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

0,8

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).

Links

Table of n, a(n) for n=0..30.

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)).

Example

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.

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]

Crossrefs

Cf. A000073, A077947, A103143.

Sequence in context: A022631 A027346 A325218 * A294958 A047134 A184698

Adjacent sequences: A373442 A373443 A373444 * A373446 A373447 A373448

Keyword

nonn

Author

Greg Dresden and Xiaoyuan Wang, Jun 05 2024

Status

approved