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A319143 G.f. A(x) satisfies: [x^(n-1)] (1+x)^(n^3) / A(x)^(n^2) = 0 for n>1.

%I A319143 #6 Sep 18 2018 19:03:42
%S A319143 1,2,5,190,24444,6189050,2551526428,1545212826174,1288051774444110,
%T A319143 1412705106844118046,1971892031185697252554,3413903325218336008192250,
%U A319143 7181500581229611492081984526,18048175081484797766245697300090,53425898749319275351535742806432314,184046820557885265127311961578368691278,730192327635057505047728578528016106455194
%N A319143 G.f. A(x) satisfies: [x^(n-1)] (1+x)^(n^3) / A(x)^(n^2) = 0 for n>1.
%e A319143 G.f.: A(x) = 1 + 2*x + 5*x^2 + 190*x^3 + 24444*x^4 + 6189050*x^5 + 2551526428*x^6 + 1545212826174*x^7 + 1288051774444110*x^8 + ...
%e A319143 The table of coefficients of x^k in (1+x)^(n^3) / A(x)^(n^2) begins:
%e A319143 n=1: [1, -1, -3, -179, -23881, -6115379, -2532879873, ...];
%e A319143 n=2: [1, 0, -16, -728, -96144, -24548304, -10154627640, ...];
%e A319143 n=3: [1, 9, 0, -1878, -231876, -57206466, -23347565964, ...];
%e A319143 n=4: [1, 32, 432, 0, -472008, -111871136, -43940424080, ...];
%e A319143 n=5: [1, 75, 2675, 55475, 0, -199916560, -76768966500, ...];
%e A319143 n=6: [1, 144, 10152, 460056, 13896684, 0, -126293662512, ...];
%e A319143 n=7: [1, 245, 29694, 2364152, 137272471, 5735706025, 0, ...]; ...
%e A319143 in which the n-th term in row n forms a diagonal of zeros after an initial '1'.
%o A319143 (PARI) {a(n) = my(A=[1]); for(m=1,n+1, A=concat(A,0); A[m] = Vec( (1+x +x*O(x^n))^(m^3)/Ser(A)^(m^2) )[m]/m^2 ); A[n+1]}
%o A319143 for(n=0,30, print1(a(n),", "))
%Y A319143 Cf. A303060.
%K A319143 nonn
%O A319143 0,2
%A A319143 _Paul D. Hanna_, Sep 18 2018