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 1-10 of 10 results.

A360950 Expansion of g.f. A(x) satisfying A(x) = Sum_{n>=0} d^n/dx^n x^(2*n) * A(x)^n / n!.

Original entry on oeis.org

1, 2, 12, 108, 1240, 16932, 264740, 4631320, 89270316, 1875586380, 42610756408, 1040307155304, 27157913296228, 754950111249488, 22267948484559720, 694746226969477744, 22863695087986373968, 791675941860401322852, 28776089467457429038620, 1095679176790207081120360
Offset: 0

Views

Author

Paul D. Hanna, Feb 26 2023

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 12*x^2 + 108*x^3 + 1240*x^4 + 16932*x^5 + 264740*x^6 + 4631320*x^7 + 89270316*x^8 + 1875586380*x^9 + ...
where
A(x) = 1 + (d/dx x^2*A(x)) + (d^2/dx^2 x^4*A(x)^2)/2! + (d^3/dx^3 x^6*A(x)^3)/3! + (d^4/dx^4 x^8*A(x)^4)/4! + (d^5/dx^5 x^10*A(x)^5)/5! + (d^6/dx^6 x^12*A(x)^6)/6! + ... + (d^n/dx^n x^(2*n)*A(x)^n)/n! + ...
Related series.
Let B(x) = Series_Reversion(x - x^2*A(x)), which begins
B(x) = x + x^2 + 4*x^3 + 27*x^4 + 248*x^5 + 2822*x^6 + 37820*x^7 + 578915*x^8 + 9918924*x^9 + 187558638*x^10 + ... + A229619(n)*x^n + ...
then A(x) = B'(x) and
B(x) = x * exp( x*A(x) + (d/dx x^3*A(x)^2)/2! + (d^2/dx^2 x^5*A(x)^3)/3! + (d^3/dx^3 x^7*A(x)^4)/4! + (d^4/dx^4 x^9*A(x)^5)/5! + (d^5/dx^5 x^11*A(x)^6)/6! + ... + (d^(n-1)/dx^(n-1) x^(2*n-1)*A(x)^n)/n! + ... ).
		

Crossrefs

Programs

  • PARI
    {Dx(n, F) = my(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n) = my(A=1); for(i=1, n, A = sum(m=0, n, Dx(m, x^(2*m)*A^m/m!)) +O(x^(n+1))); polcoeff(A, n)}
    for(n=0, 25, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) A(x) = Sum_{n>=0} d^n/dx^n x^(2*n) * A(x)^n / n!.
(2) A(x) = d/dx Series_Reversion(x - x^2*A(x)).
(3) B(x - x^2*A(x)) = x where B(x) = x * exp( Sum_{n>=1} d^(n-1)/dx^(n-1) x^(2*n-1) * A(x)^n / n! ) is the g.f. of A229619.
(4) a(n) = (n+1) * A229619(n+1) for n >= 0.
a(n) ~ c * n! * n^alfa / LambertW(1)^n, where alfa = 3*LambertW(1) + 1/(1 + LambertW(1)) = 2.33953361459... and c = 0.1926079501120681239... - Vaclav Kotesovec, Feb 27 2023

A360973 Expansion of g.f. A(x) satisfying A(x) = Sum_{n>=0} d^n/dx^n x^(3*n) * A(x)^n / n!.

Original entry on oeis.org

1, 3, 30, 462, 9243, 223014, 6232239, 196780545, 6905085381, 266134485876, 11167349876424, 506653305313320, 24713399794830075, 1289888647516181583, 71744797404818298768, 4237233894492578488668, 264870390088867267319274, 17473793269024825938396135
Offset: 0

Views

Author

Paul D. Hanna, Feb 27 2023

Keywords

Examples

			G.f.: A(x) = 1 + 3*x^2 + 30*x^4 + 462*x^6 + 9243*x^8 + 223014*x^10 + 6232239*x^12 + 196780545*x^14 + 6905085381*x^16 + ... + a(n)*x^(2*n) + ...
where
A(x) = 1 + (d/dx x^3*A(x)) + (d^2/dx^2 x^6*A(x)^2)/2! + (d^3/dx^3 x^9*A(x)^3)/3! + (d^4/dx^4 x^12*A(x)^4)/4! + (d^5/dx^5 x^15*A(x)^5)/5! + (d^6/dx^6 x^18*A(x)^6)/6! + ... + (d^n/dx^n x^(3*n)*A(x)^n)/n! + ...
Related series.
Let B(x) = Series_Reversion(x - x^3*A(x)), which begins
B(x) = x + x^3 + 6*x^5 + 66*x^7 + 1027*x^9 + 20274*x^11 + 479403*x^13 + 13118703*x^15 + 406181493*x^17 + ... + A360976(n)*x^(2*n-1) + ...
then A(x) = B'(x) and
B(x) = x * exp( x^2*A(x) + (d/dx x^5*A(x)^2)/2! + (d^2/dx^2 x^8*A(x)^3)/3! + (d^3/dx^3 x^11*A(x)^4)/4! + (d^4/dx^4 x^14*A(x)^5)/5! + (d^5/dx^5 x^17*A(x)^6)/6! + ... + (d^(n-1)/dx^(n-1) x^(3*n-1)*A(x)^n)/n! + ... ).
		

