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.

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A157308 G.f. A(x) satisfies the condition that both A(x) and F(x) = A(x*F(x)) = g.f. of A155585 have zeros for every other coefficient after initial terms; g.f. of dual sequence A157309 satisfies the same condition.

Original entry on oeis.org

1, 1, -1, 0, 3, 0, -38, 0, 947, 0, -37394, 0, 2120190, 0, -162980012, 0, 16330173251, 0, -2070201641498, 0, 324240251016266, 0, -61525045423103316, 0, 13913915097436287598, 0, -3698477457114061621492, 0
Offset: 0

Views

Author

Paul D. Hanna, Mar 11 2009

Keywords

Comments

After initial 2 terms, reversing signs yields A157310.
Conjecture: a(m) == 1 (mod 2) iff m is a power of 2 or m=0. [Paul D. Hanna, Mar 17 2009]

Examples

			G.f.: A(x) = 1 + x - x^2 + 3*x^4 - 38*x^6 + 947*x^8 - 37394*x^10 +-...
RELATED FUNCTIONS.
If F(x) = A(x*F(x)) then F(x) = o.g.f. of A155585:
A155585 = [1,1,0,-2,0,16,0,-272,0,7936,0,-353792,0,...];
...
If G(x) = A(x*G(x))/(1+x) then G(x) = o.g.f. of A122045:
A122045 = [1,0,-1,0,5,0,-61,0,1385,0,-50521,0,2702765,0,...];
...

Crossrefs

Cf. A157309, A157310, A157304, A157305, A155585, A122045 (Euler numbers).
Cf. A158119. [Paul D. Hanna, Mar 17 2009]

Programs

  • Mathematica
    terms = 28;
F[x_] = Sum[n! x^n/Product[(1 + 2k x), {k, 1, n}], {n, 0, terms+1}] + O[x]^(terms+1);
A[x_] = x/InverseSeries[x F[x]];
CoefficientList[A[x], x][[1 ;; terms]] (* Jean-François Alcover, Jul 26 2018 *)
  • PARI
    {a(n)=local(A=[1, 1]); for(i=1, n, if(#A%2==0, A=concat(A, 0);); if(#A%2==1, A=concat(A, t); A[ #A]=-subst(Vec(x/serreverse(x*Ser(A)))[ #A], t, 0))); Vec(x/serreverse(x*Ser(A)))[n+1]}

Formula

Let F(x) = o.g.f. of A155585, then o.g.f. A(x) satisfies:
A(x) = x/serreverse(x*F(x));
A(x) = 2x + F( -x/(A(x) - 2x) );
A(x) = F(x/A(x));
F(x) = A(x*F(x));
where A155585 is defined by e.g.f. exp(x)/cosh(x).
...
Let G(x) = o.g.f. of A122045, then o.g.f. A(x) satisfies:
A(x) = x + x/serreverse(x*G(x));
A(x) = x + G( x/(A(x) - x) );
G(x) = A(x*G(x))/(1+x);
where A122045 is the Euler numbers.
...
O.g.f.: A(x) = 2*(1+x) - H(x) where H(x) = g.f. of A157310.

A157310 G.f. A(x) satisfies the condition that both A(x) and F(x) = A(x/F(x)) = o.g.f. of A157309 have zeros for every other coefficient after initial terms; g.f. of dual sequence A155585 satisfies the same condition.

Original entry on oeis.org

1, 1, 1, 0, -3, 0, 38, 0, -947, 0, 37394, 0, -2120190, 0, 162980012, 0, -16330173251, 0, 2070201641498, 0, -324240251016266, 0, 61525045423103316, 0, -13913915097436287598, 0, 3698477457114061621492, 0
Offset: 0

Views

Author

Paul D. Hanna, Mar 11 2009

Keywords

Comments

After initial 2 terms, reversing signs yields A157308.

Examples

			G.f.: A(x) = 1 + x + x^2 - 3*x^4 + 38*x^6 - 947*x^8 + 37394*x^10 -+...
RELATED FUNCTIONS.
If F(x) = A(x/F(x)) then F(x) = o.g.f. of A157309:
A157309 = [1,1,0,-1,0,9,0,-176,0,5693,0,-272185,0,...];
...
If G(x) = (2 - A(x*G(x)))/(1-x) then G(x) = o.g.f. of A122045:
A122045 = [1,0,-1,0,5,0,-61,0,1385,0,-50521,0,2702765,0,...];
...
Let H(x) = A(x*H(-x)) = o.g.f. of A155585:
A155585 = [1,1,0,-2,0,16,0,-272,0,7936,0,-353792,0,...];
...

Crossrefs

Cf. A157308, A157309, A155585, A157304, A157305, A122045 (Euler numbers).

