OFFSET
0,2
COMMENTS
4th row of A082039, case k = 3 of family T(n,k) = k^2*n^2 + k*n + 1.
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
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
4*a(n) - 3 is a square. - Muniru A Asiru, May 24 2018
LINKS
Muniru A Asiru, Table of n, a(n) for n = 0..5000
P. Manuel, R. Bharati, I. Rajasingh, and Chris Monica M, On minimum metric dimension of honeycomb networks, Journal of Discrete Algorithms, Vol. 6 (2008), pp. 20-27.
Leo Tavares, Snowflake illustration.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = 18*n + a(n-1) - 6 with n > 0, a(0)=1. - Vincenzo Librandi, Aug 08 2010
From Elmo R. Oliveira, Oct 23 2024: (Start)
G.f.: (1 + 10*x + 7*x^2)/(1 - x)^3.
E.g.f.: (1 + 12*x + 9*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)
MAPLE
seq(9*n^2+3*n+1, n=0..50); # Muniru A Asiru, May 21 2018
PROG
(PARI) a(n)=9*n^2+3*n+1 \\ Charles R Greathouse IV, Jun 17 2017
(GAP) List([0..50], n->9*n^2+3*n+1); # Muniru A Asiru, May 21 2018
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Paul Barry, Apr 02 2003
STATUS
approved