A343916 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.

4, 5, 6, 7, 8, 10, 11, 14, 16, 19, 23, 26, 31, 36, 41, 47, 54, 61, 68, 76, 85, 94, 103, 113, 124, 135, 146, 158, 171, 184, 197, 211, 226, 241, 256, 272, 289, 306, 323, 341, 360, 379, 398, 418, 439, 460, 481, 503, 526, 549, 572, 596, 621, 646, 671, 697, 724, 751, 778

Offset

3,1

Comments

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).

Links

Stefano Spezia, Table of n, a(n) for n = 3..10000

R. W. Maffucci, Constructing certain families of 3-polytopal graphs, arXiv:2105.00022 [math.CO], 2021.

Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Formula

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

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

From Greg Dresden, Jun 19 2021: (Start)

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).

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

Example

If we have an n-gonal face we need at least n+1 faces in total.
For n=3 we need at least 4 faces, and the tetrahedron has 4, so a(3)=4.
For n=4, e.g., the square pyramid has at least one triangle and one quadrilateral, so a(4)=5.
For n=5, there exists a polyhedron with 6 faces, comprising at least one triangle, one quadrilateral, and one pentagon: a(5)=6.

Mathematica

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 *)

Crossrefs

Sequence in context: A301592 A194414 A182204 * A367187 A121541 A182305

Adjacent sequences: A343913 A343914 A343915 * A343917 A343918 A343919

Keyword

nonn,easy

Author

Riccardo Maffucci, May 04 2021

Status

approved