Crossrefs

Programs

  • PARI
    {Dx(n, F) = my(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n) = my(A=1); for(i=1, n, A = sum(m=0, n, Dx(m, x^(3*m)*A^m/m!)) +O(x^(2*n+1))); polcoeff(A, 2*n)}
    for(n=0, 25, print1(a(n), ", "))
    
  • PARI
    /* Using series reversion (faster) */
    {a(n) = my(A=1); for(i=1, n, A = deriv( serreverse(x - x^3*A +O(x^(2*n+3))))); polcoeff(A, 2*n)}
    for(n=0, 25, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^(2*n) may be defined by the following.
(1) A(x) = Sum_{n>=0} d^n/dx^n x^(3*n) * A(x)^n / n!.
(2) A(x) = d/dx Series_Reversion(x - x^3*A(x)).
(3) B(x - x^3*A(x)) = x where B(x) = x * exp( Sum_{n>=1} d^(n-1)/dx^(n-1) x^(3*n-1) * A(x)^n / n! ) is the g.f. of A360976.
(4) a(n) = (2*n+1) * A360976(n+1) for n >= 0.
a(n) ~ c * n! * n^alfa * 2^n / LambertW(1)^n, where alfa = 1.50465021094584... and c = 0.36053267424501... - Vaclav Kotesovec, Feb 28 2023
alfa = (9*LambertW(1) - 1 + 3/(1 + LambertW(1)))/4. - Vaclav Kotesovec, Mar 13 2023

A360974 Expansion of g.f. A(x) satisfying A(x) = Sum_{n>=0} d^n/dx^n x^(2*n) * A(x)^(2*n) / n!.

Original entry on oeis.org

1, 2, 18, 260, 4890, 110124, 2844772, 82196424, 2613699450, 90450874860, 3379153837180, 135445714293720, 5796441493971284, 263784018974675416, 12721572505160772840, 648250134428292640272, 34809708051186914034730, 1965040180185473309749788, 116359823755204505172646204
Offset: 0

Views

Author

Paul D. Hanna, Feb 27 2023

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 18*x^2 + 260*x^3 + 4890*x^4 + 110124*x^5 + 2844772*x^6 + 82196424*x^7 + 2613699450*x^8 + ... + a(n)*x^n + ...
where
A(x) = 1 + (d/dx x^2*A(x)^2) + (d^2/dx^2 x^4*A(x)^4)/2! + (d^3/dx^3 x^6*A(x)^6)/3! + (d^4/dx^4 x^8*A(x)^8)/4! + (d^5/dx^5 x^10*A(x)^10)/5! + (d^6/dx^6 x^12*A(x)^12)/6! + ... + (d^n/dx^n x^(2*n)*A(x)^(2*n))/n! + ...
Related series.
Let B(x) = Series_Reversion(x - x^2*A(x)^2), which begins
B(x) = x + x^2 + 6*x^3 + 65*x^4 + 978*x^5 + 18354*x^6 + 406396*x^7 + 10274553*x^8 + 290411050*x^9 + ... + A360977(n)*x^n + ...
then A(x) = B'(x) and
B(x) = x * exp( x*A(x)^2 + (d/dx x^3*A(x)^4)/2! + (d^2/dx^2 x^5*A(x)^6)/3! + (d^3/dx^3 x^7*A(x)^8)/4! + (d^4/dx^4 x^9*A(x)^10)/5! + (d^5/dx^5 x^11*A(x)^12)/6! + ... + (d^(n-1)/dx^(n-1) x^(2*n-1)*A(x)^(2*n))/n! + ... ).
		