Programs

  • PARI
    {a(n)=local(A=[1, 1]); for(i=1, n, if(#A%2==1, A=concat(A, 0);); if(#A%2==0, A=concat(A, t); A[ #A]=-subst(Vec(x/serreverse(x*Ser(A)))[ #A], t, 0))); A[n+1]}

Formula

Let F(x) = A(x/F(x)) = o.g.f. of A157309, then F(x) satisfies:
A(x) = Series_Reversion(x/F(x))/x;
A(x) = F(x*A(x));
F(x) = A(x/F(x));
where A157309 has zeros for every other term after initial [1,1].
...
Let G(x) = o.g.f. of A122045, then o.g.f. A(x) satisfies:
A(x) = 2+x - x/Series_Reversion(x*G(x));
A(x) = 2+x - G( x/(2+x - A(x)) );
G(x) = (2 - A(x*G(x)))/(1-x);
where A122045 is the Euler numbers.
...
Let H(x) = o.g.f. of A155585, then o.g.f. A(x) satisfies:
A(x) = 2(1+x) - x/Series_Reversion(x*H(x));
A(x) = 2 - H( -x/(2 - A(x)) );
A(x) = H(-x/A(x));
H(x) = A(x*H(-x));
where A155585 is defined by e.g.f. exp(x)/cosh(x).
...
O.g.f.: A(x) = 2*(1+x) - B(x) where B(x) = g.f. of A157308.

A158119 Unsigned bisection of A157308 and A157310.

Original entry on oeis.org

1, 1, 3, 38, 947, 37394, 2120190, 162980012, 16330173251, 2070201641498, 324240251016266, 61525045423103316, 13913915097436287598, 3698477457114061621492, 1141824214469896983332508
Offset: 0

Views

Author

Paul D. Hanna, Mar 12 2009

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 38*x^3 + 947*x^4 + 37394*x^5 + 2120190*x^6 + 162980012*x^7 + 16330173251*x^8 + 2070201641498*x^9 + 324240251016266*x^10 +...
RELATED FUNCTIONS.
G.f. of A157308, B(x) = x + A(-x^2), satisfies the condition
that both B(x) and F(x) = B(x*F(x)) = o.g.f. of A155585
have zeros for every other coefficient after initial terms:
A157308 = [1,1,-1,0,3,0,-38,0,947,0,-37394,0,2120190,0,...];
A155585 = [1,1,0,-2,0,16,0,-272,0,7936,0,-353792,0,...].
...
G.f. of A157310, C(x) = 2+x - A(-x^2), satisfies the condition
that both C(x) and G(x) = C(x/G(x)) = o.g.f. of A157309
have zeros for every other coefficient after initial terms:
A157310 = [1,1,1,0,-3,0,38,0,-947,0,37394,0,-2120190,0,...];
A157309 = [1,1,0,-1,0,9,0,-176,0,5693,0,-272185,0,...].
...

Links

Crossrefs

Cf. A157308, A157310, A155585, A157309, A158120.

Programs

  • Mathematica
    terms = 30;
F[x_] = Sum[n! x^n/Product[(1 + 2 k x), {k, 1, n}], {n, 0, terms+1}] + O[x]^(terms+1);
A[x_] = x/InverseSeries[x F[x]];
Partition[CoefficientList[A[x], x][[1 ;; terms]], 2][[All, 1]] // Abs (* Jean-François Alcover, Jul 27 2018 *)
  • PARI
    {a(n)=local(A=[1, 1]); for(i=1, 2*n, if(#A%2==0, A=concat(A, 0);); if(#A%2==1, A=concat(A, t); A[ #A]=-subst(Vec(x/serreverse(x*Ser(A)))[ #A], t, 0))); (-1)^n*Vec(x/serreverse(x*Ser(A)))[2*n+1]}
  • PARI
    {a(n) = my(A=[1],CF=1); for(i=1,n, A=concat(A,0); for(i=1,#A, CF = Ser(A) - (#A-i+1)^2*x/CF ); A[#A] = -polcoeff(CF,#A-1) );A[n+1] }
for(n=0,20,print1(a(n),", ")) \\ Paul D. Hanna, Nov 04 2020

Formula

G.f. A(x) satisfies: 1 = A(x) - x/(A(x) - 2^2*x/(A(x) - 3^2*x/(A(x) - 4^2*x/(A(x) - 5^2*x/(A(x) - 6^2*x/(A(x) - ...)))))), a continued fraction. - Paul D. Hanna, Nov 04 2020
Conjecture: a(m) == 1 (mod 2) iff m is a power of 2 or m=0. [Paul D. Hanna, Mar 16 2009]
a(n) ~ 2^(4*n + 3) * n^(2*n + 1/2) / (Pi^(2*n + 1/2) * exp(2*n)). - Vaclav Kotesovec, Nov 12 2020
Showing 1-3 of 3 results.

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