This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A327896 #31 Oct 23 2019 16:38:54 %S A327896 9,14,19,23,27,31,35,39,43,47,51,54,58,62,65,69,73,76,80,83,87,90,94, %T A327896 97,101,104,108,111,115,118,122,125,129,132,135,139,142,146,149,152, %U A327896 156,159,163,166,169,173,176,179,183,186,189,193,196,199,203,206,209,213 %N A327896 a(n) is the minimum number of tiles needed for constructing a polyiamond with n holes. %C A327896 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. %C A327896 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. %H A327896 Greg Malen and Érika Roldán, <a href="https://arxiv.org/abs/1906.08447">Polyiamonds Attaining Extremal Topological Properties</a>, arXiv:1906.08447 [math.CO], 2019. %F A327896 a(n) = 3*(n + k(n)) + 1 + ceiling(2*n/k(n)), where k(n) = floor((3 + sqrt(3*(3+8*n)))/6). %p A327896 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) %t A327896 k[n_]:=Floor[(3+Sqrt[3*(3+8n)])/6]; a[n_]:=3(n+k[n])+1+Ceiling[2n/k[n]]; Array[a,58] %Y A327896 Cf. A000217, A000577, A067628, A069813, A070764, A161680. %K A327896 nonn %O A327896 1,1 %A A327896 _Stefano Spezia_, Sep 29 2019