Crossrefs

Programs

  • PARI
    {Dx(n, F) = my(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n) = my(A=1); for(i=1, n, A = sum(m=0, n, Dx(m, x^(2*m)*A^(2*m)/m!)) +O(x^(n+1))); polcoeff(A, n)}
    for(n=0, 25, print1(a(n), ", "))
    
  • PARI
    /* Using series reversion (faster) */
    {a(n) = my(A=1); for(i=1, n, A = deriv( serreverse(x - x^2*A^2 +O(x^(n+2))))); polcoeff(A, n)}
    for(n=0, 25, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n may be defined by the following.
(1) A(x) = Sum_{n>=0} d^n/dx^n x^(2*n) * A(x)^(2*n) / n!.
(2) A(x) = d/dx Series_Reversion(x - x^2*A(x)^2).
(3) B(x - x^2*A(x)^2) = x where B(x) = x * exp( Sum_{n>=1} d^(n-1)/dx^(n-1) x^(2*n-1) * A(x)^(2*n) / n! ) is the g.f. of A360977.
(4) a(n) = (n+1) * A360977(n+1) for n >= 0.
a(n) ~ c * n! * n^alfa / LambertW(1/2)^n, where alfa = 2.498459235192... and c = 0.0920029178453... - Vaclav Kotesovec, Feb 28 2023
alfa = 5*LambertW(1/2) + 1/(1 + LambertW(1/2)). - Vaclav Kotesovec, Mar 13 2023

A361046 Expansion of g.f. A(x) satisfying A(x) = Sum_{n>=0} d^n/dx^n x^(3*n) * A(x)^(2*n) / n!.

Original entry on oeis.org

1, 3, 45, 1113, 36459, 1448568, 66726309, 3469988835, 200242815669, 12670449226269, 871389659249424, 64693985439491127, 5156607707368927875, 439261264283443326927, 39831856169938193953827, 3831650468281643037364389, 389807188331526942149375433
Offset: 0

Views

Author

Paul D. Hanna, Mar 03 2023

Keywords

Examples

			G.f.: A(x) = 1 + 3*x^2 + 45*x^4 + 1113*x^6 + 36459*x^8 + 1448568*x^10 + 66726309*x^12 + 3469988835*x^14 + ... + a(n)*x^(2*n) + ...
where
A(x) = 1 + (d/dx x^3*A(x)^2) + (d^2/dx^2 x^6*A(x)^4)/2! + (d^3/dx^3 x^9*A(x)^6)/3! + (d^4/dx^4 x^12*A(x)^8)/4! + (d^5/dx^5 x^15*A(x)^10)/5! + ... + (d^n/dx^n x^(3*n)*A(x)^(2*n))/n! + ...
Related series.
Let B(x) = Series_Reversion(x - x^3*A(x)^2), which begins
B(x) = x + x^3 + 9*x^5 + 159*x^7 + 4051*x^9 + 131688*x^11 + 5132793*x^13 + 231332589*x^15 + 11778989157*x^17 + ... + A361047(n)*x^(2*n-1) + ...
then A(x) = B'(x) and
B(x) = x * exp( x^2*A(x)^2 + (d/dx x^5*A(x)^4)/2! + (d^2/dx^2 x^8*A(x)^6)/3! + (d^3/dx^3 x^11*A(x)^8)/4! + (d^4/dx^4 x^14*A(x)^10)/5! + ... + (d^(n-1)/dx^(n-1) x^(3*n-1)*A(x)^(2*n))/n! + ... ).
		

Crossrefs

Programs

  • Mathematica
    nt = 40; (* number of terms to produce *)
    A[_] = 0;
    Do[A[x_] = D[InverseSeries[x - x^3*A[x]^2 + O[x]^k] // Normal, x], {k, 1, 2*nt}];
    CoefficientList[A[x^(1/2)], x] (* Jean-François Alcover, Mar 04 2023 *)
  • PARI
    {Dx(n, F) = my(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n) = my(A=1); for(i=1, n, A = sum(m=0, n, Dx(m, x^(3*m)*A^(2*m)/m!)) +O(x^(2*n+1))); polcoeff(A, 2*n)}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    /* Using series reversion (faster) */
    {a(n) = my(A=1); for(i=1, n, A = deriv( serreverse(x - x^3*A^2 +O(x^(2*n+3))))); polcoeff(A, 2*n)}
    for(n=0, 20, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^(2*n) may be defined by the following.
(1) A(x) = Sum_{n>=0} d^n/dx^n x^(3*n) * A(x)^(2*n) / n!.
(2) A(x) = d/dx Series_Reversion(x - x^3*A(x)^2).
(3) B(x - x^3*A(x)^2) = x where B(x) = x * exp( Sum_{n>=1} d^(n-1)/dx^(n-1) x^(3*n-1) * A(x)^(2*n) / n! ) is the g.f. of A361047.
(4) a(n) = (2*n+1) * A361047(n+1) for n >= 0.
a(n) == 0 (mod 3) for n > 0.
a(n) ~ c * 2^n * n! * n^alfa / LambertW(1/2)^n, where alfa = 1.623844426394406... and c = 0.18597481905555548924712403113114... - Vaclav Kotesovec, Mar 04 2023
alfa = (15*LambertW(1/2) - 1 + 3/(1 + LambertW(1/2)))/4. - Vaclav Kotesovec, Mar 15 2023

A360578 Expansion of g.f. A(x) satisfying A(x) = Series_Reversion( x - x*A'(x)*A(x) ).

