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Expansion of A(x) satisfying [x^n] A(x) / (1 + x*A(x)^(n+2)) = 0 for n > 0.
4

%I #9 Mar 13 2023 04:26:27

%S 1,1,4,29,294,3727,55748,950898,18094313,378363501,8600306451,

%T 210773059751,5534376088000,154911828439188,4603267204022882,

%U 144710918709587399,4798300212740184379,167370947204751098624,6127130537038980726113,234905895680130694945861,9413383171884998924237972

%N Expansion of A(x) satisfying [x^n] A(x) / (1 + x*A(x)^(n+2)) = 0 for n > 0.

%H Paul D. Hanna, <a href="/A360584/b360584.txt">Table of n, a(n) for n = 0..300</a>

%F a(n) ~ c * n! * n^(4*LambertW(1) - 1 + 2/(1 + LambertW(1))) / LambertW(1)^n, where c = 0.02048373460253911846... - _Vaclav Kotesovec_, Mar 13 2023

%e G.f.: A(x) = 1 + x + 4*x^2 + 29*x^3 + 294*x^4 + 3727*x^5 + 55748*x^6 + 950898*x^7 + 18094313*x^8 + 378363501*x^9 + 8600306451*x^10 + ...

%e The table of coefficients in the successive powers of g.f. A(x) begins:

%e n = 1: [1, 1, 4, 29, 294, 3727, 55748, ...];

%e n = 2: [1, 2, 9, 66, 662, 8274, 122143, ...];

%e n = 3: [1, 3, 15, 112, 1116, 13776, 200827, ...];

%e n = 4: [1, 4, 22, 168, 1669, 20384, 293654, ...];

%e n = 5: [1, 5, 30, 235, 2335, 28266, 402710, ...];

%e n = 6: [1, 6, 39, 314, 3129, 37608, 530334, ...];

%e n = 7: [1, 7, 49, 406, 4067, 48615, 679140, ...];

%e ...

%e The table of coefficients in A(x)/(1 + x*A(x)^(n+2)) begins:

%e n = 1: [1, 0, 1, 13, 166, 2391, 38776, 699060, ...];

%e n = 2: [1, 0, 0, 7, 119, 1911, 32823, 612983, ...];

%e n = 3: [1, 0, -1, 0, 64, 1358, 26039, 515774, ...];

%e n = 4: [1, 0, -2, -8, 0, 724, 18356, 406634, ...];

%e n = 5: [1, 0, -3, -17, -74, 0, 9702, 284785, ...];

%e n = 6: [1, 0, -4, -27, -159, -824, 0, 149478, ...];

%e n = 7: [1, 0, -5, -38, -256, -1759, -10833, 0, ...];

%e ...

%e in which the diagonal of all zeros illustrates that

%e [x^n] A(x) / (1 + x*A(x)^(n+2)) = 0 for n > 0.

%o (PARI) {a(n) = my(A=[1]); for(i=1,n, A = concat(A,0);

%o A[#A] = -polcoeff( Ser(A)/(1 + x*Ser(A)^(#A+1)), #A-1) );A[n+1]}

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A360582, A360583.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Mar 12 2023