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Triangle read by rows: T(n,k) = B(n,k) - B(n,0) + 1 where B(n,k) = 2 * (5*binomial(n,k)*(n+1)!*k!*(n-k)! + (n+k+1)!*(n-k)! + (2*n-k+1)!*k!) / ((n+1)!*k!*(n-k)!).
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%I #7 Mar 16 2026 15:57:34

%S 1,1,1,1,5,1,1,-11,-11,1,1,-99,-119,-99,1,1,-451,-611,-611,-451,1,1,

%T -1783,-2561,-2763,-2561,-1783,1,1,-6787,-10007,-11211,-11211,-10007,

%U -6787,1,1,-25651,-38231,-43627,-45071,-43627,-38231,-25651,1

%N Triangle read by rows: T(n,k) = B(n,k) - B(n,0) + 1 where B(n,k) = 2 * (5*binomial(n,k)*(n+1)!*k!*(n-k)! + (n+k+1)!*(n-k)! + (2*n-k+1)!*k!) / ((n+1)!*k!*(n-k)!).

%C Triangle is symmetric.

%e Triangle begins:

%e {1},

%e {1, 1},

%e {1, 5, 1},

%e {1, -11, -11, 1},

%e {1, -99, -119, -99, 1},

%e {1, -451, -611, -611, -451, 1},

%e {1, -1783, -2561, -2763, -2561, -1783, 1},

%e ...

%t t[n_, m_, q_] = 12*(Binomial[n, m]*(1 - q) + (((n + m + 1)!/((n + 1)!* m!)) + ((2*n - m + 1)!/((n + 1)!*(n - m)!)))*q);

%t Table[Flatten[Table[Table[t[ n, m, q] - t[n, 0, q] + 1, {m, 0, n}], {n, 0, 10}]], {q, 0, 1, 1/12}]

%K sign,tabl,less

%O 0,5

%A _Roger L. Bagula_, Apr 02 2010

%E Edited by _Sean A. Irvine_, Mar 16 2026