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.

A352383 G.f. A(x) satisfies: A(x) = (1 + 2*x*A(x)^3)^2 / (1 + x*A(x)^3).

Original entry on oeis.org

1, 3, 28, 350, 5020, 78023, 1278340, 21740636, 380161308, 6792111260, 123448657904, 2275311657814, 42427160829508, 798933055335618, 15171376583787800, 290199619787772728, 5586346847185229596, 108141141737193646020
Offset: 0

Views

Author

Paul D. Hanna, Mar 14 2022

Keywords

Comments

Equals a bisection of A351771.
Self-convolution cube equals A352384.

Examples

			G.f.: A(x) = 1 + 3*x + 28*x^2 + 350*x^3 + 5020*x^4 + 78023*x^5 + 1278340*x^6 + 21740636*x^7 + 380161308*x^8 + 6792111260*x^9 + ...
where A(x) satisfies
A(x) = (1 + 2*x*A(x)^3)^2 / (1 + x*A(x)^3).
Related series.
The cube of the g.f. A(x) equals the g.f. of A352384:
A(x)^3 = 1 + 9*x + 111*x^2 + 1581*x^3 + 24468*x^4 + 399735*x^5 + 6784186*x^6 + 118444293*x^7 + ... + A352384(n)*x^n + ...
and equals (1/x) * Series_Reversion(  x*(1+x)^3/(1+2*x)^6 ).

Crossrefs

Cf. A352384, A351771, A188110.

Programs

  • Mathematica
    CoefficientList[(InverseSeries[Series[x*(1 + x)^3/(1 + 2*x)^6, {x, 0, 20}], x]/x)^(1/3), x] (* Vaclav Kotesovec, Mar 15 2022 *)
  • PARI
    /* Using Series Reversion */
{a(n) = my(A = ((1/x)*serreverse( x*(1+x)^3/(1+2*x +x^2*O(x^n))^6 ))^(1/3)); polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) satisfies:
(1) A(x) = (1 + 2*x*A(x)^3)^2 / (1 + x*A(x)^3).
(2) 0 = 4*A(x)^6*x^2 + (4 - A(x))*A(x)^3*x + (1 - A(x)).
(3) A(x) = 4 + 8*x*A(x)^3 - sqrt( A(x)^2 + 8*A(x) ).
(4) x = ( A(x) - 4 + sqrt( A(x)^2 + 8*A(x) ) ) / (8*A(x)^3).
(5) A( x*(1+x)^3/(1+2*x)^6 ) = (1+2*x)^2/(1+x).
(6) A(x)^3 = (1/x) * Series_Reversion( x*(1+x)^3/(1+2*x)^6 ).
a(n) ~ sqrt((221 - 29*sqrt(13))/78) * 2^(3*n) * (587 - 143*sqrt(13))^n / (sqrt(Pi) * n^(3/2) * 3^(3*n+1)). - Vaclav Kotesovec, Mar 15 2022

A352384 G.f. A(x) satisfies: A(x) = (1 + 2*x*A(x))^6 / (1 + x*A(x))^3.

Original entry on oeis.org

1, 9, 111, 1581, 24468, 399735, 6784186, 118444293, 2113587804, 38377421060, 706774205943, 13170180868299, 247862354439196, 4704490506021162, 89949748461476772, 1730889637195688117, 33495746280466024908
Offset: 0

Views

Author

Paul D. Hanna, Mar 14 2022

Keywords

Comments

Self-convolution cube root yields A352383.

Examples

			G.f.: A(x) = 1 + 9*x + 111*x^2 + 1581*x^3 + 24468*x^4 + 399735*x^5 + 6784186*x^6 + 118444293*x^7 + 2113587804*x^8 + 38377421060*x^9 + ...
where
A(x)^(1/3) = (1 + 2*x*A(x))^2/(1 + x*A(x)) = 1 + 3*x + 28*x^2 + 350*x^3 + 5020*x^4 + 78023*x^5 + 1278340*x^6 + ... + A352383(n)*x^n + ...

Crossrefs

Cf. A352383, A351771.

Programs

  • Mathematica
    CoefficientList[(InverseSeries[Series[x*(1 + x)^3/(1 + 2*x)^6, {x, 0, 20}], x]/x), x] (* Vaclav Kotesovec, Mar 15 2022 *)
  • PARI
    /* Using Series Reversion */
{a(n) = my(A = (1/x)*serreverse( x*(1+x)^3/(1+2*x +x^2*O(x^n))^6 )); polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) satisfies:
(1) A(x) = (1 + 2*x*A(x))^6 / (1 + x*A(x))^3.
(2) A( x*(1+x)^3/(1+2*x)^6 ) = (1+2*x)^6/(1+x)^3.
(3) A(x) = (1/x) * Series_Reversion( x*(1+x)^3/(1+2*x)^6 ).
(4) x = ( A(x)^(1/3) - 4 + sqrt( A(x)^(2/3) + 8*A(x)^(1/3) ) ) / (8*A(x)).
(5) 0 = 4*A(x)^2*x^2 + (4 - A(x)^(1/3))*A(x)*x + (1 - A(x)^(1/3)).
a(n) ~ sqrt(278513 - 1003421/sqrt(13)) * 2^(3*n + 3/2) * (587 - 143*sqrt(13))^n / (sqrt(Pi) * n^(3/2) * 3^(3*n + 5/2)). - Vaclav Kotesovec, Mar 15 2022
D-finite with recurrence 3*n *(3*n+2) *(2*n+3) *(3*n+1) *(5114882323*n -3577270936)*(n+1)*a(n) -8*n*(12009974028164*n^5 +15575274162434*n^4 +20452834455*n^3 -7391009529770*n^2 -2779978786544*n -151626455514) *a(n-1) +64*(29840008960856*n^6 -7909817331616*n^5 -21378617546230*n^4 -22395081360175*n^3 +39992783684339*n^2 -16585158398179*n +2497181632755) *a(n-2) -4608*(6*n-11) *(6*n-7) *(3*n-4) *(2*n-3) *(3*n-5) *(4265440*n -810084569)*a(n-3)=0. - R. J. Mathar, Jul 20 2023

