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 1-2 of 2 results.

A120955 G.f. A(x) satisfies: Series_Reversion( x/A(x) ) / x = 2*A(x) - (1+x).

Original entry on oeis.org

1, 1, 1, 4, 25, 206, 2060, 23920, 314065, 4582300, 73393490, 1278859176, 24073541260, 486806278752, 10525038917720, 242318610557760, 5919811842140945, 152974724047702626, 4169576527021400852
Offset: 0

Views

Author

Paul D. Hanna, Jul 19 2006

Keywords

Comments

The g.f. for A120956 = Series_Reversion( x/A(x) ) / x = 2*A(x) - (1+x), where A(x) is the g.f. of this sequence.

Examples

			A(x) = 1 + x + x^2 + 4*x^3 + 25*x^4 + 206*x^5 + 2060*x^6 +...
The g.f. of A120956 is:
series_reversion(x/A(x))/x = 1 + x + 2*x^2 + 8*x^3 + 50*x^4 + 412*x^5 +...
Compare terms to see that A120956(n) = 2*a(n) for n>=2.
The g.f. satisfies the series:
A(x) = 1+x + x*d/dx (A(x)-1)^2/2! + x^2*d^2/dx^2 (A(x)-1)^3/3! + x^3*d^3/dx^3 (A(x)-1)^4/4! + x^4*d^4/dx^4 (A(x)-1)^5/5! +...

Links

Crossrefs

Cf. A120956, A120970 (variant), A249933.

Programs

  • PARI
    a(n)=local(A=[1,1]);for(i=1,n,A=concat(A,t); A[ #A]=subst(Vec(serreverse(x/Ser(A)))[ #A],t,0));A[n+1]
for(n=0, 25, print1(a(n), ", "))
  • PARI
    Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D
{a(n)=local(A=1+x*O(x^n)); for(i=1, n, A=1+x+sum(m=1, n, x^m*Dx(m, (A-1+x*O(x^n))^(m+1)/(m+1)!) )); polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Dec 27 2012

Formula

G.f. A(x) satisfies:
(1) A( x*(2*A(x) - 1-x) ) = 2*A(x) - 1-x.
(2) A(x) = 2*A(x/A(x)) - 1 - x/A(x).
(3) A(x) = F(x/A(x)) and F(x) = A(x*F(x)) where F(x) = g.f. of A120956.
(4) A(x) = 1+x + Sum{n>=1} x^n * d^n/dx^n (A(x)-1)^(n+1) / (n+1)!. - Paul D. Hanna, Dec 27 2012
a(n) = A120956(n)/2 for n>=2.
a(n) = [x^n] A(x)^n / (2*n) for n>1; i.e., a(n) equals the coefficient of x^n in A(x)^n divided by 2*n.
a(n) ~ c * n^(n + 1/2 + log(2)) / (exp(n) * (log(2))^n), where c = 0.33794865962155... . - Vaclav Kotesovec, Aug 10 2014

A326604 G.f. A(x) satisfies x / Series_Reversion(x*A(x)) = (2*A(x) + 1+x)/3.

Original entry on oeis.org

1, 1, 3, 21, 231, 3333, 58167, 1175877, 26827623, 679078677, 18844334727, 568229240901, 18491559492999, 645850960844469, 24099045218945031, 956889503377128261, 40291822946545245927, 1793614919867776690389, 84177429562216608349959, 4154548653801498090246597, 215137302566817565660007367, 11664210072689092804458508533
Offset: 0

Views

Author

Paul D. Hanna, Sep 21 2019

Keywords

Comments

a(k) = 6 (mod 9) when k = A191107(n) for n > 1 (conjecture).
a(n) = 3*A249933(n) for n > 1.

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 21*x^3 + 231*x^4 + 3333*x^5 + 58167*x^6 + 1175877*x^7 + 26827623*x^8 + 679078677*x^9 + 18844334727*x^10 +...
such that
x/Series_Reversion(x*A(x)) = (2*A(x) + 1+x)/3 = 1 + x + 2*x^2 + 14*x^3 + 154*x^4 + 2222*x^5 + 38778*x^6 + 783918*x^7 + 17885082*x^8 + 452719118*x^9 + ...
ITERATIONS OF x*A(x).
Let G(x) = x*A(x), then
A(x) = 1 + G(x)/3 + G(G(x))*2/3^2 + G(G(G(x)))*2^2/3^3 + G(G(G(G(x))))*2^3/3^4 + G(G(G(G(G(x)))))*2^4/3^5 +...
The table of coefficients in the iterations of x*A(x) begin:
[1, 1, 3, 21, 231, 3333, 58167, 1175877, 26827623, ...];
[1, 2, 8, 58, 630, 8958, 154530, 3096330, 70161318, ...];
[1, 3, 15, 117, 1285, 18167, 310735, 6177745, 139076385, ...];
[1, 4, 24, 204, 2308, 32800, 559124, 11053668, 247451528, ...];
[1, 5, 35, 325, 3835, 55365, 946623, 18671961, 416326935, ...];
[1, 6, 48, 486, 6026, 89158, 1539350, 30423134, 677231222, ...];
[1, 7, 63, 693, 9065, 138383, 2427943, 48304893, 1076756889, ...];
[1, 8, 80, 952, 13160, 208272, 3733608, 75127944, 1682704256, ...];
[1, 9, 99, 1269, 18543, 305205, 5614887, 114768093, 2592154167, ...]; ...
in which the following sum along column k equals a(k+1):
a(2) = 3 = 1/3 + 2*2/9 + 3*4/27 + 4*8/81 + 5*16/243 + 6*32/729 +...
a(3) = 21 = 3/3 + 8*2/9 + 15*4/27 + 24*8/81 + 35*16/243 + 48*32/729 + ...
a(4) = 231 = 21/3 + 58*2/9 + 117*4/27 + 204*8/81 + 325*16/243 + 486*32/729 +...
a(5) = 3333 = 231*2/3 + 630*2/9 + 1285*4/27 + 2308*8/81 + 3835*16/243 + 6026*32/729 +...

Links

Crossrefs

Cf. A120956, A249933.

Programs

  • Mathematica
    nmax = 21; sol = {a[0] -> 1}; Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x] - (1 + 2 A[x A[x] + O[x]^(n+1)])/(3-x), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 03 2019 *)
  • PARI
    /* Prints N terms using x/Series_Reversion(x*A(x)) = (2*A(x) + 1+x)/3 */
N = 30; {A=[1, 1]; for(i=1, N, A = concat(A, -3*Vec(x/serreverse(x*Ser(concat(A, 0))))[#A+1]); print1(i, ", ") ); A}

Formula

G.f. A(x) satisfies:
(1) A( 3*x/(2*A(x) + 1+x) ) = (2*A(x) + 1+x)/3.
(2) A(x) = (1 + 2*A(x*A(x))) / (3-x).
(3) A(x) = 1 + Sum_{n>=1} G^n(x) * 2^(n-1)/3^n where G(x) = x*A(x) and G^n(x) = G^{n-1}(x*A(x)) denotes iteration with G^0(x) = x.
Showing 1-2 of 2 results.

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