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a(0) = a(1) = a(2) = 1, for n > 2, a(n) = a(n-1) + a(n-k) + k with k = 3.
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%I #6 Apr 19 2023 02:47:46

%S 1,1,1,5,9,13,17,25,37,53,73,101,141,197,273,377,521,721,997,1377,

%T 1901,2625,3625,5005,6909,9537,13165,18173,25085,34625,47793,65969,

%U 91057,125685,173481,239453,330513,456201,629685,869141,1199657,1655861,2285549,3154693

%N a(0) = a(1) = a(2) = 1, for n > 2, a(n) = a(n-1) + a(n-k) + k with k = 3.

%C Called Leonardo 3-numbers in the Tan-Leung paper.

%H Elif Tan and Ho-Hon Leung, <a href="https://doi.org/10.5281/zenodo.7569221">On Leonardo p-Numbers</a>, Integers (2023) Vol. 23, #A7. See p. 2.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,1,-1).

%t LinearRecurrence[{2, -1, 0, 1, -1}, {1, 1, 1, 5, 9}, 44] (* or *)

%t With[{p = 3}, Nest[Append[#, #[[-1]] + #[[-p - 1]] + p] &, {1, 1, 1, 5}, 40] ]

%Y Cf. A001595, A111314, A362255.

%K nonn,easy

%O 0,4

%A _Michael De Vlieger_, Apr 13 2023