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a(n) = a(1)*a(n-1) + a(2)*a(n-2) + ... + a(n-5)*a(5) for n >= 6, with a(1)=0 and a(2)=a(3)=a(4)=a(5)=1.
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%I #22 Mar 16 2023 14:09:46

%S 0,1,1,1,1,0,1,1,2,3,4,8,11,20,31,52,88,143,247,408,700,1184,2017,

%T 3462,5909,10196,17518,30281,52365,90704,157556,273742,476893,831298,

%U 1451603,2537736,4441262,7782934,13650555,23969794,42126241,74105773,130476070

%N a(n) = a(1)*a(n-1) + a(2)*a(n-2) + ... + a(n-5)*a(5) for n >= 6, with a(1)=0 and a(2)=a(3)=a(4)=a(5)=1.

%C Shifts left 5 places under the INVERT transform.

%H J. Conrad, <a href="/A361313/b361313.txt">Table of n, a(n) for n = 1..3000</a>

%e a(11) = a(1)*a(10) + a(2)*a(9) + a(3)*a(8) + a(4)*a(7) + a(5)*a(6) = 0*3 + 1*2 + 1*1 + 1*1 + 1*0 = 4.

%o (Python)

%o def A361313(l):

%o s = [0, 1, 1, 1, 1][:l]

%o for n in range(5, l):

%o s.append(sum([s[k] * s[n-k-1] for k in range(n-4)]))

%o return s

%Y Cf. A025250.

%K nonn,eigen

%O 1,9

%A _J. Conrad_, Mar 08 2023