Original entry on oeis.org

1, 1, 5, 42, 471, 6422, 101439, 1803949, 35459549, 760744865, 17651187689, 439893743313, 11711735210140, 331666197753372, 9954249177284539, 315638779480717743, 10545365878475964736, 370309787453143694246, 13637805276205022293179, 525684316153586923528166
Offset: 1

Views

Author

Paul D. Hanna, Feb 21 2023

Keywords

Examples

			G.f.: A(x) = x + x^2 + 5*x^3 + 42*x^4 + 471*x^5 + 6422*x^6 + 101439*x^7 + 1803949*x^8 + 35459549*x^9 + 760744865*x^10 + ...
such that A( x - x*A'(x)*A(x) ) = x.
Related series.
Series_Reversion(A(x)) = x - x^2 - 3*x^3 - 22*x^4 - 235*x^5 - 3153*x^6 - 49721*x^7 - 888784*x^8 - 17615520*x^9 + ...
A'(x)*A(x) = x + 3*x^2 + 22*x^3 + 235*x^4 + 3153*x^5 + 49721*x^6 + 888784*x^7 + 17615520*x^8 + ...
A(A(x)) = x + 2*x^2 + 12*x^3 + 110*x^4 + 1294*x^5 + 18127*x^6 + 290620*x^7 + 5206800*x^8 + 102633591*x^9 + ...
A'(A(x)) = 1 + 2*x + 17*x^2 + 208*x^3 + 3108*x^4 + 53328*x^5 + 1018948*x^6 + 21297818*x^7 + 481458997*x^8 + ...
A'(A(x))*A(A(x)) = x + 4*x^2 + 33*x^3 + 376*x^4 + 5242*x^5 + 84625*x^6 + 1534652*x^7 + 30682881*x^8 + ...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=x); for(i=1, n, A=serreverse(x - x*A'*A +x*O(x^n))); polcoeff(A, n)}
    for(n=1, 25, print1(a(n), ", "))
    
  • PARI
    {Dx(n, F) = my(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n) = my(A=x); for(i=1, n, A = x + sum(m=1, n, Dx(m-1, x^m*(A')^m*A^m/m!)) +O(x^(n+1))); polcoeff(A, n)}
    for(n=1, 25, print1(a(n), ", "))
    
  • PARI
    {Dx(n, F) = my(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n)=local(A=x); for(i=1, n, A = x*exp(sum(m=1, n, Dx(m-1, x^(m-1)*(A')^m*A^m/m!)) +O(x^(n+1)))); polcoeff(A, n)}
    for(n=1, 25, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies:
(1) A( x - x*A'(x)*A(x) ) = x.
(2) A(x) = x + A(x) * A'(A(x)) * A(A(x)).
(3) A(x) = x + Sum_{n>=1} d^(n-1)/dx^(n-1) x^n * A'(x)^n * A(x)^n / n!.
(4) A(x) = x * exp( Sum_{n>=1} d^(n-1)/dx^(n-1) x^(n-1) * A'(x)^n * A(x)^n / n! ).
a(n) ~ c * n! * n^alfa / LambertW(1)^n, where alfa = 1.5447806483693... and c = 0.02888888614196289496..., conjecture: alfa = 2*(2*LambertW(1) - 1 + 1/(1 + LambertW(1))). - Vaclav Kotesovec, Feb 22 2023

A303062 G.f. A(x) satisfies: [x^(n-1)] (1 + x*A(x)^(n-1))^n / A(x)^n = 0 for n>1.

Original entry on oeis.org

1, 1, 2, 8, 60, 643, 8564, 133890, 2376261, 46832442, 1009739331, 23564025488, 590503218735, 15793704933899, 448695132962248, 13487808514722460, 427624923581550100, 14260707306806609885, 499071020445057149835, 18290961984686434723480, 700757535935308305865473, 28017787701624063252219677
Offset: 0

