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a(n) is equal to the number of shaded cells in a regular hexagon with side n drawn on a hexagonal grid.
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%I #34 Jul 01 2024 13:12:58

%S 1,7,13,19,31,49,67,85,109,139,169,199,235,277,319,361,409,463,517,

%T 571,631,697,763,829,901,979,1057,1135,1219,1309,1399,1489,1585,1687,

%U 1789,1891,1999,2113,2227,2341,2461,2587,2713,2839,2971,3109,3247,3385,3529

%N a(n) is equal to the number of shaded cells in a regular hexagon with side n drawn on a hexagonal grid.

%C On a hexagonal grid, cells are colored as follows: one cell and all those located along three straight lines passing through the center of the original cell and forming six 60° angles between each other are painted. In each of these corners, cells are painted over so that a V-shaped arrangement of cells repeats ad infinitum. The number of shaded cells in regular hexagons centered on the starting cell determines the sequence a(n).

%H Paolo Xausa, <a href="/A373584/b373584.txt">Table of n, a(n) for n = 1..10000</a>

%H Nicolay Avilov, <a href="/A373584/a373584.jpg">Members of the sequence a(1) - a(7)</a>.

%H Nicolay Avilov, <a href="https://www.diofant.ru/problem/4471/">Problem 2663. Snowflakes</a> (in Russian).

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

%H Nicolay Avilov, <a href="/A373584/a373584_1.jpg">Illustration a(13) and a(16)</a>

%F a(n+4) = a(n) + 12*n + 18.

%F a(n) = 6*ceiling(n*(n - 1)/4) + 1.

%F a(n) = A003215(n) - 6*A011848(n+1).

%F a(n) = 6*A054925(n) + 1.

%F G.f.: (1 + 4*x - 4*x^2 + 4*x^3 + x^4)/((1 - x)^3*(1 + x^2)). - _Stefano Spezia_, Jun 11 2024

%e a(3) = 19 - 6*1 = 13;

%e a(4) = 37 - 6*3 = 19.

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%e 1 7 13 19 31

%t Table[6*Ceiling[n*(n - 1)/4] + 1, {n, 100}] (* _Paolo Xausa_, Jul 01 2024 *)

%Y Cf. A003215, A011848, A054925.

%K nonn,easy

%O 1,2

%A _Nicolay Avilov_, Jun 10 2024