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a(n) = (a(n-1) * a(n-5) + a(n-3)^3) / a(n-6), a(0) = a(1) = ... = a(5) = 1.
2

%I #19 Apr 21 2023 17:56:34

%S 1,1,1,1,1,1,2,3,4,12,39,142,1077,21209,779449,106636837,245010524697,

%T 3336696488691229,1125981890791313205482,

%U 693480182652378523758257457499,47660918720485535883730945247863294175948,13387114027268508450553229985503810242341235794343085252

%N a(n) = (a(n-1) * a(n-5) + a(n-3)^3) / a(n-6), a(0) = a(1) = ... = a(5) = 1.

%H Seiichi Manyama, <a href="/A276530/b276530.txt">Table of n, a(n) for n = 0..30</a>

%t RecurrenceTable[{a[n] == (a[n - 1] a[n - 5] + a[n - 3]^3)/a[n - 6], a[0] == a[1] == a[2] == a[3] == a[4] == a[5] == 1}, a, {n, 0, 21}] (* _Michael De Vlieger_, Nov 16 2016 *)

%t nxt[{a_,b_,c_,d_,e_,f_}]:={b,c,d,e,f,(f b+d^3)/a}; NestList[nxt,{1,1,1,1,1,1},25][[;;,1]] (* _Harvey P. Dale_, Apr 21 2023 *)

%o (Ruby)

%o def A(k, m, n)

%o a = Array.new(2 * k, 1)

%o ary = [1]

%o while ary.size < n + 1

%o i = a[-1] * a[1] + a[k] ** m

%o break if i % a[0] > 0

%o a = *a[1..-1], i / a[0]

%o ary << a[0]

%o end

%o ary

%o end

%o def A276530(n)

%o A(3, 3, n)

%o end

%Y Cf. A276529, A275173, A102276.

%K nonn

%O 0,7

%A _Seiichi Manyama_, Nov 16 2016