A327896 a(n) is the minimum number of tiles needed for constructing a polyiamond with n holes.

9, 14, 19, 23, 27, 31, 35, 39, 43, 47, 51, 54, 58, 62, 65, 69, 73, 76, 80, 83, 87, 90, 94, 97, 101, 104, 108, 111, 115, 118, 122, 125, 129, 132, 135, 139, 142, 146, 149, 152, 156, 159, 163, 166, 169, 173, 176, 179, 183, 186, 189, 193, 196, 199, 203, 206, 209, 213

Offset

1,1

Comments

For n > 0, it is easy to prove that k(n) = floor((3 + sqrt(3*(3+8*n)))/6) is the unique integer that satisfies the inequalities 3*binomial(k,2) <= n <= 3*binomial(k+1,2) of Theorem 1.1 in Malen and Roldán.

Proof: solving in k the above inequalities for n > 0, one gets that x - 1 <= k <= x, where x = (3 + sqrt(3*(3+8*n)))/6. Since 3*(3+8*n) is never a perfect square, it follows that x is not an integer and k = floor(x). QED.

Links

Table of n, a(n) for n=1..58.

Greg Malen and Érika Roldán, Polyiamonds Attaining Extremal Topological Properties, arXiv:1906.08447 [math.CO], 2019.

Formula

a(n) = 3*(n + k(n)) + 1 + ceiling(2*n/k(n)), where k(n) = floor((3 + sqrt(3*(3+8*n)))/6).

Maple

k:=n->floor((3+sqrt(3*(3+8*n)))/6): a:=n->3*(n+k(n))+1+ceil(2*n/k(n)): seq(a(n), n = 1 .. 58)

Mathematica

k[n_]:=Floor[(3+Sqrt[3*(3+8n)])/6]; a[n_]:=3(n+k[n])+1+Ceiling[2n/k[n]]; Array[a, 58]

Crossrefs

Cf. A000217, A000577, A067628, A069813, A070764, A161680.

Sequence in context: A328246 A244466 A171123 * A302056 A173792 A332588

Adjacent sequences: A327893 A327894 A327895 * A327897 A327898 A327899

Keyword

nonn

Author

Stefano Spezia, Sep 29 2019

Status

approved