Views

Author

Paul D. Hanna, Apr 17 2018

Keywords

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 8*x^3 + 60*x^4 + 643*x^5 + 8564*x^6 + 133890*x^7 + 2376261*x^8 + 46832442*x^9 + 1009739331*x^10 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients in (1 + x*A(x)^(n-1))^n / A(x)^n begins:
n=1: [1, 0, -2, -6, -50, -565, -7731, -122983, ...];
n=2: [1, 0, -2, -10, -89, -1030, -14307, -230054, ...];
n=3: [1, 0, 0, -9, -111, -1380, -19677, -320958, ...];
n=4: [1, 0, 4, 0, -94, -1520, -23388, -392776, ...];
n=5: [1, 0, 10, 20, 0, -1210, -24030, -436250, ...];
n=6: [1, 0, 18, 54, 225, 0, -18345, -427944, ...];
n=7: [1, 0, 28, 105, 651, 2835, 0, -316344, ...];
n=8: [1, 0, 40, 176, 1364, 8360, 41976, 0, ...]; ...
in which the main diagonal equals all zeros after the initial term, illustrating that [x^(n-1)] (1 + x*A(x)^(n-1))^n / A(x)^n = 0 for n>1.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(m=1,n+1, A=concat(A,0); A[m] = Vec( (1 + x*Ser(A)^(m-1))^m/Ser(A)^m )[m]/m ); A[n+1]}
    for(n=0,30, print1(a(n),", "))

Formula

a(n) ~ c * n! * n^(2*LambertW(1)) / LambertW(1)^n, where c = 0.153879081661359639962985708... - Vaclav Kotesovec, Aug 11 2021

A360581 Expansion of A(x) satisfying [x^n] A(x)^n / (1 + x*A(x)^n)^n = 0 for n > 0.

Original entry on oeis.org

1, 1, 3, 17, 131, 1204, 12587, 149131, 2036675, 32358153, 587313706, 11761213199, 252859744189, 5785648936988, 141627609404793, 3737907237793369, 106414467836076985, 3241492594168333618, 104522041356412895455, 3541554178675758259947, 125782730912626755808358
Offset: 0

Views

Author

Paul D. Hanna, Mar 12 2023

Keywords

Examples

			G.f. A(x) = 1 + x + 3*x^2 + 17*x^3 + 131*x^4 + 1204*x^5 + 12587*x^6 + 149131*x^7 + 2036675*x^8 + 32358153*x^9 + 587313706*x^10 + ...
The table of coefficients in the successive powers of g.f. A(x) begins:
n = 1: [1, 1,  3,  17,  131,  1204,  12587,  149131, ...];
n = 2: [1, 2,  7,  40,  305,  2772,  28657,  335114, ...];
n = 3: [1, 3, 12,  70,  531,  4782,  48936,  565245, ...];
n = 4: [1, 4, 18, 108,  819,  7324,  74272,  848064, ...];
n = 5: [1, 5, 25, 155, 1180, 10501, 105650, 1193530, ...];
n = 6: [1, 6, 33, 212, 1626, 14430, 144208, 1613214, ...];
n = 7: [1, 7, 42, 280, 2170, 19243, 191254, 2120511, ...];
n = 8: [1, 8, 52, 360, 2826, 25088, 248284, 2730872, ...];
...
The table of coefficients in A(x)^n/(1 + x*A(x)^n)^n begins:
n = 1: [1, 0,   2,   12,  100,  955, 10258, 124565, ...];
n = 2: [1, 0,   2,   18,  161, 1606, 17757, 220834, ...];
n = 3: [1, 0,   0,   15,  168, 1806, 21000, 272856, ...];
n = 4: [1, 0,  -4,    0,  114, 1504, 19220, 270692, ...];
n = 5: [1, 0, -10,  -30,    0,  800, 12970, 215445, ...];
n = 6: [1, 0, -18,  -78, -165,    0,  4797, 123990, ...];
n = 7: [1, 0, -28, -147, -364, -329,     0,  32767, ...];
n = 8: [1, 0, -40, -240, -572,  696,  7472,      0, ...];
...
in which the diagonal of all zeros illustrates that
[x^n] A(x)^n / (1 + x*A(x)^n)^n = 0 for n > 0.
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1,n, A = concat(A,0);
    A[#A] = -polcoeff( Ser(A)^(#A)/(1 + x*Ser(A)^(#A))^(#A), #A-1)/(#A) );A[n+1]}
    for(n=0,30,print1(a(n),", "))

Formula

From Vaclav Kotesovec, Mar 13 2023: (Start)
a(n) ~ c * n! * n^alpha / LambertW(1)^n, where alpha = 0.33953... and c = 0.1881608377753...
Conjecture: alpha = 3*LambertW(1) - 2 + 1/(1 + LambertW(1)) = 0.33953361459446... (End)

A365095 Expansion of g.f. A(x) satisfying [x^(n-1)] (1 + (n-1)*x*A(x)^2)^n / A(x)^n = 0 for n > 1.

