A145349 - OEIS

%I #14 Jan 11 2016 11:08:56

%S 1,1,4,34,416,6319,111124,2177346,46440184,1061938195,25762345804,

%T 658072997702,17600573291712,490770914734054,14219015899154068,

%U 426904437068035200,13252855203929697200,424634035832800883743,14020984603588221319072,476460483567807053745450

%N G.f. satisfies: A(x/A(x)^3) = 1 + x*A(x).

%C More generally, if g.f. A(x) satisfies: A(x/A(x)^k) = 1 + x*A(x)^m, then

%C A(x) = 1 + x*G(x)^(m+k) where G(x) = A(x*G(x)^k) and G(x/A(x)^k) = A(x);

%C thus a(n) = [x^(n-1)] ((m+k)/(m+k*n))*A(x)^(m+k*n) for n>=1 with a(0)=1.

%F G.f.: A(x) = 1 + x*G(x)^4 where G(x) = A(x*G(x)^3) and A(x) = G(x/A(x)^3).

%F a(n) = [x^(n-1)] 4*A(x)^(3*n+1)/(3*n+1) for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in 4*A(x)^(3*n+1)/(3*n+1) for n>=1 (see comment).

%e G.f.: A(x) = 1 + x + 4*x^2 + 34*x^3 + 416*x^4 + 6319*x^5 +...

%e A(x)^3 = 1 + 3*x + 15*x^2 + 127*x^3 + 1512*x^4 + 22419*x^5 +...

%e A(x/A(x)^3) = 1 + x + x^2 + 4*x^3 + 34*x^4 + 416*x^5 + 6319*x^6 +...

%e A(x) = 1 + x*G(x)^4 where G(x) = A(x*G(x)^3):

%e G(x) = 1 + x + 7*x^2 + 82*x^3 + 1239*x^4 + 21942*x^5 + 434746*x^6 +...

%e G(x)^3 = 1 + 3*x + 24*x^2 + 289*x^3 + 4377*x^4 + 77097*x^5 +...

%e To illustrate the formula a(n) = [x^(n-1)] 4*A(x)^(3*n+1)/(3*n+1),

%e form a table of coefficients in A(x)^(3*n+1) as follows:

%e A^4: [(1), 4, 22, 188, 2217, 32516, 555972, ...];

%e A^7: [1, (7), 49, 441, 5131, 73248, 1220457, ...];

%e A^10: [1, 10, (85), 820, 9590, 134482, 2191060, ...];

%e A^13: [1, 13, 130, (1352), 16107, 223886, 3582072, ...];

%e A^16: [1, 16, 184, 2064, (25276), 351072, 5541912, ...];

%e A^19: [1, 19, 247, 2983, 37772, (527839), 8260174, ...];  ...

%e in which the main diagonal forms the initial terms of this sequence:

%e [4/4*(1), 4/7*(7), 4/10*(85), 4/13*(1352), 4/16*(25276), 4/19*(527839), ...].

%e ALTERNATE GENERATING METHOD.

%e This sequence forms column zero in the follow array.

%e Let A denote this sequence, and A^3 the self-convolution cube of A.

%e Start in row zero with this sequence, A, after prepending an initial '1', then repeat: drop the initial term and perform convolution with A^3 and the remaining terms in a given row to obtain the next row:

%e [1, 1, 1, 4, 34, 416, 6319, 111124, 2177346, 46440184, 1061938195, ...];

%e [1, 4, 22, 188, 2217, 32516, 555972, 10655628, 223313220, 5034249556, ...];

%e [4, 34, 314, 3619, 50829, 833591, 15417781, 313704516, 6900409869, ...];

%e [34, 416, 5071, 70714, 1131649, 20377616, 404581945, 8712077584, ...];

%e [416, 6319, 92167, 1472688, 26106282, 508663862, 10756749655, ...];

%e [6319, 111124, 1843974, 32709364, 630332565, 13142389012, 294045605744, ...];

%e [111124, 2177346, 39908146, 770233015, 15926231353, 352361228127, ...];

%e [46440184, 1061938195, 22576531219, 497127293708, 11532665230502, ...]; ...

%o (PARI) {a(n)=local(A=1+x,G);for(i=0,n,G=(serreverse(x/(A+x*O(x^n))^3)/x)^(1/3);A=1+x*G^4);polcoeff(A,n)}

%o (PARI) /* This sequence is generated when k=3, m=1: A(x/A(x)^k) = 1 + x*A(x)^m */

%o {a(n, k=3, m=1)=local(A=sum(i=0, n-1, a(i, k, m)*x^i)+x*O(x^n)); if(n==0, 1, polcoeff((m+k)/(m+k*n)*A^(m+k*n), n-1))}

%o for(n=0,20,print1(a(n),", "))

%o (PARI) /* Prints terms 0..30 */

%o {A=[1];

%o for(m=1,30,

%o   B=A;

%o   for(i=1,m-1, C=Vec(Ser(A)^3*Ser(B)); B=vector(#C-1,n,C[n+1]) );

%o   A=concat(A,0);A[#A]=B[1]

%o );

%o A} \\ _Paul D. Hanna_, Jan 10 2016

%Y Cf. variants: A002293, A145347, A145348, A120974,  A145345, A145350, A147664.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Nov 11 2008