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A373445
Triple convolution of the three tribonacci-like sequences A000073(n), A077947(n-2), and A103143(n).
0
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).
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
KEYWORD
nonn
AUTHOR
Greg Dresden and Xiaoyuan Wang, Jun 05 2024
STATUS
approved