Original entry on oeis.org

1, 1, 4, 27, 256, 3118, 46114, 797049, 15671350, 343712542, 8287906284, 217309849772, 6143454613682, 186012988954448, 5999891924386246, 205262374717093101, 7420869162700453174, 282640364822610119566, 11310634300879858185320, 474456517209788353301282, 20818983374432724237753352
Offset: 0

Views

Author

Paul D. Hanna, Sep 03 2023

Keywords

Comments

Related identities for the Catalan function C(x) = 1 + x*C(x)^2 (cf. A000108):
(1) [x^(n-1)] (1 + (n-1)*x*C(x))^n / C(x)^n = 0 for n > 1.
(2) [x^(n-1)] (1 + n*x*C(x)^2)^n / C(x)^(2*n) = 0 for n > 1.
(3) [x^(n-1)] (1 + n*x*C(x))^n / C(x)^n = n^(n-1) for n >= 1.
(4) [x^(n-1)] (1 + (n+1)*x*C(x)^2)^n / C(x)^(2*n) = n^(n-1) for n >= 1.

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 27*x^3 + 256*x^4 + 3118*x^5 + 46114*x^6 + 797049*x^7 + 15671350*x^8 + 343712542*x^9 + 8287906284*x^10 + ...
RELATED SERIES.
(1) The power series B(x) = A(x/B(x)) where A(x) = B(x*A(x)) begins
B(x) = 1 + x + 3*x^2 + 17*x^3 + 151*x^4 + 1817*x^5 + 27041*x^6 + 472297*x^7 + 9377293*x^8 + 207254037*x^9 + 5025044843*x^10 + ...
and appears to have only odd coefficients.
(2) The power series D(x) = A(x/D(x)^2) where A(x) = D(x*A(x)^2) begins
D(x) = 1 + x + 2*x^2 + 10*x^3 + 90*x^4 + 1106*x^5 + 16684*x^6 + 293796*x^7 + 5860280*x^8 + 129807560*x^9 + 3149052896*x^10 + ...
modulo 16 of which appears to equal the series
D(x) (mod 16) = (1 + x + 2*x^2 + 10*x^3 + 10*x^4 + 2*x^5 + 12*x^6 + 4*x^7 + 7*x^8 + 7*x^9 - 2*x^10 - 2*x^11 - 10*x^12 + 6*x^13 - 12*x^14 + 4*x^15 - 8*x^16 - 8*x^17) / (1 - x^8).
Explicitly,
D(x) (mod 16) = 1 + x + 2*x^2 + 10*x^3 + 10*x^4 + 2*x^5 + 12*x^6 + 4*x^7 + 8*x^8 + 8*x^9 + 8*x^11 + 8*x^13 + 8*x^15 + 8*x^19 + 8*x^21 + 8*x^23 + ...
RELATED TABLES.
The table of coefficients of x^k in (1 + (n-1)*x*A(x)^2)^n/A(x)^n begins:
n=1: [1, -1,   -3,   -20,  -197,  -2504,  -38396, ...];
n=2: [1,  0,   -4,   -32,  -336,  -4432,  -69620, ...];
n=3: [1,  3,    0,   -41,  -501,  -6795, -107500, ...];
n=4: [1,  8,   24,     0,  -640, -10112, -163272, ...];
n=5: [1, 15,   95,   310,     0, -13027, -246265, ...];
n=6: [1, 24,  252,  1520,  5448,      0, -321580, ...];
n=7: [1, 35,  546,  5033, 30534, 119728,       0, ...]; ...
in which the main diagonal equals all zeros after the initial term, illustrating that [x^(n-1)] (1 + (n-1)*x*A(x)^2)^n / A(x)^n = 0 for n > 1.
The table of coefficients of x^k in A(x)^n begins:
n=1: [1, 1,  4,  27,  256,  3118,  46114, ...];
n=2: [1, 2,  9,  62,  582,  6964, 101241, ...];
n=3: [1, 3, 15, 106,  990, 11667, 166861, ...];
n=4: [1, 4, 22, 160, 1493, 17372, 244658, ...];
n=5: [1, 5, 30, 225, 2105, 24241, 336540, ...];
n=6: [1, 6, 39, 302, 2841, 32454, 444660, ...];
n=7: [1, 7, 49, 392, 3717, 42210, 571438, ...]; ...
		

Crossrefs

Cf. A303063.

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0); m=#A;
    A[#A] = polcoeff( (1 + (m-1)*x*Ser(A)^2)^m / Ser(A)^m , m-1)/m ); A[n+1]}
    for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) [x^(n-1)] (1 + (n-1)*x*A(x)^2)^n / A(x)^n = 0.
(2) [x^(n-1)] (1 + (k*n-1)*x*A(x)^2)^n / A(x)^n is divisible by n^2 for n > 0 and all integer k (conjecture).

