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%I #21 Jul 10 2016 22:57:04
%S 2,3,2,4,4,2,5,8,5,2,6,12,12,6,2,7,18,21,16,7,2,8,24,36,32,20,8,2,9,
%T 32,54,60,45,24,9,2,10,40,80,100,90,60,28,10,2,11,50,110,160,165,126,
%U 77,32,11,2,12,60,150,240,280,252,168,96,36,12,2,13,72,195,350,455,448,364,216,117,40,13,2
%N Triangle read by rows: T(n,k) (n>=3, 0<=k<=n-3) = number of n-sequences of 0's and 1's with exactly one pair of adjacent 0's and exactly k pairs of adjacent 1's.
%F T(n,k) = (k+1)*(binomial(floor((n+k-2)/2),k+1)+binomial(floor((n+k-3)/2),k+1))+2*binomial(floor((n+k-3)/2),k).
%F T(n,k) = (k+1)*A073044(n-2,k+1) + 2*A046854(n-3,k).
%F T(n,k) = A274742(n,k)+A274742(n-1,k)+A046854(n-3,k).
%e n=3 => 100, 001 -> T(3,0) = 2.
%e n=4 => 0010, 0100, 1001 -> T(4,0) = 3; 0011, 1100 -> T(4,1) = 2.
%e Triangle starts:
%e 2,
%e 3, 2,
%e 4, 4, 2,
%e 5, 8, 5, 2,
%e 6, 12, 12, 6, 2,
%e 7, 18, 21, 16, 7, 2,
%e 8, 24, 36, 32, 20, 8, 2,
%e 9, 32, 54, 60, 45, 24, 9, 2,
%e 10, 40, 80, 100, 90, 60, 28, 10, 2,
%e 11, 50, 110, 160, 165, 126, 77, 32, 11, 2,
%e 12, 60, 150, 240, 280, 252, 168, 96, 36, 12, 2,
%e 13, 72, 195, 350, 455, 448, 364, 216, 117, 40, 13, 2,
%e ...
%t Table[(k + 1) (Binomial[Floor[(n + k - 2)/2], k + 1] + Binomial[Floor[(n + k - 3)/2], k + 1]) + 2 Binomial[Floor[(n + k - 3)/2], k], {n, 3, 14}, {k, 0, n - 3}] // Flatten (* _Michael De Vlieger_, Jun 16 2016 *)
%o (PARI) T(n,k) = (k+1)*(binomial((n+k-2)\2,k+1)+binomial((n+k-3)\2,k+1))+2*binomial((n+k-3)\2,k); \\ _Michel Marcus_, Jun 17 2016
%Y Row sums give A001629.
%Y Cf. A073044.
%Y Columns of table:
%Y T(n,0)=A000027(n-1)
%Y T(n,1)=A007590(n-1)
%Y T(n,2)=A080838(n-1)
%Y T(n,3)=A032091(n)
%K nonn,tabl
%O 3,1
%A _Jeremy Dover_, Jun 14 2016
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