A373445 Triple convolution of the three tribonacci-like sequences A000073(n), A077947(n-2), and A103143(n).

0, 0, 0, 0, 0, 0, 1, 3, 9, 28, 75, 195, 498, 1229, 2978, 7115, 16756, 39031, 90089, 206228, 468795, 1059197, 2380257, 5323610, 11856514, 26306896, 58172254, 128246136, 281957282, 618367332, 1353112803

Offset: 0

Author

Greg Dresden and Xiaoyuan Wang, Jun 05 2024

Keywords

• nonn

Comments

If we set b(n)=A000073(n), c(n)=A077947(n-2) with c(0)=c(1)=0, and d(n)=A103143(n), then all three sequences b(n), c(n), and d(n) start with the terms 0,0,1,1,2 and have signatures {1,1,1}, {1,1,2}, and {1,1,3} respectively. The triple convolution is defined as a(n) = Sum_{i+j+k=n} b(i)*c(j)*d(k).

Examples

			For n=7 the triple convolution of the three sequences b(n)=A000073(n), c(n)=A077947(n-2) with c(0)=c(1)=0, and d(n)=A103143(n) has only three nonzero terms in the sum: b(2)*c(2)*d(3), b(2)*c(3)*d(2), and b(3)*c(2)*c(2). All three terms are 1, so the triple convolution adds up to 3. Hence, a(7) = 3.
		

Crossrefs

Cf. A000073, A077947, A103143.

Programs

Mathematica

CoefficientList[Series[x^6/((1-x-x^2-x^3)(1-x-x^2-2x^3)(1-x-x^2-3x^3)), {x, 0, 30}], x]

Formula

a(n) = (A000073(n+2) + A103143(n+2))/2 - A077947(n).

a(n) = 3*a(n-1) + a(n-3) - 12*a(n-4) - 3*a(n-5) + 2*a(n-6) + 17*a(n-7) + 11*a(n-8) + 6*a(n-9).

G.f.: x^6/((1 - 2*x)*(1 + x + x^2)*(1 - x - x^2 - x^3)*(1 - x - x^2 - 3*x^3)).