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%I #33 Dec 13 2022 02:07:37
%S 1,1,1,2,3,5,8,12,17,23,30,38,47,57,68,80,93,107,122,138,155,173,192,
%T 212,233,255,278,302,327,353,380,408,437,467,498,530,563,597,632,668,
%U 705,743,782,822,863,905,948,992,1037,1083,1130,1178,1227
%N a(n) = n-2 + Sum_{i = 1..n-2} (a(i+1) mod a(i)) for n >= 3 with a(1) = a(2) = 1.
%C Essentially triangular numbers + 2, but with three extra initial terms.
%H G. C. Greubel, <a href="/A022856/b022856.txt">Table of n, a(n) for n = 1..5000</a>
%H Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>, 2011. [Cached copy]
%H Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F For n > 3, a(n) = (n^2 - 7*n + 16)/2 = A027689(n-4)/2 = A000217(n-4) + 2 = A000124(n-4) + 1. - _Henry Bottomley_, Jun 27 2000
%F a(n) = Sum_{k=0..2} A007318(n-k-2, k) for n > 3. - _Johannes W. Meijer_, Aug 11 2013
%F Sum_{n>=1} 1/a(n) = 3 + 2*Pi*tanh(sqrt(15)*Pi/2)/sqrt(15). - _Amiram Eldar_, Dec 13 2022
%t a[n_] := If[n<4, 1, (n^2-7n+16)/2]; Array[a, 60] (* _Jean-François Alcover_, Mar 08 2017 *)
%o (PARI) for(n=1,100, print1(if(n<4, 1, (n^2 - 7*n +16)/2), ", ")) \\ _G. C. Greubel_, Jul 13 2017
%Y Cf. A000124, A000217, A007318, A027689.
%K nonn,easy
%O 1,4
%A _Clark Kimberling_
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