A343916 - OEIS

%I #36 Jul 02 2021 16:44:52

%S 4,5,6,7,8,10,11,14,16,19,23,26,31,36,41,47,54,61,68,76,85,94,103,113,

%T 124,135,146,158,171,184,197,211,226,241,256,272,289,306,323,341,360,

%U 379,398,418,439,460,481,503,526,549,572,596,621,646,671,697,724,751,778

%N a(n) is the minimal total number of faces of a polyhedron with at least one i-gonal face for every i in 3..n.

%C a(n) is also the minimal number of vertices of a polyhedral (planar, 3-connected) graph with at least one vertex of degree i for every i in 3..n (proven by duality).

%H Stefano Spezia, <a href="/A343916/b343916.txt">Table of n, a(n) for n = 3..10000</a>

%H R. W. Maffucci, <a href="https://arxiv.org/abs/2105.00022">Constructing certain families of 3-polytopal graphs</a>, arXiv:2105.00022 [math.CO], 2021.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,-4,4,-3,1).

%F For n >= 14, a(n) = ceiling((n^2 - 11*n + 62)/4).

%F a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) for n > 18. - _Stefano Spezia_, May 05 2021

%F From _Greg Dresden_, Jun 19 2021: (Start)

%F For n >= 14, a(n) = (n^2 - 11*n + 62)/4 for n == 1,2 (mod 4), and a(n) = 1/2 + (n^2 - 11*n + 62)/4 for n == 0,3 (mod 4).

%F a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12) for n >= 26. (End)

%e If we have an n-gonal face we need at least n+1 faces in total.

%e For n=3 we need at least 4 faces, and the tetrahedron has 4, so a(3)=4.

%e For n=4, e.g., the square pyramid has at least one triangle and one quadrilateral, so a(4)=5.

%e For n=5, there exists a polyhedron with 6 faces, comprising at least one triangle, one quadrilateral, and one pentagon: a(5)=6.

%t LinearRecurrence[{3, -4, 4, -3, 1}, {4, 5, 6, 7, 8, 10, 11, 14, 16, 19, 23, 26, 31, 36, 41, 47}, 40] (* _Greg Dresden_, Jun 19 2021 *)

%K nonn,easy

%O 3,1

%A _Riccardo Maffucci_, May 04 2021