*x^4 + 1269*x^5 + 5061*x^6 +...
where A(x) = (1-x^2)*A(x)^2 - x*A(x)^3 - x.
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + 1/A(x)^3)*x*A(x) + (1 + 2^2/A(x)^3 + 1/A(x)^6)*x^2*A(x)^2/2 +
(1 + 3^2/A(x)^3 + 3^2/A(x)^6 + 1/A(x)^9)*x^3*A(x)^3/3 +
(1 + 4^2/A(x)^3 + 6^2/A(x)^6 + 4^2/A(x)^9 + 1/A(x)^12)*x^4*A(x)^4/4 +
(1 + 5^2/A(x)^3 + 10^2/A(x)^6 + 10^2/A(x)^9 + 5^2/A(x)^12 + 1/A(x)^15)*x^5*A(x)^5/5 +...
		

Crossrefs

Programs

  • Maple
    S:= series(RootOf(x+y+x^2*y^2-y^2+x*y^3, y, 1), x, 41):
    seq(coeff(S,x,j),j=0..40); # Robert Israel, Jul 10 2017
  • 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^2)*AGF^2 - x*AGF^3 - x - AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}]; Flatten[{1,aa}] (* Vaclav Kotesovec, Sep 18 2013 *)
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=(1 + x*A^2)*(1 + x/(A+x*O(x^n)))); polcoeff(A, n)}
    
  • PARI
    {a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2/A^(3*j))*x^m*A^m/m))); polcoeff(A, n)}
    for(n=0, 31, print1(a(n), ", "))

Formula

G.f. satisfies:
A(x) = exp( Sum_{n>=1} x^n*A(x)^n/n * Sum_{k=0..n} C(n,k)^2 / A(x)^(3*k) ).
The formal inverse of the g.f. A(x) is (sqrt(1 - 2*x^3 + 4*x^4 + x^6) - (1+x^3))/(2*x^2).
Recurrence: n*(n+1)*(1241*n^5 - 21306*n^4 + 135203*n^3 - 381522*n^2 + 435524*n - 104880)*a(n) = 6*n*(1201*n^4 - 19476*n^3 + 114613*n^2 - 287442*n + 255364)*a(n-1) + 2*(12410*n^7 - 237880*n^6 + 1771109*n^5 - 6388366*n^4 + 11032829*n^3 - 6363274*n^2 - 3856020*n + 4157712)*a(n-2) + 6*(2482*n^7 - 51299*n^6 + 419427*n^5 - 1705769*n^4 + 3477465*n^3 - 2797370*n^2 - 637684*n + 1410288)*a(n-3) + 2*(4964*n^7 - 110044*n^6 + 983093*n^5 - 4442260*n^4 + 10160177*n^3 - 8790970*n^2 - 4722180*n + 9233280)*a(n-4) - 6*(2482*n^7 - 58745*n^6 + 553921*n^5 - 2617109*n^4 + 6255337*n^3 - 6022682*n^2 - 1392300*n + 4289616)*a(n-5) + 60*(n-7)*(2*n - 11)*(n^3 - 40*n^2 + 280*n - 552)*a(n-6) + 2*(n-8)*(2*n - 13)*(1241*n^5 - 15101*n^4 + 62389*n^3 - 91339*n^2 - 930*n + 64260)*a(n-7). - Vaclav Kotesovec, Sep 18 2013
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = 4.77053998540509708... is the root of the equation -4 + 12*d^2 - 8*d^3 - 12*d^4 - 20*d^5 + d^7 = 0 and c = 1.27852844884923435863262213680985089152... - Vaclav Kotesovec, Sep 18 2013
In closed form, c = (-4 + (1 + sqrt(1+8/d^2))*d^2) * sqrt((d^3*(1 + sqrt(1+8/d^2) + (4*(4 + d^2*(-3-sqrt(1+8/d^2) + d*(4+d))))/d^6)) / (1 + 1/64*(1 + sqrt(1+8/d^2)-4/d^2)^3*d^3)) / (32*d). - Vaclav Kotesovec, Aug 18 2014
From Peter Bala, Sep 10 2024: (Start)
For n not of the form 3*m + 1, we conjecture that a(n) = Sum_{k = 0..n} binomial(-n+3*k+1, k)*binomial(-n+3*k+1, n-k)/(-n+3*k+1).
Define a sequence operator R: {u(n): n >= 0} -> {v(n): n >= 0} by Sum_{n >= 0} v(n)*x^n = (1/x) * series_reversion(x/Sum_{n >= 0} u(n)*x^n). Then R({a(n)}) = A364336, R^2({a(n)}) = A215623 and R^3({a(n)}) = A364331. Cf. A073157. (End)

A364335 G.f. satisfies A(x) = (1 + x*A(x)^3) * (1 + x*A(x)^5).

Original entry on oeis.org

1, 2, 17, 204, 2852, 43489, 701438, 11767095, 203223146, 3589167533, 64524575635, 1176860764416, 21723084076739, 405038036077647, 7617437252889030, 144328483391622298, 2752414654270742784, 52790626691557217602, 1017655117382823639414, 19706520281177438174530
Offset: 0

Views

Author

Seiichi Manyama, Jul 18 2023

Keywords

Crossrefs

Programs

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

Formula

a(n) = Sum_{k=0..n} binomial(3*n+2*k+1,k) * binomial(3*n+2*k+1,n-k) / (3*n+2*k+1).
Showing 1-5 of 5 results.