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%I #10 Jun 07 2024 14:30:45
%S 0,0,0,0,0,0,1,3,9,28,75,195,498,1229,2978,7115,16756,39031,90089,
%T 206228,468795,1059197,2380257,5323610,11856514,26306896,58172254,
%U 128246136,281957282,618367332,1353112803
%N Triple convolution of the three tribonacci-like sequences A000073(n), A077947(n-2), and A103143(n).
%C 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).
%F a(n) = (A000073(n+2) + A103143(n+2))/2 - A077947(n).
%F 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).
%F G.f.: x^6/((1 - 2*x)*(1 + x + x^2)*(1 - x - x^2 - x^3)*(1 - x - x^2 - 3*x^3)).
%e 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.
%t 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]
%Y Cf. A000073, A077947, A103143.
%K nonn
%O 0,8
%A _Greg Dresden_ and Xiaoyuan Wang, Jun 05 2024
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