A317353 G.f. satisfies: A(x) = 1 + x*A(x) * ( d/dx x*A(x)^3 ).

1, 1, 7, 85, 1391, 27852, 650161, 17204220, 506849175, 16418826823, 579491009098, 22126506842782, 908765624923517, 39955891685644210, 1872870160239756648, 93252093750481971120, 4916210973540632697015, 273627175793531458018185, 16036039197194394497206285, 987180880355901292395754555, 63693966375385851829434069094

Offset: 0

Author

Paul D. Hanna, Jul 26 2018

Keywords

• nonn

Examples

			O.g.f.: A(x) = 1 + x + 7*x^2 + 85*x^3 + 1391*x^4 + 27852*x^5 + 650161*x^6 + 17204220*x^7 + 506849175*x^8 + 16418826823*x^9 + ...
where A(x) = 1 + x*A(x)^4 + 3*x^2*A(x)^3*A'(x).
RELATED TABLE.
The table of coefficients of x^k/k! in exp( n*x*A(x)^3 ) / A(x) begins:
n=1: [1, 0, -7, -326, -23619, -2514244, -367699955, -70373452722, ...];
n=2: [1, 1, 0, -184, -17344, -2038008, -313614848, -61905013280, ...];
n=3: [1, 2, 9, 0, -9555, -1469682, -250952499, -52305865308, ...];
n=4: [1, 3, 20, 232, 0, -795496, -178628288, -41455820640, ...];
n=5: [1, 4, 33, 518, 11597, 0, -95428955, -29221983590, ...];
n=6: [1, 5, 48, 864, 25536, 934056, 0, -15457466592, ...];
n=7: [1, 6, 65, 1276, 42141, 2025842, 109167637, 0, ...];
n=8: [1, 7, 84, 1760, 61760, 3296568, 233751616, 17329568992, 0, ...]; ...
in which the coefficient of x^n in row n forms a diagonal of zeros.
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..376

Crossrefs

Cf. A088716, A317352, A317354.

Programs

PARI

{a(n) = my(A=1); for(i=1,n, A = 1 + x*A*(x*A^3)' +x*O(x^n)); polcoeff(A,n)}
for(n=0,30, print1(a(n),", "))

PARI

{a(n) = my(A=[1]); for(i=0,n, A=concat(A,0); m=#A-1; A[#A] = -Vec( exp(-m * x*Ser(A)^3 ) * (2 - 1/Ser(A)))[#A] );H=A; A[n+1]}
for(n=0,30, print1(a(n),", "))

PARI

{a(n) = my(A=[1]); for(i=0,n, A=concat(A,0); m=#A-1; A[#A] = -Vec( exp(-m^2 * x*Ser(A)^3 ) * (m+1 - m/Ser(A)))[#A]/m );H=A; A[n+1]}
for(n=0,30, print1(a(n),", "))

Formula

O.g.f. A(x) satisfies:

(1) A(x) = 1 + x*A(x) * ( d/dx x*A(x)^3 ).

(2) [x^n] exp( n * x*A(x)^3 ) / A(x) = 0 for n>0.

(3.a) [x^n] exp(-n * x*A(x)^3) * (2 - 1/A(x)) = 0 for n >= 1.

(3.b) [x^n] exp(-n^2 * x*A(x)^3) * (n + 1 - n/A(x)) = 0 for n >= 1.

(3.c) [x^n] exp(-n^(p+1) * x*A(x)^3) * (n^p + 1 - n^p/A(x)) = 0 for n>=1 and for fixed integer p >= 0.

a(n) ~ c * 3^n * n! * n^(4/3), where c = 0.147653317889315463510794506... - Vaclav Kotesovec, Aug 09 2018