A155927 -id:A155927 - cp's OEIS frontend

A103365 First column of triangle A103364, which equals the matrix inverse of the Narayana triangle (A001263).

1, -1, 2, -7, 39, -321, 3681, -56197, 1102571, -27036487, 810263398, -29139230033, 1238451463261, -61408179368043, 3513348386222286, -229724924077987509, 17023649385410772579, -1419220037471837658603, 132236541042728184852942, -13690229149108218523467549

Offset: 1

Paul D. Hanna, Feb 02 2005

sign

			From _Paul D. Hanna_, Jan 31 2009: (Start)
G.f.: A(x) = 1 - x + 2*x^2/3 - 7*x^3/18 + 39*x^4/180 - 321*x^5/2700 +...
G.f.: A(x) = 1/B(x) where:
B(x) = 1 + x + x^2/3 + x^3/18 + x^4/180 + x^5/2700 +...+ x^n/[n!*(n+1)!/2^n] +... (End)

G. C. Greubel, Table of n, a(n) for n = 1..250

Cf. A001263, A103364, A103366, A103367.
Cf. A155926, A155927. [Paul D. Hanna, Jan 31 2009]
Cf. A238390, A180874, A115369.

Table[(-1)^((n-1)/2) * (CoefficientList[Series[x/BesselJ[1,2*x],{x,0,40}],x])[[n]] * ((n+1)/2)! * ((n-1)/2)!,{n,1,41,2}] (* Vaclav Kotesovec, Mar 01 2014 *)

a(n)=if(n<1,0,(matrix(n,n,m,j,binomial(m-1,j-1)*binomial(m,j-1)/j)^-1)[n,1])

{a(n)=local(B=sum(k=0,n,x^k/(k!*(k+1)!/2^k))+x*O(x^n));polcoeff(1/B,n)*n!*(n+1)!/2^n} \\ Paul D. Hanna, Jan 31 2009

From Paul D. Hanna, Jan 31 2009: (Start)
G.f.: A(x) = 1/B(x) where A(x) = Sum_{n>=0} (-1)^n*a(n)*x^n/[n!*(n+1)!/2^n] and B(x) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n].
G.f. satisfies: A(x) = 1/F(x*A(x)) and F(x) = 1/A(x*F(x)) where F(x) = Sum_{n>=0} A155926(n)*x^n/[n!*(n+1)!/2^n].
G.f. satisfies: A(x) = 1/G(x/A(x)) and G(x) = 1/A(x/G(x)) where G(x) = Sum_{n>=0} A155927(n)*x^n/[n!*(n+1)!/2^n]. (End)
a(n) ~ (-1)^(n+1) * c * n! * (n-1)! * d^n, where d = 4/BesselJZero[1, 1]^2 = 0.2724429913055159309179376055957891881897555639652..., and c = 9.11336321311226744479181866135367355200240221549667284076... = BesselJZero[1, 1]^2 / (4*BesselJ[2, BesselJZero[1, 1]]). - Vaclav Kotesovec, Mar 01 2014, updated Apr 01 2018

A155926 G.f. satisfies: A(x) = B(xA(x)) where A(x) = Sum_{n>=0} a(n)x^n/[n!(n+1)!/2^n] and B(x) = Sum_{n>=0} x^n/[n!(n+1)!/2^n].

Original entry on oeis.org

1, 1, 4, 37, 621, 16526, 640207, 34039027, 2379382609, 211619306134, 23337543447296, 3125553148981176, 499716551101393705, 94016487294245251308, 20561796731966531616954, 5172827581575899147920471

Offset: 0

Paul D. Hanna, Jan 30 2009

nonn

			G.f.: A(x) = 1 + x + 4*x^2/3 + 37*x^3/18 + 621*x^4/180 + 16526*x^5/2700 +...+ a(n)*x^n/[n!*(n+1)!/2^n] +...
B(x) = 1 + x + 1/3*x^2 + 1/18*x^3 + 1/180*x^4 +...+ x^n/[n!*(n+1)!/2^n] +... where
A(x) = B(x*A(x)) and B(x) = A(x/B(x)) ;
1/B(x) = 1 - x + 2*x^2/3 - 7*x^3/18 + 39*x^4/180 - 321*x^5/2700 +...+ (-1)^n*A103365(n)*x^n/[n!*(n+1)!/2^n] +...
Also, A(x) = C(x*A(x)^2) where:
C(x) = 1 + x - 2*x^2/3 + 19*x^3/18 - 379*x^4/180 + 12726*x^5/2700 +...+ A155927(n)*x^(n+1)/[n!*(n+1)!/2^n] +...
A(x)^2 = 1 + 2*x + 11*x^2/3 + 122*x^3/18 + 2302*x^4/180 + 66482*x^5/2700 +...

Cf. A105558, A105556, A001263, A103365, A006472, A090181, A155927, A155928 (A^2).

{a(n)=local(F=sum(k=0,n,x^k/(k!*(k+1)!/2^k))+x*O(x^n));polcoeff(serreverse(x/F)/x,n)*n!*(n+1)!/2^n}

{a(n)=local(N=matrix(n+1, n+1, m, j, if(m>=j, binomial(m-1, j-1)*binomial(m, j-1)/j))); sum(j=0, n, (N^n)[n+1, j+1])/(n+1)}

a(n) = A105558(n)/(n+1) = A105556(2n,n)/(n+1) = [N^(n+1)](n+1,1)/(n+1) for n>=0, where N^(n+1) is the (n+1)-th matrix power of the Narayana triangle N=A001263.
G.f.: A(x) = Series_Reversion[x/B(x)]/x where B(x) = A(x/B(x)) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n].
G.f. satisfies: A(x) = C(x*A(x)^2) and C(x) = A(x/C(x)^2) where C(x) is the g.f. of A155927.

cp's OEIS Frontend

A103365 First column of triangle A103364, which equals the matrix inverse of the Narayana triangle (A001263).

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A155926 G.f. satisfies: A(x) = B(xA(x)) where A(x) = Sum_{n>=0} a(n)x^n/[n!(n+1)!/2^n] and B(x) = Sum_{n>=0} x^n/[n!(n+1)!/2^n].

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cp's OEIS Frontend

A103365 First column of triangle A103364, which equals the matrix inverse of the Narayana triangle (A001263).

Views

Author

Keywords

Examples

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Programs

Mathematica

PARI

PARI

Formula

A155926 G.f. satisfies: A(x) = B(x*A(x)) where A(x) = Sum_{n>=0} a(n)*x^n/[n!*(n+1)!/2^n] and B(x) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n].

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Author

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PARI

PARI

Formula

A155926 G.f. satisfies: A(x) = B(xA(x)) where A(x) = Sum_{n>=0} a(n)x^n/[n!(n+1)!/2^n] and B(x) = Sum_{n>=0} x^n/[n!(n+1)!/2^n].