A352850 - OEIS

%I #5 Apr 07 2022 12:12:35

%S 1,1,-1,3,-18,150,-1590,20790,-321930,5751270,-116263350,2628687600,

%T -65723427000,1800165767400,-53590470733800,1722471401851200,

%U -59462981199121500,2194469988757546500,-86216322855144982500,3592621186732990912500,-158253385898375658891000

%N E.g.f. A(x) satisfies: [x^k] A(x)^n = 0 for k = n*(n-1)/2 + 1 through n*(n+1)/2, for n > 1.

%e E.g.f.: A(x) = 1 + x - x^2/2! + 3*x^3/3! - 18*x^4/4! + 150*x^5/5! - 1590*x^6/6! + 20790*x^7/7! - 321930*x^8/8! + 5751270*x^9/9! - 116263350*x^10/10! + ...

%e The series of reduced fractional coefficients begins:

%e A(x) = 1 + x - 1/2*x^2 + 1/2*x^3 - 3/4*x^4 + 5/4*x^5 - 53/24*x^6 + 33/8*x^7 - 511/64*x^8 + 3043/192*x^9 - 4101/128*x^10 + ...

%e Related table.

%e The table of coefficients of x^k/k! in A(x)^n begins:

%e n=1: [1, 1, -1, 3, -18, 150, -1590, 20790, -321930, 5751270, ...];

%e n=2: [1, 2, 0, 0, -6, 60, -660, 9240, -149100, 2729160, ...];

%e n=3: [1, 3, 3, -3, 0, 0, 0, 1260, -28350, 595350, -13012650, ...];

%e n=4: [1, 4, 8, 0, -12, 0, 120, 0, 0, 0, 0, 30353400, -1166734800, ...];

%e n=5: [1, 5, 15, 15, -30, -30, 150, 1050, -1050, -47250, 1445850, 0, 0, 0, 0, 0, -141443655853500, ...];

%e ...

%e which illustrates that A(x)^n has n contiguous zeros for coefficients of x^k starting at k = n*(n-1)/2 + 1 through k = n*(n+1)/2, for n > 1.

%o (PARI) {a(n) = my(A=1+x,m=floor(sqrt(2*n)+1/2));

%o for(k=1,m, A = ( truncate(A^k) + x*O(x^(k*(k+1)/2)) )^(1/k));

%o n!*polcoeff(H=A,n)}

%o for(n=0,21,print1(a(n),", "))

%K sign

%O 0,4

%A _Paul D. Hanna_, Apr 05 2022