A373308 Expansion of Product_{n>=0} (1 - x^(2^n))^3.

1, -3, 0, 8, -9, 3, 8, -24, 15, 19, -24, 0, 17, -27, 0, 64, -57, -21, 64, -72, 33, 53, -72, 24, 17, -51, 24, 64, -81, 27, 64, -192, 135, 107, -192, 120, 1, -171, 120, 152, -183, -27, 152, -192, 87, 163, -192, 0, 89, -75, 0, 136, -129, -21, 136, -216, 111, 179, -216, 0, 145, -219, 0, 512

Offset: 0

Author

Paul D. Hanna, Jun 20 2024

Keywords

• sign
• look

Comments

Equals the self-convolution cube of the signed Thue-Morse sequence A106400: [1, -1, -1, 1, -1, 1, 1, -1, ...].

Conjecture: a(3*n) == A001285(n) (mod 3) for n >= 0, where A001285 is the Thue-Morse sequence: [1, 2, 2, 1, 2, 1, 1, 2, ...].

Conjecture: a(3*n+1) and a(3*n+2) are divisible by 3 for n >= 0.

Conjecture: a(n) = 0 iff n = 3*A000695(k) - 1 for k >= 1, where A000695 lists sums of distinct powers of 4.

Examples

			G.f.: A(x) = 1 - 3*x + 8*x^3 - 9*x^4 + 3*x^5 + 8*x^6 - 24*x^7 + 15*x^8 + 19*x^9 - 24*x^10 + 17*x^12 - 27*x^13 + 64*x^15 - 57*x^16 - 21*x^17 + 64*x^18 + ...
where A(x) = (1-x)^3 * (1-x^2)^3 * (1-x^4)^3 * (1-x^8)^3 * (1-x^16)^3 * ...
Notice that a(n) = 0 at n = [2, 11, 14, 47, 50, 59, 62, 191, 194, 203, 206, 239, 242, 251, 254, 767, ...] which appears to equal 3*A000695(k) - 1 for k >= 1.
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..16400

Crossrefs

Cf. A373309, A106400, A106407, A001285, A000695.

Programs

PARI

{a(n) = my(A = prod(k=0,#binary(n), (1 - x^(2^k))^3 +x*O(x^n))); polcoeff(A, n)}
for(n=0,60, print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.

(1) A(x) = Product_{n>=0} (1 - x^(2^n))^3.

(2) A(x) = (1-x)^3 * A(x^2).

(3) a(n) = Sum_{k=0..n} A106400(k) * A106407(n-k+1) for n >= 0.