A352701 G.f. (1/x)*Series_Reversion( x*(1-x)*(3 - 2*x + sqrt(1+4*x-4*x^2))/4 ).

Original entry on oeis.org

1, 1, 3, 7, 28, 79, 350, 1075, 5020, 16180, 78023, 259417, 1278340, 4343642, 21740636, 75065787, 380161308, 1328887420, 6792111260, 23975385148, 123448657904, 439228736887, 2275311657814, 8148868193557, 42427160829508
Offset: 0

Views

Author

Paul D. Hanna, Mar 29 2022

Keywords

Comments

Essentially an unsigned version of A351771 (after dropping the initial term).
a(2*n) = A352383(n) for n >= 0.

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 7*x^3 + 28*x^4 + 79*x^5 + 350*x^6 + 1075*x^7 + 5020*x^8 + 16180*x^9 + 78023*x^10 + 259417*x^11 + ...
such that A(x) = (1/x)*Series_Reversion(x*G(x)) and A(x*G(x)) = 1/G(x),
where G(x) = (1-x)*(3 - 2*x + sqrt(1+4*x-4*x^2))/4, which starts
G(x) = 1 - x - x^2 + 3*x^3 - 8*x^4 + 26*x^5 - 92*x^6 + 344*x^7 - 1336*x^8 + 5336*x^9 - 21776*x^10 + ...
Let B(x) =  Series_Reversion( x*(1-x^2)/(1+x^2)^3 ),
B(x) = x + 4*x^3 + 39*x^5 + 496*x^7 + 7180*x^9 + 112236*x^11 + 1846082*x^13 + 31485120*x^15 + ...,
then A(x) = 1 + x*A(x)^2 + B(x)^2, where
B(x)^2 = x^2 + 8*x^4 + 94*x^6 + 1304*x^8 + 19849*x^10 + 320600*x^12 + 5396108*x^14 + 93615864*x^16 + ...

Crossrefs

Cf. A351771, A352383.

Programs

  • PARI
    {a(n) = my(A = (1/x)*serreverse( x*(1-x)*(3 - 2*x + sqrt(1+4*x-4*x^2 +x*O(x^n) ))/4 ));
polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

Formula

The g.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion( x*(1-x)*(3 - 2*x + sqrt(1+4*x-4*x^2))/4 );
(2) (A(x) - A(-x))/2 = x*(A(x)^2 + A(-x)^2)/2;
(3) ((A(x) + A(-x))/2)^3 = F(x^2), where F(x) = (1/x)*Series_Reversion( x*(1+x)^3/(1+2*x)^6 );
(4) 1 - x*(A(x) - A(-x))/2 = x/Series_Reversion( x - x*(C(x) + C(-x))/2 ), where C(x) = (1 - sqrt(1-4*x))/2 is the Catalan function (A000108);
(5a) (1/A(x) + 1/A(-x))/2 = ( 1 - x*(A(x) - A(-x))/2 )^2;
(5b) (1/A(x) - 1/A(-x))/2 = (-x)/(1 - 2*x*(A(x) - A(-x))/2);
(6a) (1/A(x)^2 + 1/A(-x)^2)/2 = ( 1 - x*(A(x) - A(-x))/2 )^3.
(6b) (1/A(x)^2 - 1/A(-x)^2)/2 = -2*x*(1 - x*(A(x) - A(-x))/2)^2/( 1 - x*(A(x) - A(-x)) ).
Let B(x) = Series_Reversion( x*(1-x^2)/(1+x^2)^3 ), then
(7) A(x) = (1 - sqrt(1 - 4*x - 4*x*B(x)^2))/(2*x);
(8) A(x) - x*A(x)^2 = A(-x) + x*A(-x)^2 = 1 + B(x)^2;
(9) 1 - x*(A(x) - A(-x))/2 = 1/(1 + B(x)^2);
(10) 1/A(x) = 1/(1 + B(x)^2)^2 - x*(1 + B(x)^2)/(1 - B(x)^2);
(10a) (1/A(x) + 1/A(-x))/2 = 1/(1 + B(x)^2)^2;
(10b) (1/A(x) - 1/A(-x))/2 = (-x)*(1 + B(x)^2)/(1 - B(x)^2);
(11) 1/A(x)^2 = 1/(1 + B(x)^2)^3 - 2*x/(1 - B(x)^4);
(11a) (1/A(x)^2 + 1/A(-x)^2)/2 = 1/(1 + B(x)^2)^3;
(11b) (1/A(x)^2 - 1/A(-x)^2)/2 = -2*x/(1 - B(x)^4).
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.