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Independence numbers for KT_4 knight on hexagonal board.
3

%I #27 Mar 18 2024 12:53:17

%S 1,3,3,4,7,9,12,15,19,22,25,28,37,40,46,51,61,66,71,76,91,96,103,111,

%T 127

%N Independence numbers for KT_4 knight on hexagonal board.

%H J.-P. Bode and H. Harborth, <a href="https://doi.org/10.1016/S0012-365X(03)00181-X">Independence for knights on hexagon and triangle boards</a>, Discrete Math., 272 (2003), 27-35.

%H Andy Huchala, <a href="/A087329/a087329_1.py.txt">Python program</a>.

%F a(4*n+1) = 3*n^2 + 3*n + 1 = A003215(n) from Bode and Harborth. - _Andy Huchala_, Mar 16 2024

%Y Cf. A003215, A087327, A087328.

%K nonn,more

%O 1,2

%A _N. J. A. Sloane_, Oct 21 2003

%E a(17)-a(25) from _Andy Huchala_, Mar 16 2024