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a(n) = a(n-1) + 3*a(n-2) + 3*a(n-3) + a(n-4), where a(0) = a(1) = a(2) = a(3) = 1.
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%I #8 Apr 09 2022 13:29:04

%S 1,1,1,1,8,15,43,113,295,778,2045,5377,14141,37185,97784,257139,

%T 676187,1778141,4675903,12296026,32334345,85028273,223595289,

%U 587979169,1546184200,4065935847,10692021243,28116360553,73936416023,194427497258,511277848229

%N a(n) = a(n-1) + 3*a(n-2) + 3*a(n-3) + a(n-4), where a(0) = a(1) = a(2) = a(3) = 1.

%H Clark Kimberling, <a href="/A275277/b275277.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,3,1).

%F a(n) = a(n-1) + 3*a(n-2) + 3*a(n-3) + a(n-4), where a(0) = a(1) = a(2) = a(3) = 1.

%F G.f.: (-1 + 3 x^2 + 6 x^3)/(-1 + x + 3 x^2 + 3 x^3 + x^4).

%t LinearRecurrence[{1, 3, 3, 1}, {1, 1, 1, 1}, 50]

%t RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==1,a[n]==a[n-1]+3a[n-2]+ 3a[n-3]+ a[n-4]},a,{n,30}] (* _Harvey P. Dale_, Apr 09 2022 *)

%Y Cf. A099234.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Aug 11 2016