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.

A120593 G.f. satisfies: 5*A(x) = 4 + x + A(x)^4, starting with [1,1,6].

Original entry on oeis.org

1, 1, 6, 76, 1201, 21252, 402892, 8001412, 164321982, 3461110532, 74358814838, 1623152780808, 35897318940028, 802620009567628, 18112759482614328, 412020809942451504, 9437537418826749369, 217486633306640519124
Offset: 0

Views

Author

Paul D. Hanna, Jun 16 2006, Jan 24 2008

Keywords

Comments

See comments in A120588 for conditions needed for an integer sequence to satisfy a functional equation of the form: r*A(x) = c + b*x + A(x)^n.

Examples

			A(x) = 1 + x + 6*x^2 + 76*x^3 + 1201*x^4 + 21252*x^5 +...
A(x)^4 = 1 + 4*x + 30*x^2 + 380*x^3 + 6005*x^4 + 106260*x^5 +...

Crossrefs

Cf. A120588 - A120592, A120594 - A120607; A244856.

Programs

  • Mathematica
    CoefficientList[1 + InverseSeries[Series[1+5*x - (1+x)^4, {x, 0, 20}], x], x] (* Vaclav Kotesovec, Nov 28 2017 *)
  • PARI
    {a(n)=local(A=1+x+6*x^2+x*O(x^n));for(i=0,n,A=A-5*A+4+x+A^4);polcoeff(A,n)}

Formula

G.f. satisfies:
(1) A(x) = 1 + Series_Reversion(1+5*x - (1+x)^4).
(2) A(x) = Sum_{n>=0} C(4*n,n)/(3*n+1) * (4+x)^(3*n+1)/5^(4*n+1), by Lagrange Inversion.
(3) A(x) = F(x/A(x)) and F(x) = A(x*F(x)) where F(x) = (4 + F(x)^4)/(5-x) is the g.f. of A244856. - Paul D. Hanna, Jul 09 2014
a(n) ~ 2^(-7/3 + 3*n) * (-32 + 15*10^(1/3))^(1/2 - n) / (5^(1/3) * n^(3/2) * sqrt(3*Pi)). - Vaclav Kotesovec, Nov 28 2017

A245043 G.f. satisfies: A(x) = (12 + A(x)^4) / (13 - 27*x).

Original entry on oeis.org

1, 3, 15, 117, 1158, 12930, 154986, 1947582, 25317009, 337610451, 4592807895, 63488144109, 889226772132, 12592147132572, 179982549300948, 2593187073716460, 37622924436008574, 549178914689660106, 8059539548880228138, 118846096104074358942, 1760035125442960123992
Offset: 0

Views

Author

Paul D. Hanna, Jul 10 2014

Keywords

Examples

			G.f.: A(x) =  1 + 3*x + 15*x^2 + 117*x^3 + 1158*x^4 + 12930*x^5 +...
Compare A(x)^4 to (13-27*x)*A(x):
A(x)^4 = 1 + 12*x + 114*x^2 + 1116*x^3 + 11895*x^4 + 136824*x^5 +...
(13-27*x)*A(x) = 13 + 12*x + 114*x^2 + 1116*x^3 + 11895*x^4 + 136824*x^5 +...

Crossrefs

Cf. A120595, A245009, A244627, A244594, A244856.

Programs

  • Mathematica
    CoefficientList[1 + InverseSeries[Series[(1+13*x - (1+x)^4)/(27*(1+x)), {x, 0, 20}], x], x] (* Vaclav Kotesovec, Nov 27 2017 *)
  • PARI
    {a(n)=polcoeff(1 + serreverse( (1+13*x - (1+x)^4)/(27*(1+x +x*O(x^n)))), n)}
for(n=0, 30, print1(a(n), ", "))
  • PARI
    {a(n)=local(A=[1], Ax=1+x); for(i=1, n, A=concat(A, 0); Ax=Ser(A); A[#A]=Vec( ( Ax^4 - (13-27*x)*Ax )/9 )[#A]); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. satisfies:
(1) A(x) = 1 + Series_Reversion( (1+13*x - (1+x)^4)/(27*(1+x)) ).
(2) A(x) = Sum_{n>=0} C(4*n,n)/(3*n+1) * (12 + 27*x*A(x))^(3*n+1) / 13^(4*n+1).
(3) A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where G(x) = (12+27*x + G(x)^4)/13 is the g.f. of A120595.
a(n) ~ 3^(3*n - 7/2) / (2^(7/4) * sqrt(Pi) * n^(3/2) * (13 - 8*sqrt(2))^(n - 3/2)). - Vaclav Kotesovec, Nov 27 2017

A245009 G.f. satisfies: A(x) = (7 + A(x)^4) / (8 - 8*x).

Original entry on oeis.org

1, 2, 10, 88, 978, 12200, 163156, 2286448, 33138874, 492657384, 7470940300, 115115319376, 1797128902132, 28364816229008, 451870965523368, 7256283996155360, 117333885356923274, 1908844190372949224, 31221135850863938268, 513100005743085437328, 8468653781083527106012, 140314257925457275837488
Offset: 0

Views

Author

Paul D. Hanna, Jul 09 2014

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 10*x^2 + 88*x^3 + 978*x^4 + 12200*x^5 +...
Compare A(x)^4 to 8*(1-x)*A(x):
A(x)^4 = 1 + 8*x + 64*x^2 + 624*x^3 + 7120*x^4 + 89776*x^5 +...
8*(1-x)*A(x) = 8 + 8*x + 64*x^2 + 624*x^3 + 7120*x^4 + 89776*x^5 +...

Crossrefs

Cf. A120594, A244627, A244594, A244856, A245043.

Programs

  • Mathematica
    CoefficientList[1 + InverseSeries[Series[(1+8*x - (1+x)^4)/(8*(1+x)), {x, 0, 20}], x], x] (* Vaclav Kotesovec, Nov 27 2017 *)
  • PARI
    {a(n)=polcoeff(1 + serreverse((1+8*x - (1+x)^4)/(8*(1+x +x*O(x^n)))), n)}
for(n=0, 30, print1(a(n), ", "))
  • PARI
    {a(n)=local(A=[1], Ax=1+2*x); for(i=1, n, A=concat(A, 0); Ax=Ser(A); A[#A]=Vec( ( Ax^4 - 8*(1-x)*Ax )/4 )[#A]); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Formula

G.f. satisfies:
(1) A(x) = 1 + Series_Reversion( (1+8*x - (1+x)^4)/(8*(1+x)) ).
(2) A(x) = Sum_{n>=0} C(4*n,n)/(3*n+1) * (7 + 8*x*A(x))^(3*n+1) / 8^(4*n+1).
(3) A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where G(x) = (7+8*x + G(x)^4)/8 is the g.f. of A120594.
a(n) ~ 3^(3*(n-1)/4) * 7^((n-1)/4) / (sqrt(Pi) * n^(3/2) * (3^(3/4)*7^(1/4) - 7/2)^(n - 1/2)). - Vaclav Kotesovec, Nov 27 2017
Showing 1-3 of 3 results.

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