

A279681


Irregular triangle read by rows: possible numbers of diagonals of convex polyhedra having n vertices.


3



0, 0, 1, 0, 1, 2, 3, 0, 1, 2, 3, 4, 5, 6, 0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 0, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28
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OFFSET

4,6


COMMENTS

Let n>4 denote the number of vertices. The set of possible numbers of diagonals is the union of sets {(k1)(nk4), ..., (k1)(n(k+6)/2)}, where 1 <= k <= floor((sqrt(8n15)5)/2), and the set {(k1)(nk4), ..., (n3)(n4)/2}, where k = floor((sqrt(8n15)3)/2). Note that cardinalities of all sets of this union excluding the last one are consecutive triangular numbers.


LINKS

Table of n, a(n) for n=4..85.
Vladimir Letsko, Table of rows of a(n)


EXAMPLE

Triangle begins:
4  0;
5  0, 1;
6  0, 1, 2, 3;
7  0, 1, 2, 3, 4, 5, 6;
8  0, 2, 3, 4, 5, 6, 7, 8, 9, 10;
9  0, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15;
10  0, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21;


MAPLE

dm:=(n, k)>simplify((n1)*n/2(2*nk1)(nk)*(nk3)/22*(k1)(k+2)*(k1)/2);
dM:=(n, k)>simplify((n1)*n/22*nk+3(nk)*(nk3)/2);
Dv:=proc(n) local k, DD; DD:={0}:for k from 2 to n/21 do
DD:=DD union {seq(i, i=dm(n, k)..dM(n, k))} od:
DD:=DD union {seq(i, i=dm(n, k1)..(n3)*(n4)/2)}:
DD end;


CROSSREFS

Row lengths are in A023536.
Cf. A000217, A279015, A279019, A279022, A279620, A279647, A279679.
Sequence in context: A293497 A278164 A328480 * A053645 A212598 A274650
Adjacent sequences: A279678 A279679 A279680 * A279682 A279683 A279684


KEYWORD

nonn,tabf


AUTHOR

Vladimir Letsko, Dec 16 2016


STATUS

approved