A365516 Expansion of g.f. A(x) satisfying A(x) = (1 + 2*x*A(x))^2 / (1 + 2*x*A(x) - 2*x^3*A(x)^3).

Original entry on oeis.org

1, 2, 4, 10, 32, 112, 400, 1464, 5520, 21296, 83456, 331136, 1328320, 5379200, 21959936, 90271904, 373347840, 1552438016, 6486311680, 27217331456, 114649525760, 484640538112, 2055185596416, 8740711936000, 37273693649920, 159340373710848, 682708771254272, 2931290431277056
Offset: 0

Views

Author

Paul D. Hanna, Sep 07 2023

Keywords

Comments

Related identities for the Catalan function C(x) = 1 + x*C(x)^2 (A000108):
(1) [x^(n-1)] (1 + n*x*C(x))^n / C(x)^n = n^(n-1) for n >= 1.
(2) [x^(n-1)] (1 + (n+1)*x*C(x)^2)^n / C(x)^(2*n) = n^(n-1) for n >= 1.
Related identity: [x^(n-1)] (1 + (n+1)*x*B(x))^n / B(x)^n = n*(n-1)^(n-1) for n >= 1 when B(x) = 1/(1 - 2*x).
Related identity: F(x) = (2/x) * Sum{n>=1} n*(n+1)^(n-2) * x^n * F(x)^n / (1 + (n+1)*x*F(x))^(n+1), which holds formally for all Maclaurin series F(x). - Paul D. Hanna, Oct 03 2023

Examples

			G.f.: A(x) = 1 + 2*x + 4*x^2 + 10*x^3 + 32*x^4 + 112*x^5 + 400*x^6 + 1464*x^7 + 5520*x^8 + 21296*x^9 + 83456*x^10 + ...
where A(x) = (1 + 2*x*A(x))^2/(1 + 2*x*A(x) - 2*x^3*A(x)^3).
RELATED SERIES.
B(x) = A(x/B(x)) where B(x) = (1 + 2*x)^2/(1 + 2*x - 2*x^3) begins
B(x) = 1 + 2*x + 2*x^3 + 4*x^6 - 8*x^7 + 16*x^8 - 24*x^9 + 32*x^10 - 32*x^11 + 16*x^12 + 32*x^13 - 128*x^14 + ...
RELATED TABLE.
The table of coefficients of x^k in (1 + (n+1)*x*A(x))^n/A(x)^n begins:
n=1: [1,  0,    0,    -2,     -8,    -24,     -76,    -272, ...];
n=2: [1,  2,    1,    -4,    -20,    -64,    -196,    -664, ...];
n=3: [1,  6,   12,     2,    -48,   -192,    -600,   -1896, ...];
n=4: [1, 12,   54,   100,    -23,   -600,   -2224,   -6944, ...];
n=5: [1, 20,  160,   630,   1080,   -696,   -8660,  -30960, ...];
n=6: [1, 30,  375,  2488,   9027,  14406,  -15371, -144852, ...];
n=7: [1, 42,  756,  7546,  44800, 153552,  229376, -342728, ...];
n=8: [1, 56, 1372, 19192, 167222, 921400, 3026332, 4251528, ...]; ...
in which the main diagonal equals n*(n+1)^(n-2) for n > 1.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Module[{A = {1}, m, x}, Do[AppendTo[A, 0]; m = Length[A]; A[[-1]] = Coefficient[(1 + (m + 1)*x*SeriesData[x, 0, A, 0, m + 1, 1])^m/SeriesData[x, 0, A, 0, m + 1, 1]^m, x, m - 1]/m - (m + 1)^(m - 2), {i, 1, n}]; A[[n + 1]]]; Table[a[n], {n, 0, 27}] (* Robert P. P. McKone, Sep 07 2023 *)
  • PARI
    /* Using series reversion */
    {a(n) = my(A = (1/x)*serreverse(x*(1 + 2*x - 2*x^3)/(1 + 2*x +x*O(x^n))^2));
    polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    /* Formula [x^(n-1)] (1 + (n+1)*x*A(x))^n/A(x)^n = n*(n+1)^(n-2) */
    {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0); m=#A;
    A[#A] = polcoeff( (1 + (m+1)*x*Ser(A))^m / Ser(A)^m , m-1)/m - (m+1)^(m-2) );A[n+1]}
    for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) A(x) = (1 + 2*x*A(x))^2 / (1 + 2*x*A(x) - 2*x^3*A(x)^3).
(2) A(x/B(x)) = B(x) where B(x) = (1 + 2*x)^2 / (1 + 2*x - 2*x^3).
(3) A(x) = (1/x) * Series_Reversion( x*(1 + 2*x - 2*x^3)/(1 + 2*x)^2 ).
(4) [x^(n-1)] (1 + (n+1)*x*A(x))^n / A(x)^n = n*(n+1)^(n-2) for n > 1.
(5) [x^(n-1)] (1 + (n-1)*x*A(x))^n / A(x)^n = -n*(n-3)^(n-2) for n > 1.
(6) [x^(n-1)] (1 + n*x*A(x))^n / A(x)^n = (n^(n-1) - n*(n-2)^(n-2))/2 for n >= 1.
(7) A(x) = (2/x) * Sum{n>=1} n*(n+1)^(n-2) * x^n * A(x)^n / (1 + (n+1)*x*A(x))^(n+1). - Paul D. Hanna, Oct 03 2023

