A157310 -id:A157310 - cp's OEIS frontend

A158119 Unsigned bisection of A157308 and A157310.

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

Offset: 0

Paul D. Hanna, Mar 12 2009

nonn

			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,...].
...

Vaclav Kotesovec, Table of n, a(n) for n = 0..200 (terms 0..50 from Paul D. Hanna)
M. Bozejko and T. Hasebe, On free infinite divisibility for classical Meixner distributions, arXiv preprint arXiv:1302.4885 [math.PR], 2013-2014.

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

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 *)

{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]}

{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

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

A157309 G.f. A(x) satisfies the condition that both A(x) and F(x) = A(x*F(x)) = o.g.f. of A157310 have zeros for every other coefficient after initial terms; g.f. of dual sequence A157308 satisfies the same condition.

Original entry on oeis.org

1, 1, 0, -1, 0, 9, 0, -176, 0, 5693, 0, -272185, 0, 18043492, 0, -1587355800, 0, 179258676373, 0, -25305967691715, 0, 4370075849887361, 0, -906689353191842372, 0, 222613537277330398444, 0, -63850898347335510126988

Offset: 0

Paul D. Hanna, Mar 11 2009

sign

			G.f.: A(x) = 1 + x - x^3 + 9*x^5 - 176*x^7 + 5693*x^9 -+...
RELATED FUNCTIONS.
If F(x) = A(x*F(x)) then F(x) = o.g.f. of A157310:
A157310 = [1,1,1,0,-3,0,38,0,-947,0,37394,0,-2120190,0,...];
has zeros for every other coefficient after initial terms.
...
O.g.f. A(x) has similar properties as o.g.f. of A157308:
A157308 = [1,1,-1,0,3,0,-38,0,947,0,-37394,0,2120190,0,...].

Cf. A157308, A157310, A157307, A157304, A157305.

{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))); Vec(x/serreverse(x*Ser(A)))[n+1]}

Let F(x) = o.g.f. of A157310, 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 A157310 has zeros for every other term after initial [1,1,1].

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

Paul D. Hanna, Mar 11 2009

sign

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]

			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,...];
...

Cf. A157309, A157310, A157304, A157305, A155585, A122045 (Euler numbers).

Cf. A158119. [Paul D. Hanna, Mar 17 2009]

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 *)

{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]}

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.

A172396 G.f. satisfies: A(x) = G(x/A(x)) where o.g.f. G(x) = A(xG(x)) = Sum_{n>=0} A003701(n)x^n.

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

Paul D. Hanna, Feb 07 2010

nonn

The e.g.f. of A003701 is exp(x)/cos(x) = Sum_{n>=0} A003701(n)*x^n/n!.

Compare to A157308 and A157310.

			G.f.: A(x) = 1 + x + x^2 + 3*x^4 + 38*x^6 + 947*x^8 + 37394*x^10 +...
where G(x) = A(x*G(x)) is the o.g.f. of A003701:
G(x) = 1 + x + 2*x^2 + 4*x^3 + 12*x^4 + 36*x^5 + 152*x^6 + 624*x^7 +...
while the e.g.f. of A003701 is given by:
exp(x)/cos(x) = 1 + x + 2*x^2/2! + 4*x^3/3! + 12*x^4/4! + 36*x^5/5! +...

Cf. A003701, A158119, A157308, A157310.

{a(n)=local(X=x+x*O(x^n),G=sum(m=0,n,m!*polcoeff(exp(X)/cos(X),m)*x^m)+x*O(x^n)); polcoeff(x/serreverse(x*G),n)}

a(n) = |A157308(n)| = |A157310(n)| for n>=0.
a(2n) = A158119(n) for n>=0; a(2n-1) = 0 for n>=2, with a(1)=1.
G.f. A = A(x) satisfies: A(x) = 1/(1-x/A - (x/A)^2/(1-x/A - 2^2*(x/A)^2/(1-x/A - 3^2*(x/A)^2/(1-x/A - 4^2*(x/A)^2/(1-x/A - 5^2*(x/A)^2/(1-x/A -...)))))), a recursive continued fraction. [From Paul D. Hanna, Jan 05 2012]

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A158119 Unsigned bisection of A157308 and A157310.

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A157309 G.f. A(x) satisfies the condition that both A(x) and F(x) = A(x*F(x)) = o.g.f. of A157310 have zeros for every other coefficient after initial terms; g.f. of dual sequence A157308 satisfies the same condition.

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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.

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A172396 G.f. satisfies: A(x) = G(x/A(x)) where o.g.f. G(x) = A(x*G(x)) = Sum_{n>=0} A003701(n)*x^n.

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A172396 G.f. satisfies: A(x) = G(x/A(x)) where o.g.f. G(x) = A(xG(x)) = Sum_{n>=0} A003701(n)x^n.