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%I #17 Oct 09 2020 12:13:02
%S 7,13,13,24,23,24,45,40,40,45,85,71,66,71,85,162,127,112,112,127,162,
%T 311,230,192,183,192,230,311,601,421,334,303,303,334,421,601,1168,779,
%U 588,510,487,510,588,779,1168,2281,1456,1048,869,798,798,869,1048,1456,2281
%N T(n,k) is the number of (n+1) X (k+1) 0..1 arrays with each 2 X 2 subblock having clockwise pattern 0000 0011 or 0101.
%C Table starts
%C 7 13 24 45 85 162 311 601 1168 2281 4473 8802 17371
%C 13 23 40 71 127 230 421 779 1456 2747 5227 10022 19345
%C 24 40 66 112 192 334 588 1048 1890 3448 6360 11854 22308
%C 45 71 112 183 303 510 869 1499 2616 4619 8251 14910 27249
%C 85 127 192 303 487 798 1325 2227 3784 6499 11283 19806 35161
%C 162 230 334 510 798 1278 2078 3422 5694 9566 16222 27774 48030
%C 311 421 588 869 1325 2078 3319 5377 8804 14545 24225 40670 68843
%C 601 779 1048 1499 2227 3422 5377 8591 13888 22655 37231 61598 102589
%C 1168 1456 1890 2616 3784 5694 8804 13888 22210 35872 58368 95550 157276
%C 2281 2747 3448 4619 6499 9566 14545 22655 35872 57455 92767 150686 245965
%C Each row (and each column, by symmetry) has a rational generating function (and therefore a linear recurrence with constant coefficients) because the growth from an array to the next larger one is described by the transfer matrix method. - _R. J. Mathar_, Oct 09 2020
%H R. H. Hardin, <a href="/A259222/b259222.txt">Table of n, a(n) for n = 1..480</a>
%F Empirical for diagonal and column k (k=3..7 recurrences work also for k=1,2):
%F diagonal: a(n) = 6*a(n-1) - 10*a(n-2) - 2*a(n-3) + 16*a(n-4) - 6*a(n-5) - 5*a(n-6) + 2*a(n-7).
%F k=1: a(n) = 3*a(n-1) - a(n-2) - 2*a(n-3)
%F k=2: a(n) = 3*a(n-1) - a(n-2) - 2*a(n-3)
%F k=3: a(n) = 4*a(n-1) - 4*a(n-2) - a(n-3) + 2*a(n-4)
%F k=4: a(n) = 4*a(n-1) - 4*a(n-2) - a(n-3) + 2*a(n-4)
%F k=5: a(n) = 4*a(n-1) - 4*a(n-2) - a(n-3) + 2*a(n-4)
%F k=6: a(n) = 4*a(n-1) - 4*a(n-2) - a(n-3) + 2*a(n-4)
%F k=7: a(n) = 4*a(n-1) - 4*a(n-2) - a(n-3) + 2*a(n-4)
%F Empirical: T(n,k) = 2^(k+1) + 2^(n+1) + F(n+3)*F(k+3) - 2*F(n+3) - 2*F(k+3) + 2 = 2^(n+1) + A001911(k)*F(n+3) + A234933(k+1) = A234933(n+1) + A234933(k+1) + A143211(n+3,k+3) - 2, F=A000045. - _Ehren Metcalfe_, Dec 27 2018
%e Some solutions for n=4, k=4:
%e 0 0 1 0 1 1 1 1 0 1 0 0 0 0 1 0 0 0 1 0
%e 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0
%e 1 1 0 1 0 0 0 0 1 0 1 1 1 1 0 0 0 0 1 0
%e 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 1 1 1 0 1
%e 1 1 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0
%Y Cf. A259215, A259216, A259217, A259218, A259219, A259220, A259221.
%K nonn,tabl
%O 1,1
%A _R. H. Hardin_, Jun 21 2015
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