A155928
G.f. satisfies: A(x) = F(x)^2 where F(x) = Sum_{n>=0} A155926(n)*x^n/[n!*(n+1)!/2^n] and A(x) = Sum_{n>=0} a(n)*x^n/[n!*(n+1)!/2^n].
Original entry on oeis.org
1, 2, 11, 122, 2302, 66482, 2735721, 152359874, 11048880926, 1012437290342, 114445632250776, 15649612498128050, 2546878326578431588, 486567378291992448726, 107845834421517755737817
Offset: 0
G.f.: A(x) = 1 + 2*x + 11*x^2/3 + 122*x^3/18 + 2302*x^4/180 + 66482*x^5/2700 +...
G.f.: A(x) = F(x)^2 where:
F(x) = 1 + x + 4*x^2/3 + 37*x^3/18 + 621*x^4/180 + 16526*x^5/2700 +...+ A155926(n)*x^n/[n!*(n+1)!/2^n] +...
G.f. satisfies: A(x) = B( x*sqrt(A(x)) )^2 where:
B(x) = 1 + x + x^2/3 + x^3/18 + x^4/180 + x^5/2700 +...+ x^n/[n!*(n+1)!/2^n] +...
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{a(n)=local(B=sum(k=0,n,x^k/(k!*(k+1)!/2^k))+x*O(x^n));polcoeff((serreverse(x/B)/x)^2,n)*n!*(n+1)!/2^n}
A103365
First column of triangle A103364, which equals the matrix inverse of the Narayana triangle (A001263).
Original entry on oeis.org
1, -1, 2, -7, 39, -321, 3681, -56197, 1102571, -27036487, 810263398, -29139230033, 1238451463261, -61408179368043, 3513348386222286, -229724924077987509, 17023649385410772579, -1419220037471837658603, 132236541042728184852942, -13690229149108218523467549
Offset: 1
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)
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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 *)
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a(n)=if(n<1,0,(matrix(n,n,m,j,binomial(m-1,j-1)*binomial(m,j-1)/j)^-1)[n,1])
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{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
A105556
Triangle, read by rows, such that column n equals the row sums of A001263^n, which is the n-th matrix power of the Narayana triangle A001263, for n>=0.
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 1, 5, 3, 1, 1, 14, 12, 4, 1, 1, 42, 57, 22, 5, 1, 1, 132, 303, 148, 35, 6, 1, 1, 429, 1743, 1144, 305, 51, 7, 1, 1, 1430, 10629, 9784, 3105, 546, 70, 8, 1, 1, 4862, 67791, 90346, 35505, 6906, 889, 92, 9, 1, 1, 16796, 448023, 885868, 444225, 99156
Offset: 0
Triangle begins:
1;
1, 1;
1, 2, 1;
1, 5, 3, 1;
1, 14, 12, 4, 1;
1, 42, 57, 22, 5, 1;
1, 132, 303, 148, 35, 6, 1;
1, 429, 1743, 1144, 305, 51, 7, 1;
1, 1430, 10629, 9784, 3105, 546, 70, 8, 1;
1, 4862, 67791, 90346, 35505, 6906, 889, 92, 9, 1;
...
From _Paul D. Hanna_, Feb 01 2009: (Start)
G.f. for rows n=0..3 are:
B(x) = 1 + x + x^2/3 + x^3/18 + x^4/180 + x^5/2700 + ... + x^n/[n!*(n+1)!/2^n] + ...
B(x)^2 = 1 + 2*x + 5*x^2/3 + 14*x^3/18 + 42*x^4/180 + ... + A000108(n)*x^n/[n!*(n+1)!/2^n] + ...
B(x)^3 = 1 + 3*x +12*x^2/3 + 57*x^3/18 +303*x^4/180 + ... + A103370(n)*x^n/[n!*(n+1)!/2^n] + ...
B(x)^4 = 1 + 4*x +22*x^2/3 +148*x^3/18+1144*x^4/180 + 9784*x^5/2700 + 90346*x^5/56700 + ... (End)
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{T(n,k)=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-k,(N^k)[n-k+1,j+1])}
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{T(n,k)=local(B=sum(j=0,n-k,x^j/(j!*(j+1)!/2^j))+x*O(x^(n-k))); polcoeff(B^(k+1),n-k)*(n-k)!*(n-k+1)!/2^(n-k)} \\ Paul D. Hanna, Feb 01 2009
A155927
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) = A(x*B(x)) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n].
Original entry on oeis.org
1, 1, -2, 19, -379, 12726, -641465, 45181627, -4232016719, 508271819428, -76108505872794, 13896010073569130, -3038043685025188631, 783439451518414509612, -235289860249420249309140
Offset: 0
G.f.: A(x) = 1 + x - 2*x^2/3 + 19*x^3/18 - 379*x^4/180 + 12726*x^5/2700 +...+ a(n)*x^n/[n!*(n+1)!/2^n] +...
G.f. satisfies: A(x) = B(x/A(x)) and B(x) = A(x*B(x)) where:
B(x) = 1 + x + 1/3*x^2 + 1/18*x^3 + 1/180*x^4 +...+ x^n/[n!*(n+1)!/2^n] +...
G.f. satisfies: A(x) = F(x/A(x)^2) and F(x) = A(x*F(x)^2) where:
F(x) = 1 + x + 4*x^2/3 + 37*x^3/18 + 621*x^4/180 + 16526*x^5/2700 +...+ A155926(n)*x^(n+1)/[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) = 1 - x + 2*x^2/3 - 7*x^3/18 + 39*x^4/180 - 321*x^5/2700 +...+ A103365(n)*x^(n+1)/[n!*(n+1)!/2^n] +...
A105558
Central terms in even-indexed rows of triangle A105556 and thus equals the n-th row sum of the n-th matrix power of the Narayana triangle A001263.
Original entry on oeis.org
1, 2, 12, 148, 3105, 99156, 4481449, 272312216, 21414443481, 2116193061340, 256712977920256, 37506637787774112, 6496315164318118165, 1316230822119433518312, 308426950979497974254310
Offset: 0
Terms a(n) divided by (n+1) begin:
1,1,4,37,621,16526,640207,34039027,2379382609,211619306134,...
Contribution from _Paul D. Hanna_, Jan 31 2009: (Start)
G.f.: A(x) = 1 + 2*x + 12*x^2/3 + 148*x^3/18 + 3105*x^4/180 +...+ a(n)*x^n/[n!*(n+1)!/2^n] +...
G.f.: A(x) = d/dx x*F(x) where F(x) = B(x*F(x)) and:
F(x) = 1 + x + 4*x^2/3 + 37*x^3/18 + 621*x^4/180 + 16526*x^5/2700 +...+ A155926(n)*x^n/[n!*(n+1)!/2^n] +...
B(x) = 1 + x + x^2/3 + x^3/18 + x^4/180 +...+ x^n/[n!*(n+1)!/2^n] +... (End)
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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])
for(n=0,20,print1(a(n),", "))
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a(n)=local(F=sum(k=0,n,x^k/(k!*(k+1)!/2^k))+x*O(x^n));polcoeff(deriv(serreverse(x/F)),n)*n!*(n+1)!/2^n
for(n=0,20,print1(a(n),", ")) \\ Paul D. Hanna, Jan 31 2009
Showing 1-5 of 5 results.
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