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A302059 G.f. A(x) satisfies: [x^n] A(x)^(3*n+2) / (x*A(x)^3)' = 0 for n>1. 5

%I #16 Oct 20 2020 03:41:25

%S 1,1,7,223,12876,1041470,106367932,12976266030,1828572078504,

%T 291167594109079,51587932089790950,10054999390608521223,

%U 2137102578158115474600,491830012906134545686644,121847472306799372796495932,32335452582105155555135147936,9152510003592208592171248479432,2752729182740256406998201329266149

%N G.f. A(x) satisfies: [x^n] A(x)^(3*n+2) / (x*A(x)^3)' = 0 for n>1.

%C More generally, [x^n] G(x,k)^(k*(n+1)-1) / (x*G(x,k)^k)' = 0 is satisfied by an integer series G(x,k) when k is a fixed positive integer.

%H Paul D. Hanna, <a href="/A302059/b302059.txt">Table of n, a(n) for n = 0..300</a>

%F G.f. A(x) satisfies: [x^n] A(x)^(3*n) / (A(x) + 3*x*A'(x)) = 0 for n>1.

%F a(n) ~ c * 18^n * n! / n^(4/9), where c = 0.0140759690511059570797... - _Vaclav Kotesovec_, Oct 20 2020

%e G.f.: A(x) = 1 + x + 7*x^2 + 223*x^3 + 12876*x^4 + 1041470*x^5 + 106367932*x^6 + 12976266030*x^7 + 1828572078504*x^8 + 291167594109079*x^9 + ...

%e such that [x^n] A(x)^(3*n+2) / (x*A(x)^3)' = 0 for n>1.

%e ILLUSTRATION OF DEFINITION.

%e The table of coefficients in A(x)^(3*n+2) / (x*A(x)^3)' begins:

%e n=0: [1, -4, -33, -1902, -149243, -15230208, -1876337625, -267941179110, ...];

%e n=1: [1, -1, -21, -1385, -118455, -12695769, -1615045187, -235896844893, ...];

%e n=2: [1, 2, 0, -760, -83692, -9927612, -1334925264, -201948321520, ...];

%e n=3: [1, 5, 30, 0, -44414, -6904812, -1034743200, -165997446000, ...];

%e n=4: [1, 8, 69, 922, 0, -3604176, -713169380, -127940381076, ...];

%e n=5: [1, 11, 117, 2033, 50252, 0, -368770482, -87667218244, ...];

%e n=6: [1, 14, 174, 3360, 107125, 3936174, 0, -45061548696, ...];

%e n=7: [1, 17, 240, 4930, 171483, 8235801, 394811962, 0, ...]; ...

%e in which the main diagonal consists of all zeros after the initial terms, illustrating that [x^n] A(x)^(3*n+2) / (x*A(x)^3)' = 0 for n>1.

%e RELATED SERIES.

%e A(x)^3 = 1 + 3*x + 24*x^2 + 712*x^3 + 40134*x^4 + 3211848*x^5 + 326090932*x^6 + 39631822680*x^7 + 5570315156529*x^8 + ...

%e (x*A(x)^3)' = 1 + 6*x + 72*x^2 + 2848*x^3 + 200670*x^4 + 19271088*x^5 + 2282636524*x^6 + 317054581440*x^7 + 50132836408761*x^8 + ...

%e log(A(x)) = x + 13*x^2/2 + 649*x^3/3 + 50541*x^4/4 + 5136491*x^5/5 + 631363729*x^6/6 + 90027008835*x^7/7 + 14517298906509*x^8/8 + 2603034994183642*x^9/9 + ...

%o (PARI) {a(n) = my(A=[1, 1]); for(i=1, n, A = concat(A, 0); A[#A] = Vec( Ser(A)^(3*#A-1)/(x*Ser(A)^3)' )[#A]); A[n+1]}

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

%Y Cf. A300994, A300627, A300995, A302060.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Mar 31 2018

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