A365574 Expansion of g.f. A(x) satisfying [x^(n-1)] (1 + (n+1)*x*A(x))^n / A(x)^n = n*(n+2)^(n-2) for n > 1.

Original entry on oeis.org

1, 2, 3, 4, 16, 104, 515, 2090, 8170, 34704, 160014, 751282, 3479758, 16012684, 74362915, 350282602, 1665651094, 7952638460, 38067823370, 182874936368, 882344022104, 4274341269824, 20773195676078, 101228332620524, 494521566769160, 2421729829067636, 11886902458813596
Offset: 0

Views

Author

Paul D. Hanna, Sep 11 2023

Keywords

Comments

Conjecture 1: a(k) is odd iff k = 2^n - 2 for n >= 1.
Conjecture 2: a(2^n - 2) == 3 (mod 16) for n > 1.
Is there a closed formula for the g.f. of this sequence? Compare to the g.f. of A365516.
Related identities for the Catalan function C(x) = 1 + x*C(x)^2 (A000108):
(1) [x^(n-1)] (1 + n*x*C(x))^n / C(x)^n = n^(n-1) for n >= 1.
(2) [x^(n-1)] (1 + (n+1)*x*C(x)^2)^n / C(x)^(2*n) = n^(n-1) for n >= 1.
Related identity: F(x) = (3/x) * Sum{n>=1} n*(n+2)^(n-2) * x^n * F(x)^n / (1 + (n+2)*x*F(x))^(n+1), which holds formally for all Maclaurin series F(x). - Paul D. Hanna, Oct 03 2023

Examples

			G.f.: A(x) = 1 + 2*x + 3*x^2 + 4*x^3 + 16*x^4 + 104*x^5 + 515*x^6 + 2090*x^7 + 8170*x^8 + 34704*x^9 + 160014*x^10 + 751282*x^11 + 3479758*x^12 + ...
RELATED TABLE.
The table of coefficients of x^k in (1 + (n+1)*x*A(x))^n/A(x)^n begins:
n=1: [1,  0,    1,     0,    -11,     -54,    -182,    -594, ...];
n=2: [1,  2,    3,     2,    -21,    -130,    -494,   -1660, ...];
n=3: [1,  6,   15,    20,    -18,    -288,   -1391,   -5070, ...];
n=4: [1, 12,   58,   144,    151,    -468,   -3934,  -17376, ...];
n=5: [1, 20,  165,   720,   1715,    1274,   -8960,  -60530, ...];
n=6: [1, 30,  381,  2650,  10824,   24576,   10623, -176034, ...];
n=7: [1, 42,  763,  7812,  49084,  191016,  413343,   49818, ...];
n=8: [1, 56, 1380, 19600, 176242, 1033664, 3873296, 8000000, ...]; ...
in which the main diagonal equals n*(n+2)^(n-2) for n > 1.
		

Crossrefs

Programs

  • PARI
    /* Formula [x^(n-1)] (1 + (n+1)*x*A(x))^n / A(x)^n = n*(n+2)^(n-2) */
    {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A;
    A[#A] = polcoeff( (1 + (m+1)*x*Ser(A))^m / Ser(A)^m , m-1)/m - (m+2)^(m-2) ); A[n+1]}
    for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) [x^(n-1)] (1 + (n+1)*x*A(x))^n / A(x)^n = n*(n+2)^(n-2) for n > 1.
(2) [x^(n-1)] (1 + (n-2)*x*A(x))^n / A(x)^n = -2*n*(n-4)^(n-2) for n > 1.
(3) [x^(n-1)] (1 + n*x*A(x))^n / A(x)^n = 2*n*((n+1)^(n-2) - (n-2)^(n-2))/3 for n > 1.
(4) A(x) = (3/x) * Sum{n>=1} n*(n+2)^(n-2) * x^n * A(x)^n / (1 + (n+2)*x*A(x))^(n+1). - Paul D. Hanna, Oct 03 2023
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