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a(n) = 9*n^2 + 3*n + 1.
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%I #55 Oct 23 2024 16:52:52

%S 1,13,43,91,157,241,343,463,601,757,931,1123,1333,1561,1807,2071,2353,

%T 2653,2971,3307,3661,4033,4423,4831,5257,5701,6163,6643,7141,7657,

%U 8191,8743,9313,9901,10507,11131,11773,12433,13111,13807,14521,15253,16003,16771,17557

%N a(n) = 9*n^2 + 3*n + 1.

%C 4th row of A082039, case k = 3 of family T(n,k) = k^2*n^2 + k*n + 1.

%C a(n)^2 = 81*n^4 + 54*n^3 + 27*n^2 + 6*n + 1 = (24*((3*((3*n^2 + n)/2)^2 + ((3*n^2 + n)/2))/2) + 1). Therefore, (a(n)^2 - 1)/24 is a second pentagonal number (A005449) of index number equal to the n-th second pentagonal number. For example, a(30) = 8191 and (8191^2 - 1)/24 = (67092481 - 1)/24 = 2795520, the 1365th second pentagonal number. 1365 is the 30th second pentagonal number. - _Raphie Frank_, Sep 19 2012

%C For n >= 1, a(n) is the number of vertices in the hex derived network HDN1(n+1) from the Manuel et al. reference (see HFN1(4) in Fig. 8). - _Emeric Deutsch_, May 21 2018

%C 4*a(n) - 3 is a square. - _Muniru A Asiru_, May 24 2018

%H Muniru A Asiru, <a href="/A082040/b082040.txt">Table of n, a(n) for n = 0..5000</a>

%H P. Manuel, R. Bharati, I. Rajasingh, and Chris Monica M, <a href="https://doi.org/10.1016/j.jda.2006.09.002">On minimum metric dimension of honeycomb networks</a>, Journal of Discrete Algorithms, Vol. 6 (2008), pp. 20-27.

%H Leo Tavares, <a href="/A082040/a082040.jpg">Snowflake illustration</a>.

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

%F a(n) = 18*n + a(n-1) - 6 with n > 0, a(0)=1. - _Vincenzo Librandi_, Aug 08 2010

%F a(n) = A045945(n) + 1: subsequence of A002061. - _Muniru A Asiru_, May 26 2018

%F a(n) = A003215(n) + 6*A000290(n). - _Leo Tavares_, Jul 14 2023

%F From _Elmo R. Oliveira_, Oct 23 2024: (Start)

%F G.f.: (1 + 10*x + 7*x^2)/(1 - x)^3.

%F E.g.f.: (1 + 12*x + 9*x^2)*exp(x).

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

%p seq(9*n^2+3*n+1,n=0..50); # _Muniru A Asiru_, May 21 2018

%o (PARI) a(n)=9*n^2+3*n+1 \\ _Charles R Greathouse IV_, Jun 17 2017

%o (GAP) List([0..50], n->9*n^2+3*n+1); # _Muniru A Asiru_, May 21 2018

%Y Cf. A002061, A005449, A045945, A054569, A082039.

%Y Partial sums of A019557.

%Y Cf. A000290, A003215.

%K nonn,easy

%O 0,2

%A _Paul Barry_, Apr 02 2003