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%I #18 Jan 11 2016 11:08:09
%S 1,1,3,18,154,1632,20007,273164,4058556,64628487,1091488334,
%T 19403175105,361028420037,7000932594042,141010975529568,
%U 2942134448306481,63449975020918843,1411787024678728344,32360032648643379471,763096191377494726161,18491954778730596443088
%N G.f. satisfies: A(x/A(x)^2) = 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.
%H Vaclav Kotesovec, <a href="/A145350/b145350.txt">Table of n, a(n) for n = 0..374</a>
%F G.f.: A(x) = 1 + x*G(x)^3 where G(x) = A(x*G(x)^2) and A(x) = G(x/A(x)^2).
%F a(n) = [x^(n-1)] 3*A(x)^(2*n+1)/(2*n+1) for n>=1 with a(0)=1; i.e., a(n) equals the coefficient of x^(n-1) in 3*A(x)^(2*n+1)/(2*n+1) for n>=1.
%e G.f.: A(x) = 1 + x + 3*x^2 + 18*x^3 + 154*x^4 + 1632*x^5 +...
%e A(x)^2 = 1 + 2*x + 7*x^2 + 42*x^3 + 353*x^4 + 3680*x^5 + 44526*x^6+...
%e A(x/A(x)^2) = 1 + x + x^2 + 3*x^3 + 18*x^4 + 154*x^5 + 1632*x^6 +...
%e A(x) = 1 + x*G(x)^3 where G(x) = A(x*G(x)^2):
%e G(x) = 1 + x + 5*x^2 + 41*x^3 + 432*x^4 + 5329*x^5 + 73512*x^6 +...
%e G(x)^2 = 1 + 2*x + 11*x^2 + 92*x^3 + 971*x^4 + 11932*x^5 +...
%e To illustrate the formula a(n) = [x^(n-1)] 3*A(x)^(2*n+1)/(2*n+1),
%e form a table of coefficients in A(x)^(2*n+1) as follows:
%e A^3: [(1), 3, 12, 73, 606, 6225, 74370, 994668, ...];
%e A^5: [1, (5), 25, 160, 1315, 13191, 153930, 2017620, ...];
%e A^7: [1, 7, (42), 287, 2373, 23436, 267988, 3445835, ...];
%e A^9: [1, 9, 63, (462), 3888, 38106, 428637, 5414760, ...];
%e A^11: [1, 11, 88, 693, (5984), 58619, 651354, 8099410, ...];
%e A^13: [1, 13, 117, 988, 8801, (86697), 955656, 11723712, ...];
%e A^15: [1, 15, 150, 1355, 12495, 124398, (1365820), 16571385, ...]; ...
%e in which the main diagonal forms the initial terms of this sequence:
%e [3/3*(1), 3/5*(5), 3/7*(42), 3/9*(462), 3/11*(5984), 3/13*(86697), ...].
%e ALTERNATE GENERATING METHOD.
%e This sequence forms column zero in the follow array.
%e Let A denote this sequence, and A^2 the self-convolution square 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^2 and the remaining terms in a given row to obtain the next row:
%e [1, 1, 1, 3, 18, 154, 1632, 20007, 273164, 4058556, 64628487, ...];
%e [1, 3, 12, 73, 606, 6225, 74370, 994668, 14535285, 228349287, ...];
%e [3, 18, 118, 962, 9511, 109404, 1415942, 20128565, 309001962, ...];
%e [18, 154, 1324, 13017, 146470, 1849625, 25701033, 386747469, ...];
%e [154, 1632, 16743, 188240, 2343654, 32006379, 473572975, ...];
%e [1632, 20007, 233150, 2905879, 39290669, 573813430, 8978918475, ...];
%e [20007, 273164, 3512228, 47574771, 689590692, 10679554646, ...];
%e [273164, 4058556, 56511375, 820798718, 12635699895, ...];
%e [4058556, 64628487, 962231360, 14843336308, 241004566025, ...]; ...
%o (PARI) {a(n)=local(A=1+x,G);for(i=0,n,G=(serreverse(x/(A+x*O(x^n))^2)/x)^(1/2);A=1+x*G^3);polcoeff(A,n)}
%o (PARI) /* This sequence is generated when k=2, m=1: A(x/A(x)^k) = 1 + x*A(x)^m */
%o {a(n, k=2, 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)^2*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: A001764, A147664, A120972, A145345, A145349, A147664.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Nov 12 2008
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