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A302906 a(0) = 0; for n > 0, a(n) = a(n-1) + 5*n + 4.

%I A302906 #21 Sep 08 2022 08:46:21
%S A302906 0,9,23,42,66,95,129,168,212,261,315,374,438,507,581,660,744,833,927,
%T A302906 1026,1130,1239,1353,1472,1596,1725,1859,1998,2142,2291,2445,2604,
%U A302906 2768,2937,3111,3290,3474,3663,3857,4056,4260,4469,4683,4902,5126,5355,5589
%N A302906 a(0) = 0; for n > 0, a(n) = a(n-1) + 5*n + 4.
%C A302906 In A302717, if we count the terms added from each 4-tuple during each iteration we find that either two or three terms are added: 2, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 2, 2, 3, ... where the set of three twos (2, 2, 2) appears with decreasing frequency. a(n) is the index into A302717 where each such set begins.
%C A302906 40*a(n) + 169 is a square, whose square root belongs to A063226 and A063206. - _Bruno Berselli_, Apr 17 2018
%H A302906 Colin Barker, <a href="/A302906/b302906.txt">Table of n, a(n) for n = 0..1000</a>
%H A302906 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A302906 From _Colin Barker_, Apr 16 2018: (Start)
%F A302906 G.f.: x*(9 - 4*x)/(1 - x)^3.
%F A302906 a(n) = n*(5*n + 13)/2.
%F A302906 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2. (End)
%e A302906 In A302717:
%e A302906 with a=9  and b=10, a(20,21,22) are appended;
%e A302906 with a=10 and b=11, a(23,24) are appended;
%e A302906 with a=11 and b=12, a(25,26) are appended;
%e A302906 with a=12 and b=13, a(27,28) are appended;
%e A302906 with a=13 and b=14, a(29,30,31) are appended, so a(2) = 23, because A302717(23) is the start of three consecutively added pairs of terms.
%t A302906 LinearRecurrence[{3, -3, 1}, {0, 9, 23}, 50] (* _Vincenzo Librandi_, Apr 17 2018 *)
%o A302906 (PARI) concat(0, Vec(x*(9 - 4*x) / (1 - x)^3 + O(x^50))) \\ _Colin Barker_, Apr 16 2018
%o A302906 (Magma) [n*(5*n+13)/2: n in [0..60]]; // _Vincenzo Librandi_, Apr 17 2018
%Y A302906 Cf. A302717.
%K A302906 nonn,easy
%O A302906 0,2
%A A302906 _J. Stauduhar_, Apr 15 2018