%I #36 Nov 14 2024 10:07:54

%S 9,4,3,6,13,24,39,58,81,108,139,174,213,256,303,354,409,468,531,598,

%T 669,744,823,906,993,1084,1179,1278,1381,1488,1599,1714,1833,1956,

%U 2083,2214,2349,2488,2631,2778,2929,3084,3243,3406,3573,3744,3919,4098,4281,4468

%N a(n) = 2*n^2 - 7*n + 9.

%C If zero polygonal numbers are ignored, then for n >= 3, the a(n)-th n-gonal number is a sum of the (a(n)-1)-th n-gonal number and the (2*n-3)-th n-gonal number.

%H Michael De Vlieger, <a href="/A236257/b236257.txt">Table of n, a(n) for n = 0..10000</a>

%H W. Burrows, C. Tuffley, <a href="http://arxiv.org/abs/1502.06664">Maximising common fixtures in a round robin tournament with two divisions</a>, arXiv preprint arXiv:1502.06664 [math.CO], 2015.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F From _Colin Barker_, Jan 21 2014: (Start)

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

%F G.f.: -(18*x^2 - 23*x + 9)/(x-1)^3. (End)

%F E.g.f.: exp(x)*(9 - 5*x + 2*x^2). - _Elmo R. Oliveira_, Nov 13 2024

%e a(7)=58. This means that the 58th heptagonal number 8323 (cf. A000566) is a sum of two heptagonal numbers. We have 8323 = 8037 + 286 with indices in A000566 58,57,11.

%t Table[2 n^2 - 7 n + 9, {n, 0, 48}] (* _Michael De Vlieger_, Apr 19 2015 *)

%t LinearRecurrence[{3,-3,1},{9,4,3},50] (* _Harvey P. Dale_, Nov 24 2017 *)

%o (PARI) Vec(-(18*x^2-23*x+9)/(x-1)^3 + O(x^100)) \\ _Colin Barker_, Jan 21 2014

%Y Cf. A000217, A000290, A000326, A000384, A000566, A000567, A001106, A001107, A051624, A051682, A051865-A051876, A152948.

%K nonn,easy,changed

%O 0,1

%A _Vladimir Shevelev_, Jan 21 2014