OFFSET
0,3
COMMENTS
Conjecture: for n > 0, a(n) is odd iff n is a power of 2.
Conjecture: a(n) == 2 (mod 4) iff n/2 is a sum of two distinct powers of 2 (A018900).
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..512
FORMULA
G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) [x^n] 1/(A(x) - n^2*x) = 0 for n >= 1.
(2) [x^n] 1/A(x) = -Sum_{k=1..n} n^(2*k) * [x^(n-k)] 1/A(x)^(k+1) for n >= 1.
a(n) ~ (1 - exp(-2))^2 * n^(2*n). - Vaclav Kotesovec, Mar 15 2026
EXAMPLE
G.f.: A(x) = 1 + x + 9*x^2 + 368*x^3 + 33591*x^4 + 5229144*x^5 + 1220168698*x^6 + 395808157536*x^7 + 169851873309147*x^8 + ...
where [x^n] 1/(A(x) - n^2*x) = 0 for n >= 1.
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in 1/(A(x) - n^2*x) begins
n = 0: [1, -1, -8, -351, -32800, -5156650, ...];
n = 1: [1, 0, -9, -368, -33510, -5222520, ...];
n = 2: [1, 3, 0, -395, -35880, -5434002, ...];
n = 3: [1, 8, 55, 0, -37030, -5814352, ...];
n = 4: [1, 15, 216, 2737, 0, -5837130, ...];
n = 5: [1, 24, 567, 13024, 265050, 0, ...];
n = 6: [1, 35, 1216, 41877, 1408280, 42060590, 0, ...]; ...
in which the main diagonal, the coefficient of x^n in row n, is all zeros after the initial term.
PROG
(PARI) {a(n) = my(A=[1], m); for(i=1, n, A = concat(A, 0); m=#A-1;
A[#A] = polcoef(1/(Ser(A) - m^2*x), m) ); A[n+1]}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Mar 05 2026
STATUS
approved